Q1: A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30° (see figure).
Ans: Given: length of the rope (AC) = 20 m, ∠ACB = 30° Let the height of the pole (AB) = h metre
⇒ h/20 = 1/2⇒ h = 20/2 = 10 m Hence, height of the pole = 10 m
Q2: A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree. Ans: Let DB is a tree and AD is the broken part of it that touches the ground at C.
Given: ∠ACB = 30º and BC = 8m Let AB = x m and AD = y m ∴ Now, length of the tree = (x + y) m In Δ ABCHence, total height of the tree =
Q3: A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for older children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case? Ans: Let l1 is the length of the slide for children below the age of 5 years and l2 is the length of the slide for elder children. In ΔABC
Q4: The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower. Ans: Let h be the height of the tower ⇒ ⇒
Q5: A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string. Ans: Given: height AB = 60 m, ∠ACB = 60°, AC = length of the string
Hence, the length of the string = 40√3 m
Q6: A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building. Ans: Let AB = height of the building
The distance walked by the boy towards building DE = DF – EF
Q7: From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower. Ans: Given: AB = 20 m (Height of the building) Let AD = h m (Height of the tower)
Hence, height of the tower =
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Q8: A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal. Ans: Let the height of the pedestal AB = h m Given: height of the statue = 1.6 m, ∠ACB = 45° and ∠DCB = 60°
Hence, height of the pedestal
Q9: The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building. Ans: Given: height of the tower AB = 50 m
∠ACB = 60°, ∠DBC = 30° Let the height of the building CD = x m
Q10: Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distances of the point from the poles. Ans: Let AB = CD = h m [Height of the poles] Given: BC = 80 m [Width of the road] Let CE = x m ∴ BE = (80 – x) m In ΔCDE,
In ΔABE,
… (ii) From equation (i) and (ii), we get
Substituting h in equation (i),
80 – x = 20 m Hence, position of the point is at a distance of 60 m from pole CD and 20 m from pole AB.
Q11: A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see figure). Find the height of the tower and the width of the canal.
Ans: Let the height of the tower AB = h m and BC be the width of the canal. Given: ∠ACB = 60° and ∠ADB = 30°
⇒ ⇒ ⇒
Hence, the height of the tower = 10√3 m and width of the canal = 10 m.
Q12: From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower. Ans: Let height of the tower AB = (h + 7) m
Given: CD = 7m (height of the building), ∠ACE = 60°, and ∠ECB = 45°
⇒ ∠CBD = 45º
⇒ ⇒
Q13: As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. Ans: Given: height of the lighthouse = 75 m Let C and D are the positions of two ships. We have ∠XAD = ∠ADB = 30° and ∠XAC = ∠ACB = 45°
⇒
Hence, the distance between two ships is 54.75 m.
Q14: A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (see figure). Find the distance travelled by the balloon during the interval.
Ans: Let the first position of the balloon is A and after some time it will reach to the point D. The vertical height ED = AB = (88.2 – 1.2) m = 87 m.
Distance travelled by the balloon from A to D is BE. So, BE = CE – CB
Q15: A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point. Ans: Let the height of the tower AB = h m Given: ∠XAD = ∠ADB = 30° and ∠XAC – ∠ACB = 60° Let the speed of the car = x m/sec
Q1. In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine: (i) sin A, cos A (ii) sin C, cos C Ans: In a given triangle ABC, right angled at B = ∠B = 90° Given: AB = 24 cm and BC = 7 cm According to the Pythagoras Theorem, In a right-angled triangle, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides. By applying the Pythagoras theorem, we get AC2 = AB2 + BC2 AC2 = (24)2+72 AC2 = (576+49) AC2 = 625cm2 AC = √625 = 25 Therefore, AC = 25 cm (i) To find Sin (A), Cos (A) We know that the sine (or) Sin function is equal to the ratio of the length of the opposite side to the hypotenuse side. So it becomes Sin (A) = Opposite side /Hypotenuse = BC/AC = 7/25 Cosine or Cos function is equal to the ratio of the length of the adjacent side to the hypotenuse side and it becomes, Cos (A) = Adjacent side/Hypotenuse = AB/AC = 24/25 (ii) To find Sin (C), Cos (C) Sin (C) = AB/AC = 24/25 Cos (C) = BC/AC = 7/25
Q2. In Fig. 8.13, find tan P – cot R
Ans: In the given triangle PQR, the given triangle is right-angled at Q and the given measures are: PR = 13cm, PQ = 12cm Since the given triangle is right-angled triangle, to find the side QR, apply the Pythagoras theorem According to Pythagoras theorem, In a right-angled triangle, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides. PR2 = QR2 + PQ2 Substitute the values of PR and PQ 132 = QR2+122 169 = QR2+144 Therefore, QR2 = 169−144 QR2 = 25 QR = √25 = 5 Therefore, the side QR = 5 cm Now, tan P – cot R=? According to the trigonometric ratio, the tangent function is equal to the ratio of the length of the opposite side to the adjacent sides, the value of tan (P) becomes tan (P) = Opposite side /Adjacent side = QR/PQ = 5/12 Since cot function is the reciprocal of the tan function, the ratio of cot function becomes, Cot (R) = Adjacent side/Opposite side = QR/PQ = 5/12 Therefore, tan (P) – cot (R) = 5/12 – 5/12 = 0 Therefore, tan(P) – cot(R) = 0
Q3. If sin A = 3/4, Calculate cos A and tan A. Ans: Let us assume a right-angled triangle ABC, right-angled at B Given: Sin A = 3/4 We know that Sin function is equal to the ratio of the length of the opposite side to the hypotenuse side. Therefore, Sin A = Opposite side /Hypotenuse= 3/4 Let BC be 3k and AC will be 4k where k is a positive real number. According to the Pythagoras theorem, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides of a right-angle triangle and we get, AC2=AB2 + BC2 Substitute the value of AC and BC (4k)2=AB2 + (3k)2 16k2−9k2 =AB2 AB2=7k2 Therefore, AB = √7k Now, we have to find the value of cos A and tan A We know that, Cos (A) = Adjacent side/Hypotenuse Substitute the value of AB and AC and cancel the constant k in both numerator and denominator, we get AB/AC = √7k/4k = √7/4 Therefore, cos (A) = √7/4 tan(A) = Opposite side/Adjacent side Substitute the Value of BC and AB and cancel the constant k in both numerator and denominator, we get, BC/AB = 3k/√7k = 3/√7 Therefore, tan A = 3/√7
Q4. Given 15 cot A = 8, find sin A and sec A. Ans: Let us assume a right-angled triangle ABC, right-angled at B Given: 15 cot A = 8 So, Cot A = 8/15 We know that cot function is equal to the ratio of the length of the adjacent side to the opposite side. Therefore, cot A = Adjacent side/Opposite side = AB/BC = 8/15 Let AB be 8k and BC will be 15k Where k is a positive real number. According to the Pythagoras theorem, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides of a right-angle triangle and we get, AC2=AB2 + BC2 Substitute the value of AB and BC AC2= (8k)2 + (15k)2 AC2= 64k2 + 225k2 AC2= 289k2 Therefore, AC = 17k Now, we have to find the value of sin A and sec A We know that, Sin (A) = Opposite side /Hypotenuse Substitute the value of BC and AC and cancel the constant k in both numerator and denominator, we get Sin A = BC/AC = 15k/17k = 15/17 Therefore, sin A = 15/17 Since secant or sec function is the reciprocal of the cos function which is equal to the ratio of the length of the hypotenuse side to the adjacent side. Sec (A) = Hypotenuse/Adjacent side Substitute the Value of BC and AB and cancel the constant k in both numerator and denominator, we get, AC/AB = 17k/8k = 17/8 Therefore sec (A) = 17/8
Q5. Given sec θ = 13/12 Calculate all other trigonometric ratios Ans: We know that sec function is the reciprocal of the cos function which is equal to the ratio of the length of the hypotenuse side to the adjacent side Let us assume a right-angled triangle ABC, right-angled at B sec θ =13/12 = Hypotenuse/Adjacent side = AC/AB Let AC be 13k and AB will be 12k Where k is a positive real number. According to the Pythagoras theorem, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides of a right-angled triangle and we get, AC2=AB2 + BC2 Substitute the value of AB and AC (13k)2= (12k)2 + BC2 169k2= 144k2 + BC2 169k2= 144k2 + BC2 BC2 = 169k2 – 144k2 BC2= 25k2 Therefore, BC = 5k Now, substitute the corresponding values in all other trigonometric ratios So, Sin θ = Opposite Side/Hypotenuse = BC/AC = 5/13 Cos θ = Adjacent Side/Hypotenuse = AB/AC = 12/13 tan θ = Opposite Side/Adjacent Side = BC/AB = 5/12 Cosec θ = Hypotenuse/Opposite Side = AC/BC = 13/5 cot θ = Adjacent Side/Opposite Side = AB/BC = 12/5
Q6. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠ A = ∠ B. Ans: Let us assume the triangle ABC in which CD⊥AB Give that the angles A and B are acute angles, such that Cos (A) = cos (B) As per the angles taken, the cos ratio is written as AD/AC = BD/BC Now, interchange the terms, we get AD/BD = AC/BC Let take a constant value AD/BD = AC/BC = k Now consider the equation as AD = k BD …(1) AC = k BC …(2) By applying Pythagoras theorem in △CAD and △CBD we get, CD2 = BC2 – BD2 … (3) CD2 =AC2 −AD2 ….(4) From the equations (3) and (4) we get, AC2−AD2 = BC2−BD2 Now substitute the equations (1) and (2) in (3) and (4) K2(BC2−BD2)=(BC2−BD2) k2=1 Putting this value in equation, we obtain AC = BC ∠A=∠B (Angles opposite to equal side are equal-isosceles triangle)
Q7. If cot θ = 7/8, evaluate: (i) (1 + sin θ)(1 – sin θ)/(1+cos θ)(1-cos θ) (ii) cot2 θ Ans: Let us assume a △ABC in which ∠B = 90° and ∠C = θ Given: cot θ = BC/AB = 7/8 Let BC = 7k and AB = 8k, where k is a positive real number According to the Pythagoras theorem in △ABC we get. AC2 = AB2+BC2 AC2 = (8k)2+(7k)2 AC2 = 64k2+49k2 AC2 = 113k2 AC = √113 k According to the sine and cos function ratios, it is written as sin θ = AB/AC = Opposite Side/Hypotenuse = 8k/√113 k = 8/√113 and cos θ = Adjacent Side/Hypotenuse = BC/AC = 7k/√113 k = 7/√113 Now apply the values of sin function and cos function:
Q8. If 3 cot A = 4, check whether (1-tan2 A)/(1+tan2 A) = cos2 A – sin 2 A or not. Ans: Let △ABC in which ∠B=90° We know that cot function is the reciprocal of tan function and it is written as cot(A) = AB/BC = 4/3 Let AB = 4k and BC =3k, where k is a positive real number. According to the Pythagoras theorem, AC2=AB2+BC2 AC2=(4k)2+(3k)2 AC2=16k2+9k2 AC2=25k2 AC=5k Now, apply the values corresponding to the ratios tan(A) = BC/AB = 3/4 sin (A) = BC/AC = 3/5 cos (A) = AB/AC = 4/5 Now compare the left-hand side(LHS) with right-hand side(RHS)
Since, both the LHS and RHS = 7/25 R.H.S. = L.H.S. Hence, (1-tan2 A)/(1+tan2 A) = cos2 A – sin 2 A is proved
Q9. In triangle ABC, right-angled at B, if tan A = 1/√3 find the value of: (i) sin A cos C + cos A sin C (ii) cos A cos C – sin A sin C Ans: Let ΔABC in which ∠B=90° tan A = BC/AB = 1/√3 Let BC = 1k and AB = √3 k, Where k is the positive real number of the problem By Pythagoras theorem, in ΔABC we get: AC2=AB2+BC2 AC2=(√3 k)2+(k)2 AC2=3k2+k2 AC2=4k2 AC = 2k Now find the values of cos A, Sin A Sin A = BC/AC = 1/2 Cos A = AB/AC = √3/2 Then find the values of cos C and sin C Sin C = AB/AC = √3/2 Cos C = BC/AC = 1/2 Now, substitute the values in the given problem (i) sin A cos C + cos A sin C = (1/2) ×(1/2 )+ √3/2 ×√3/2 = 1/4 + 3/4 = 1 (ii) cos A cos C – sin A sin C = (√3/2 )(1/2) – (1/2) (√3/2 ) = 0
Q10. In ∆ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P Ans: In a given triangle PQR, right angled at Q, the following measures are PQ = 5 cm PR + QR = 25 cm Now let us assume, QR = x PR = 25-QR PR = 25- x According to the Pythagoras Theorem, PR2 = PQ2 + QR2 Substitute the value of PR as x (25- x) 2 = 52 + x2 252 + x2 – 50x = 25 + x2 625 + x2-50x -25 – x2 = 0 -50x = -600 x= -600/-50 x = 12 = QR Now, find the value of PR PR = 25- QR Substitute the value of QR PR = 25-12 PR = 13 Now, substitute the value to the given problem (1) Sin P = Opposite Side/Hypotenuse = QR/PR = 12/13 (2) Cos P= Adjacent Side/Hypotenuse = PQ/PR = 5/13 (3) Tan P =Opposite Side/Adjacent side = QR/PQ = 12/5
Q11. State whether the following are true or false. Justify your answer. (i) The value of tan A is always less than 1. (ii) sec A = 12/5 for some value of angle A. (iii) cos A is the abbreviation used for the cosecant of angle A. (iv) cot A is the product of cot and A. (v) sin θ = 4/3 for some angle θ. Ans: (i) The value of tan A is always less than 1. False Justification: In ΔMNC in which ∠N = 90∘, MN = 3, NC = 4 and MC = 5 Value of tan M = 4/3 which is greater than 1. The triangle can be formed with sides equal to 3, 4 and hypotenuse = 5 as it will follow the Pythagoras theorem. MC2 = MN2 + NC2 52 = 32 + 42 25 = 9 + 16 25=25
(ii) sec A = 12/5 for some value of angle A True Justification: Let a ΔMNC in which ∠N = 90º, MC=12k and MB=5k, where k is a positive real number. By Pythagoras theorem we get, MC2=MN2+NC2 (12k)2=(5k)2+NC2 NC2+25k2=144k2 NC2=119k2 Such a triangle is possible as it will follow the Pythagoras theorem. (iii) cos A is the abbreviation used for the cosecant of angle A. False Justification: Abbreviation used for cosecant of angle M is cosec M. Cos M is the abbreviation used for cosine of angle M. (iv) cot A is the product of cot and A. False Justification: cot M is not the product of cot and A. It is the cotangent of ∠A. (v) sin θ = 4/3 for some angle θ. False Justification: sin θ = Height/Hypotenuse We know that in a right-angled triangle, Hypotenuse is the longest side. ∴ sin θ will always be less than 1 and it can never be 4/3 for any value of θ.
Exercise 8.2
Q1. Evaluate the following: (i) sin 60° cos 30° + sin 30° cos 60° (ii) 2 tan2 45° + cos2 30° – sin2 60
Solution: (i) The values of the given trigonometric ratios: sin 30° = 1/2, cos 30° = √3/2, sin 60° = 3/2, cos 60°= 1/2 Now, substitute the values in the given problem sin 60° cos 30° + sin 30° cos 60° = √3/2 ×√3/2 + (1/2) ×(1/2 ) = 3/4+1/4 = 4/4 = 1
(ii) The values of the trigonometric ratios: sin 60° = √3/2, cos 30° = √3/2, tan 45° = 1 Substitute the values in the given problem 2 tan2 45° + cos2 30° – sin2 60 = 2(1)2 + (√3/2)2-(√3/2)2 = 2 + 0 = 2
(iii) We know that:
cos 45° = 1/√2, sec 30° = 2/√3, cosec 30° = 2
Substitute the values, we get
Now, multiply both the numerator and denominator by √2, we get
Therefore, cos 45°/(sec 30°+cosec 30°) = (3√2 – √6)/8
We know that, sin 30° = 1/2, tan 45° = 1, cosec 60° = 2/√3, sec 30° = 2/√3, cos 60° = 1/2, cot 45° = 1 Substitute the values in the given problem, we get
We know that, cos 60° = 1/2, sec 30° = 2/√3, tan 45° = 1, sin 30° = 1/2, cos 30° = √3/2 Now, substitute the values in the given problem, we get (5cos260° + 4sec230° – tan245°)/(sin2 30° + cos2 30°) = 5(1/2)2 + 4(2/√3)2 – 12/(1/2)2 + (√3/2)2 = (5/4 + 16/3 – 1)/(1/4 + 3/4) = {(15 + 64 – 12)/12}/(4/4) = 67/12
Q2. Choose the correct option and justify your choice: (i) (a) sin 60° (b) cos 60° (c) tan 60° (d) sin 30° Ans. (A) is correct Justification: Substitute tan 30° = 1/√3 in the given equation
∵ √3/2 = sin 60° The obtained solution is equivalent to the trigonometric ratio sin 60°
(a) tan 90° (b) 1 (c) sin 45° (d) 0
Ans. (D) is correct Justification:
(iii) sin 2A = 2 sin A is true when A = (a) 0° (b) 30° (c) 45° (d) 60° Ans. (A) is correct. Justification: To find the value of A, substitute the degree given in the options one by one sin 2A = 2 sin A is true when A = 0° As, sin 2A = sin 0° = 0 and 2 sin A = 2 sin 0° = 2 × 0 = 0 or, Apply the sin 2A formula, to find the degree value sin 2A = 2sin A cos A ⇒ 2sin A cos A = 2 sin A ⇒ 2cos A = 2 ⇒ cos A = 1 Now, we have to check, to get the solution as 1, which degree value has to be applied. When 0 degree is applied to cos value, i.e., cos 0 = 1 Therefore, ⇒ A = 0°
(iv) (a) cos 60° (b) sin 60° (c) tan 60° (d) sin 30°
Ans. (C) is correct. Justification:
Q3. If tan (A + B) = √3 and tan (A – B) = 1/√3 ,0° < A + B ≤ 90°; A > B, find A and B. Solution: tan (A + B) = √3 Since √3 = tan 60° Now substitute the degree value ⇒ tan (A + B) = tan 60° (A + B) = 60° … (i) The above equation is assumed as equation (i) tan (A – B) = 1/√3 Since 1/√3 = tan 30° Now substitute the degree value ⇒ tan (A – B) = tan 30° (A – B) = 30° … equation (ii) Now add the equation (i) and (ii), we get A + B + A – B = 60° + 30° Cancel the terms B 2A = 90° A= 45° Now, substitute the value of A in equation (i) to find the value of B 45° + B = 60° B = 60° – 45° B = 15° Therefore A = 45° and B = 15°
Q4. State whether the following are true or false. Justify your answer. (i) sin (A + B) = sin A + sin B. Ans. False Justification: Let us take A = 30° and B = 60°, then Substitute the values in the sin (A + B) formula, we get sin (A + B) = sin (30° + 60°) = sin 90° = 1 and, sin A + sin B = sin 30° + sin 60° = 1/2 + √3/2 = 1+√3/2 Since the values obtained are not equal, the solution is false.
(ii) The value of sin θ increases as θ increases. Ans. True Justification: According to the values obtained as per the unit circle, the values of sin are: sin 0° = 0, sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2, sin 90° = 1 Thus the value of sin θ increases as θ increases. Hence, the statement is true
(iii) The value of cos θ increases as θ increases. Ans. False Justification: According to the values obtained as per the unit circle, the values of cos are: cos 0° = 1, cos 30° = √3/2, cos 45° = 1/√2, cos 60° = 1/2, cos 90° = 0 Thus, the value of cos θ decreases as θ increases. So, the statement given above is false.
(iv) sin θ = cos θ for all values of θ. Ans. False Justification: sin θ = cos θ, when a right triangle has 2 angles of (π/4). Therefore, the above statement is false.
(v) cot A is not defined for A = 0°. Ans. True Justification: Since cot function is the reciprocal of the tan function, it is also written as: cot A = cos A/sin A Now substitute A = 0° cot 0° = cos 0°/sin 0° = 1/0 = undefined. Hence, it is true
Exercise 8.3
Q1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A. Solution: To convert the given trigonometric ratios in terms of cot functions, use trigonometric formulas We know that, cosec2A– cot2A = 1 cosec2A = 1 + cot2A Since cosec function is the inverse of sin function, it is written as 1/sin2A = 1 + cot2A Now, rearrange the terms, it becomes sin2A = 1/(1+cot2A) Now, take square roots on both sides, we get sin A = ±1/(√(1+cot2A) The above equation defines the sin function in terms of cot function Now, to express sec function in terms of cot function, use this formula sin2A = 1/ (1+cot2A) Now, represent the sin function as cos function 1 – cos2A = 1/ (1+cot2A) Rearrange the terms, cos2A = 1 – 1/(1+cot2A) ⇒cos2A = (1-1+cot2A)/(1+cot2A) Since sec function is the inverse of cos function, ⇒ 1/sec2A = cot2A/(1+cot2A) Take the reciprocal and square roots on both sides, we get ⇒ sec A = ±√ (1+cot2A)/cotA Now, to express tan function in terms of cot function tan A = sin A/cos A and cot A = cos A/sin A Since cot function is the inverse of tan function, it is rewritten as tan A = 1/cot A
Q2. Write all the other trigonometric ratios of ∠A in terms of sec A. Solution: Cos A function in terms of sec A: sec A = 1/cos A ⇒ cos A = 1/sec A sec A function in terms of sec A: cos2A + sin2A = 1 Rearrange the terms sin2A = 1 – cos2A sin2A = 1 – (1/sec2A) sin2A = (sec2A-1)/sec2A sin A = ± √(sec2A-1)/sec A cosec A function in terms of sec A: sin A = 1/cosec A ⇒cosec A = 1/sin A cosec A = ± sec A/√(sec2A-1) Now, tan A function in terms of sec A: sec2A – tan2A = 1 Rearrange the terms ⇒ tan2A = sec2A + 1 tan A = √(sec2A + 1) cot A function in terms of sec A: tan A = 1/cot A ⇒ cot A = 1/tan A cot A = ±1/√(sec2A + 1)
Q3. Choose the correct option. Justify your choice. (i) 9 sec2A – 9 tan2A = (a) 1 (b) 9 (c) 8 (d) 0 (ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ) (a) 0 (b) 1 (c) 2 (d) – 1 (iii) (sec A + tan A) (1 – sin A) = (a) sec A (b) sin A (c) cosec A (d) cos A
(iv) 1+tan2A/1+cot2A = (a) sec2 A (b) -1 (c) cot2A (d) tan2A Solution: (i) (B) is correct. Justification: Take 9 outside, and it becomes 9 sec2A – 9 tan2A = 9 (sec2A – tan2A) = 9×1 = 9 (∵ sec2 A – tan2 A = 1) Therefore, 9 sec2A – 9 tan2A = 9 (ii) (C) is correct Justification: (1 + tan θ + sec θ) (1 + cot θ – cosec θ) We know that, tan θ = sin θ/cos θ sec θ = 1/ cos θ cot θ = cos θ/sin θ cosec θ = 1/sin θ Now, substitute the above values in the given problem, we get = (1 + sin θ/cos θ + 1/ cos θ) (1 + cos θ/sin θ – 1/sin θ) Simplify the above equation, = (cos θ +sin θ+1)/cos θ × (sin θ+cos θ-1)/sin θ = (cos θ+sin θ)2-12/(cos θ sin θ) = (cos2θ + sin2θ + 2cos θ sin θ -1)/(cos θ sin θ) = (1+ 2cos θ sin θ -1)/(cos θ sin θ) (Since cos2θ + sin2θ = 1) = (2cos θ sin θ)/(cos θ sin θ) = 2 Therefore, (1 + tan θ + sec θ) (1 + cot θ – cosec θ) =2 (iii) (D) is correct. Justification: We know that, Sec A= 1/cos A Tan A = sin A / cos A Now, substitute the above values in the given problem, we get (secA + tanA) (1 – sinA) = (1/cos A + sin A/cos A) (1 – sinA) = (1+sin A/cos A) (1 – sinA) = (1 – sin2A)/cos A = cos2A/cos A = cos A Therefore, (secA + tanA) (1 – sinA) = cos A (iv) (D) is correct. Justification: We know that, tan2A =1/cot2A Now, substitute this in the given problem, we get 1+tan2A/1+cot2A = (1+1/cot2A)/1+cot2A = (cot2A+1/cot2A)×(1/1+cot2A) = 1/cot2A = tan2A So, 1+tan2A/1+cot2A = tan2A
Q4. Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (i) (cosec θ – cot θ)2 = (1-cos θ)/(1+cos θ) (ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A (iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ [Hint : Write the expression in terms of sin θ and cos θ] (iv) (1 + sec A)/sec A = sin2A/(1-cos A) [Hint : Simplify LHS and RHS separately] (v) ( cos A–sin A+1)/( cos A +sin A–1) = cosec A + cot A, using the identity cosec2A = 1+cot2A.
(vii) (sin θ – 2sin3θ)/(2cos3θ-cos θ) = tan θ (viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7+tan2A+cot2A (ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA) [Hint : Simplify LHS and RHS separately] (x) (1+tan2A/1+cot2A) = (1-tan A/1-cot A)2 =tan2A Solution: (i) (cosec θ – cot θ)2 = (1-cos θ)/(1+cos θ) To prove this, first take the Left-Hand side (L.H.S) of the given equation, to prove the Right Hand Side (R.H.S) L.H.S. = (cosec θ – cot θ)2 The above equation is in the form of (a-b)2, and expand it Since (a-b)2 = a2 + b2 – 2ab Here a = cosec θ and b = cot θ = (cosec2θ + cot2θ – 2cosec θ cot θ) Now, apply the corresponding inverse functions and equivalent ratios to simplify = (1/sin2θ + cos2θ/sin2θ – 2cos θ/sin2θ) = (1 + cos2θ – 2cos θ)/(1 – cos2θ) = (1-cos θ)2/(1 – cosθ)(1+cos θ) = (1-cos θ)/(1+cos θ) = R.H.S. Therefore, (cosec θ – cot θ)2 = (1-cos θ)/(1+cos θ) Hence proved. (ii) (cos A/(1+sin A)) + ((1+sin A)/cos A) = 2 sec A Now, take the L.H.S of the given equation. L.H.S. = (cos A/(1+sin A)) + ((1+sin A)/cos A) = [cos2A + (1+sin A)2]/(1+sin A)cos A = (cos2A + sin2A + 1 + 2sin A)/(1+sin A) cos A Since cos2A + sin2A = 1, we can write it as = (1 + 1 + 2sin A)/(1+sin A) cos A = (2+ 2sin A)/(1+sin A)cos A = 2(1+sin A)/(1+sin A)cos A = 2/cos A = 2 sec A = R.H.S. L.H.S. = R.H.S. (cos A/(1+sin A)) + ((1+sin A)/cos A) = 2 sec A Hence proved. (iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ L.H.S. = tan θ/(1-cot θ) + cot θ/(1-tan θ) We know that tan θ =sin θ/cos θ cot θ = cos θ/sin θ Now, substitute it in the given equation, to convert it in a simplified form = [(sin θ/cos θ)/1-(cos θ/sin θ)] + [(cos θ/sin θ)/1-(sin θ/cos θ)] = [(sin θ/cos θ)/(sin θ-cos θ)/sin θ] + [(cos θ/sin θ)/(cos θ-sin θ)/cos θ] = sin2θ/[cos θ(sin θ-cos θ)] + cos2θ/[sin θ(cos θ-sin θ)] = sin2θ/[cos θ(sin θ-cos θ)] – cos2θ/[sin θ(sin θ-cos θ)] = 1/(sin θ-cos θ) [(sin2θ/cos θ) – (cos2θ/sin θ)] = 1/(sin θ-cos θ) × [(sin3θ – cos3θ)/sin θ cos θ] = [(sin θ-cos θ)(sin2θ+cos2θ+sin θ cos θ)]/[(sin θ-cos θ)sin θ cos θ] = (1 + sin θ cos θ)/sin θ cos θ = 1/sin θ cos θ + 1 = 1 + sec θ cosec θ = R.H.S. Therefore, L.H.S. = R.H.S. Hence proved (iv) (1 + sec A)/sec A = sin2A/(1-cos A) First find the simplified form of L.H.S L.H.S. = (1 + sec A)/sec A Since secant function is the inverse function of cos function and it is written as = (1 + 1/cos A)/1/cos A = (cos A + 1)/cos A/1/cos A Therefore, (1 + sec A)/sec A = cos A + 1 R.H.S. = sin2A/(1-cos A) We know that sin2A = (1 – cos2A), we get = (1 – cos2A)/(1-cos A) = (1-cos A)(1+cos A)/(1-cos A) Therefore, sin2A/(1-cos A)= cos A + 1 L.H.S. = R.H.S. Hence proved (v) (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A, using the identity cosec2A = 1+cot2A. With the help of identity function, cosec2A = 1+cot2A, let us prove the above equation. L.H.S. = (cos A–sin A+1)/(cos A+sin A–1) Divide the numerator and denominator by sin A, we get = (cos A–sin A+1)/sin A/(cos A+sin A–1)/sin A We know that cos A/sin A = cot A and 1/sin A = cosec A = (cot A – 1 + cosec A)/(cot A+ 1 – cosec A) = (cot A – cosec2A + cot2A + cosec A)/(cot A+ 1 – cosec A) (using cosec2A – cot2A = 1 = [(cot A + cosec A) – (cosec2A – cot2A)]/(cot A+ 1 – cosec A) = [(cot A + cosec A) – (cosec A + cot A)(cosec A – cot A)]/(1 – cosec A + cot A) = (cot A + cosec A)(1 – cosec A + cot A)/(1 – cosec A + cot A) = cot A + cosec A = R.H.S. Therefore, (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A Hence Proved
First divide the numerator and denominator of L.H.S. by cos A,
We know that 1/cos A = sec A and sin A/ cos A = tan A and it becomes, = √(sec A+ tan A)/(sec A-tan A) Now using rationalization, we get
= (sec A + tan A)/1 = sec A + tan A = R.H.S Hence proved (vii) (sin θ – 2sin3θ)/(2cos3θ-cos θ) = tan θ L.H.S. = (sin θ – 2sin3θ)/(2cos3θ – cos θ) Take sin θ as in numerator and cos θ in denominator as outside, it becomes = [sin θ(1 – 2sin2θ)]/[cos θ(2cos2θ- 1)] We know that sin2θ = 1-cos2θ = sin θ[1 – 2(1-cos2θ)]/[cos θ(2cos2θ -1)] = [sin θ(2cos2θ -1)]/[cos θ(2cos2θ -1)] = tan θ = R.H.S. Hence proved (viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7+tan2A+cot2A L.H.S. = (sin A + cosec A)2 + (cos A + sec A)2 It is of the form (a+b)2, expand it (a+b)2 =a2 + b2 +2ab = (sin2A + cosec2A + 2 sin A cosec A) + (cos2A + sec2A + 2 cos A sec A) = (sin2A + cos2A) + 2 sin A(1/sin A) + 2 cos A(1/cos A) + 1 + tan2A + 1 + cot2A = 1 + 2 + 2 + 2 + tan2A + cot2A = 7+tan2A+cot2A = R.H.S. Therefore, (sin A + cosec A)2 + (cos A + sec A)2 = 7+tan2A+cot2A Hence proved. (ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A + cotA) First, find the simplified form of L.H.S L.H.S. = (cosec A – sin A)(sec A – cos A) Now, substitute the inverse and equivalent trigonometric ratio forms = (1/sin A – sin A)(1/cos A – cos A) = [(1-sin2A)/sin A][(1-cos2A)/cos A] = (cos2A/sin A)×(sin2A/cos A) = cos A sin A Now, simplify the R.H.S R.H.S. = 1/(tan A+cotA) = 1/(sin A/cos A +cos A/sin A) = 1/[(sin2A+cos2A)/sin A cos A] = cos A sin A L.H.S. = R.H.S. (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA) Hence proved (x) (1+tan2A/1+cot2A) = (1-tan A/1-cot A)2 =tan2A L.H.S. = (1+tan2A/1+cot2A) Since cot function is the inverse of tan function, = (1+tan2A/1+1/tan2A) = 1+tan2A/[(1+tan2A)/tan2A] Now cancel the 1+tan2A terms, we get = tan2A (1+tan2A/1+cot2A) = tan2A Similarly, (1-tan A/1-cot A)2 =tan2A Hence proved
Q1. Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2 : 3. Ans: Given, Let P(x, y) be the required point. Let A(−1, 7) and B(4, −3) m: n = 2:3 Hence x1 = −1 y1 = 7 x2 = 4 y2 = −3 By Section formula
By substituting the values in the Equation (1)
Therefore, the co-ordinates of point P are (1, 3).
Q2. Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3). Ans:
Given, Let line segment joining the points be A(4, −1) and B(−2, −3). Let P (x1, y1) and Q (x2, y2) be the points of trisection of the line segment joining the given points i.e., AP = PQ = QB By Section formula
Therefore, by observation point P divides AB internally in the ratio 1:2. Hence m: n = 1:2 By substituting the values in the Equation (1)
Therefore, Therefore, by observation point Q divides AB internally in the ratio 2:1. Hence m:n = 2:1 By substituting the values in the Equation (1)
Therefore, Hence the points of trisection are P(x1, y1) =
Q3. To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in the figure. Niharika runs 1/4 th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5 th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?Ans: From the Figure, Given,
By observation, that Niharika posted the green flag at of the distance P i.e., from the starting point of 2nd line. Therefore, the coordinates of this point P is (2, 25).
Similarly, Preet posted red flag at 1/5 of the distance from the starting point of 8th line. Therefore, the coordinates of this point Q are (8, 20)
We know that the distance between the two points is given by the Distance Formula,
To find the distance between these flags PQ by substituting the values in Equation (1),
The point at which Rashmi should post her blue flag is the mid-point of the line joining these points.
Let this point be M (x, y).
By Section formula
Therefore, Rashmi should post her blue flag at 22.5 m on 5th line
Q4. Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6). Ans: From the figure, Given,
Let the ratio in which the line segment joining A(−3, 10) and B(6, −8) is divided by point P(−1, 6) be k:1.
By Section formula
Therefore,
Hence the point P divides AB in the ratio 2:7
Q5. Find the ratio in which the line segment joining A (1, -5) and B (-4, 5) is divided by the x-axis. Also find the coordinates of the point of division. Ans: From the Figure,Given,
Let the ratio be k : 1.
Let the line segment joining A (1, −5) and B (−4, 5)
By Section formula
By substituting the values in Equation (1) Therefore, the coordinates of the point of division is We know that y-coordinate of any point on x-axis is 0.
Therefore, x-axis divides it in the ratio 1:1.
Division point
Q6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y. Ans: From the Figure,
Given,
Let A (1, 2), B (4, y), C(x, 6), and D (3, 5) are the vertices of a parallelogram ABCD.
Since the diagonals of a parallelogram bisect each other, Intersection point O of diagonal AC and BD also divides these diagonals
Therefore, O is the mid-point of AC and BD. If O is the mid-point of AC, then the coordinates of O are
If O is the mid-point of BD, then the coordinates of O are
Since both the coordinates are of the same point O,
Q7. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, -3) and B is (1, 4). Ans: From the Figure,
Given,
Let the coordinates of point A be (x, y).
Mid-point of AB is C (2, −3), which is the center of the circle.
Therefore, the coordinates of A are (3, −10)
Q8. If A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that and P lies on the line segment AB. Ans: From the Figure, Given,
The coordinates of point A and B are (−2, −2) and (2, −4) respectively.
AP = 3/7AB
Hence AB/AP = 7/3 We know that AB = AP + PB from figure,
Therefore, AP:PB = 3:4 Point P(x, y) divides the line segment AB in the ratio 3:4. Using Section Formula,
Q9. Find the coordinates of the points which divide the line segment joining A (-2, 2) and B (2, 8) into four equal parts. Ans: From the Figure,
By observation, that points P, Q, R divides the line segment A (−2, 2) and B (2, 8) into four equal parts Point P divides the line segment AQ into two equal parts Hence, Coordinates of P = = Point Q divides the line segment AB into two equal parts Coordinates of Q
Point R divides the line segment BQ into two equal parts Coordinates of R
Q10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order. [Hint: Area of a rhombus = 1/2 (product of its diagonals)] Ans: From the Figure,
Given,
Let A(3, 0), B(4, 5), C(−1, 4) and D(−2, −1) are the vertices of a rhombus ABCD.
We know that the distance between the two points is given by the Distance Formula, Therefore, distance between A (3, 0) and C (−1, 4) is given by
Therefore, distance between B (4, 5) and D (−2, −1) is given by
Area of the rhombus ABCD = 1/2 x (Product of lengths of diagonals) = 1/2 AC x BD Therefore, area of rhombus
Q1. Find the distance between the following pairs of points: (i) (2, 3), (4, 1)
Ans: Let P(2, 3) and Q(4, 1)
(ii) (-5, 7), (-1, 3)
Ans: Let P (-5, 7) and Q (-1, 3)
(iii) (a, b), (-a, -b)
Ans: Let P (a, b) and Q (-a, -b)
Q2. Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2? Ans: Let points be A (0, 0) and B (36, 15) The distance between the two points is:
The distance between towns A and B will be 39 km.
Q3. Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear. Ans: Let points be A(1, 5), B(2, 3) and C(-2, -11)
AB + BC ≠ AC Hence, the given points are not collinear.
Q4. Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle. Ans: Since two sides of any isosceles triangle are equal. To check whether the given points are vertices of an isosceles triangle, we will find the distance between all the points.
Let the points be A (5, -2), B (6, 4) and C (7, -2).
Here AB = BC ∴ Δ ABC is an isosceles.
Q5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, ‘‘Don’t you think ABCD is a square?’’ Chameli disagrees. Using distance single formula, find which of them is correct.
Ans: Points A (3, 4), B (6, 7), C (9, 4) and D (6, 1)
∴ ABCD is a square. Hence, Champa is correct.
Q6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (i) (-1, -2), (1, 0), (-1, 2), (-3, 0) Ans: Let the points be: A (-1, -2), B (1, 0), C (-1, 2) and D (-3, 0).
Here AC = BD, AB = BC = CD = AD Hence, the quadrilateral ABCD is a square. (ii) (-3, 5), (3, 1), (0, 3), (-1, – 4) Let points be A (-3, 5), B (3, 1), C (0, 3) and D (-1, -4)
It is seen that points A, B and C are collinear.
So, the given points can only form 3 sides i.e, a triangle and not a quadrilateral that has 4 sides.
Therefore, the given points do not form any quadrilateral. (iii) (4, 5), (7, 6), (4, 3), (1, 2) Let points be A (4, 5), B (7, 6), C (4, 3) and D (1, 2)
Here, AB = CD, BC = AD and AC ≠ BD ∴ The quadrilateral ABCD is a parallelogram.
Q7. Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9). Ans: Let points be A (2, -5) and B (-2, 9) Let P (x, 0) be the point on the x-axis. PA = PB Then, PA2 = PB2
Q8. Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units. Ans: Points P (2, -3), Q (10, y) and PQ = 10 units The distance between the two points is:
⇒ y – 3 = 0 or y + 9 = 0 ⇒ y = 3 or – 9
Q9. If Q (0, 1) is equidistant from P (5, -3) and R (x, 6), find the values of x. Also, find the distances QR and PR. Ans: PQ = QR
Therefore, point R is (4, 6) or (−4, 6). When point R is (4, 6),
Q10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (-3, 4). Ans: Points A (3, 6), B (-3, 4) are given and point P(x, y) is equidistant from points A and B.
Q1: Fill in the blanks using the correct word given in brackets. (i) All circles are _______(congruent, similar). (ii) All squares are _______ (similar, congruent). (iii) All _______ triangles are similar. (isosceles, equilateral). (iv)Two polygons of the same number of sides are similar, if (a) their corresponding angles are _______ and (b) their corresponding sides are _______ (equal, proportional). Ans: (i) All circles are similar. (ii) All squares are similar. (iii) All equilateral triangles are similar. (iv) Two polygons of the same number of sides are similar if: (a) Their corresponding angles are equal and (b) Their corresponding sides are proportional.
Q2: Give two different examples of pair of: (i) similar figures (ii) non-similar figures. Ans: (i). (a) Any two circles are similar figures. (b) Any two squares are similar figures. (ii). (a) A circle and a triangle are non-similar figures. (b) An isosceles triangle and a scalene triangle are non-similar figures.
Q3: State whether the following quadrilaterals are similar or not.
Ans: No, the sides of quadrilateral PQRS and ABCD are proportional but their corresponding angles are not equal. ∴ These are not similar.Exercise 6.2
Q1. In figures (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
⇒ EC = 3/1.5 EC = 3×10/15 = 2 cm Hence, EC = 2 cm.
(ii) Given, in △ ABC, DE∥BC ∴ AD/DB = AE/EC [Using Basic proportionality theorem] ⇒ AD/7.2 = 1.8 / 5.4 ⇒ AD = 1.8 ×7.2/5.4 = (18/10)×(72/10)×(10/54) = 24/10 ⇒ AD = 2.4 Hence, AD = 2.4 cm.
Q2. E and F are points on the sides PQ and PR respectively of a ΔPQR. For each of the following cases, state whether EF || QR: (i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
Sol. Given, in ΔPQR, E and F are two points on side PQ and PR respectively. See the figure below;
(i) Given, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2,4 cm Therefore, by using Basic proportionality theorem, we get, PE/EQ = 3.9/3 = 39/30 = 13/10 = 1.3 And PF/FR = 3.6/2.4 = 36/24 = 3/2 = 1.5 So, we get, PE/EQ ≠ PF/FR Hence, EF is not parallel to QR.
(ii) Given, PE = 4 cm, QE = 4.5 cm, PF = 8cm and RF = 9cm Therefore, by using Basic proportionality theorem, we get, PE/QE = 4/4.5 = 40/45 = 8/9 And, PF/RF = 8/9 So, we get here, PE/QE = PF/RF Hence, EF is parallel to QR.
(iii) Given, PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm From the figure, EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm And, FR = PR – PF = 2.56 – 0.36 = 2.20 cm So, PE/EQ = 0.18/1.10 = 18/110 = 9/55…………. (i) And, PE/FR = 0.36/2.20 = 36/220 = 9/55………… (ii) So, we get here, PE/EQ = PF/FR Hence, EF is parallel to QR.
Q3.In the figure, if LM || CB and LN || CD, prove that AM/AB = AN/AD Sol. In the given figure, we can see, LM || CB, By using basic proportionality theorem, we get, AM/AB = AL/AC……………………..(i) Similarly, given, LN || CD and using basic proportionality theorem, ∴ AN/AD = AL/AC……………………………(ii) From equation (i) and (ii), we get, AM/AB = AN/AD Hence, proved.
Q4. In the figure, DE||AC and DF||AE. Prove that BF/FE = BE/EC Sol. In ΔABC, given as, DE || AC Thus, by using Basic Proportionality Theorem, we get, ∴ BD/DA = BE/EC ………………………………………………(i) In ΔABC, given as, DF || AE Thus, by using Basic Proportionality Theorem, we get, ∴ BD/DA = BF/FE ………………………………………………(ii) From equation (i) and (ii), we get BE/EC = BF/FE Hence, proved.
Q5. In the figure, DE || OQ and DF || OR. Show that EF || QR. Sol. Given, In ΔPQO, DE || OQ So by using Basic Proportionality Theorem, PD/DO = PE/EQ……………… ..(i) Again given, in ΔPQO, DE || OQ , So by using Basic Proportionality Theorem, PD/DO = PF/FR………………… (ii) From equation (i) and (ii), we get, PE/EQ = PF/FR Therefore, by converse of Basic Proportionality Theorem, EF || QR, in ΔPQR.
Q6. In the figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR. Sol. Given here, In ΔOPQ, AB || PQ By using Basic Proportionality Theorem, OA/AP = OB/BQ…………….(i) Also given, In ΔOPR, AC || PR By using Basic Proportionality Theorem ∴ OA/AP = OC/CR……………(ii) From equation (i) and (ii), we get, OB/BQ = OC/CR Therefore, by converse of Basic Proportionality Theorem, In ΔOQR, BC || QR.
Q7. Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side. Sol. Given, in ΔABC, D is the midpoint of AB such that AD=DB. A line parallel to BC intersects AC at E as shown in above figure such that DE || BC. We have to prove that E is the mid point of AC. Since, D is the mid-point of AB. ∴ AD=DB ⇒ AD/DB = 1 ………… (i) In ΔABC, DE || BC, By using Basic Proportionality Theorem, Therefore, AD/DB = AE/EC From equation (i), we can write, ⇒ 1 = AE/EC ∴ AE = EC Hence, proved, E is the midpoint of AC.
Q8. Using Converse of basic proportionality theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. Sol. Given, in ΔABC, D and E are the mid points of AB and AC respectively, such that, AD=BD and AE=EC. We have to prove that: DE || BC. Since, D is the midpoint of AB ∴ AD=DB ⇒ AD/BD = 1…………. (i) Also given, E is the mid-point of AC. ∴ AE=EC ⇒ AE/EC = 1 From equation (i) and (ii), we get, AD/BD = AE/EC By converse of Basic Proportionality Theorem, DE || BC Hence, proved.
Q9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO. Sol. Given, ABCD is a trapezium where AB || DC and diagonals AC and BD intersect each other at O.
We have to prove, AO/BO = CO/DO From the point O, draw a line EO touching AD at E, in such a way that, EO || DC || AB In ΔADC, we have OE || DC Therefore, By using Basic Proportionality Theorem AE/ED = AO/CO ……………..(i) Now, In ΔABD, OE || AB Therefore, By using Basic Proportionality Theorem DE/EA = DO/BO…………….(ii) From equation (i) and (ii), we get, AO/CO = BO/DO ⇒ AO/BO = CO/DO Hence, proved.
Q10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is a trapezium. Sol. Given, Quadrilateral ABCD where AC and BD intersects each other at O such that, AO/BO = CO/DO.
We have to prove here, ABCD is a trapezium From the point O, draw a line EO touching AD at E, in such a way that, EO || DC || AB In ΔDAB, EO || AB Therefore, By using Basic Proportionality Theorem DE/EA = DO/OB ………(i) Also, given, AO/BO = CO/DO ⇒ AO/CO = BO/DO ⇒ CO/AO = DO/BO ⇒ DO/OB = CO/AO ………….(ii) From equation (i) and (ii), we get DE/EA = CO/AO Therefore, By using converse of Basic Proportionality Theorem, EO || DC also EO || AB ⇒ AB || DC. Hence, quadrilateral ABCD is a trapezium with AB || CD.Exercise 6.3
Q1. State which pairs of triangles in Figure, are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:
(ii) Given, in ΔABC and ΔPQR, AB/QR = BC/RP = CA/PQ By SSS similarity criterion, ΔABC ~ ΔQRP
(iii) Given, in ΔLMP and ΔDEF, LM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6 MP/DE = 2/4 = 1/2 PL/DF = 3/6 = 1/2 LM/EF = 2.7/5 = 27/50 Here , MP/DE = PL/DF ≠ LM/EF Therefore, ΔLMP and ΔDEF are not similar.
(iv) In ΔMNL and ΔQPR, it is given, MN/QP = ML/QR = 1/2 ∠M = ∠Q = 70° Therefore, by SAS similarity criterion ∴ ΔMNL ~ ΔQPR
(v) In ΔABC and ΔDEF, given that, AB = 2.5, BC = 3, ∠A = 80°, EF = 6, DF = 5, ∠F = 80° Here , AB/DF = 2.5/5 = 1/2 And, BC/EF = 3/6 = 1/2 ⇒ ∠B ≠ ∠F Hence, ΔABC and ΔDEF are not similar.
(vi) In ΔDEF, by sum of angles of triangles, we know that, ∠D + ∠E + ∠F = 180° ⇒ 70° + 80° + ∠F = 180° ⇒ ∠F = 180° – 70° – 80° ⇒ ∠F = 30° Similarly, In ΔPQR, ∠P + ∠Q + ∠R = 180 (Sum of angles of Δ) ⇒ ∠P + 80° + 30° = 180° ⇒ ∠P = 180° – 80° -30° ⇒ ∠P = 70° Now, comparing both the triangles, ΔDEF and ΔPQR, we have ∠D = ∠P = 70° ∠F = ∠Q = 80° ∠F = ∠R = 30° Therefore, by AAA similarity criterion, Hence, ΔDEF ~ ΔPQR
Q2. In the figure, ΔODC ~ ΔOBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.
Sol. As we can see from the figure, DOB is a straight line. Therefore, ∠DOC + ∠ COB = 180° ⇒ ∠DOC = 180° – 125° (Given, ∠ BOC = 125°) = 55° In ΔDOC, sum of the measures of the angles of a triangle is 180º Therefore, ∠DCO + ∠ CDO + ∠ DOC = 180° ⇒ ∠DCO + 70º + 55º = 180°(Given, ∠ CDO = 70°) ⇒ ∠DCO = 55° It is given that, ΔODC ∝ ¼ ΔOBA, Therefore, ΔODC ~ ΔOBA. Hence, Corresponding angles are equal in similar triangles ∠OAB = ∠OCD ⇒ ∠ OAB = 55° ∠OAB = ∠OCD ⇒ ∠OAB = 55°
Q3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that AO/OC = OB/OD Sol.
In ΔDOC and ΔBOA, AB || CD, thus alternate interior angles will be equal, ∴ ∠CDO = ∠ABO Similarly, ∠DCO = ∠BAO Also, for the two triangles ΔDOC and ΔBOA, vertically opposite angles will be equal; ∴ ∠DOC = ∠BOA Hence, by AAA similarity criterion, ΔDOC ~ ΔBOA Thus, the corresponding sides are proportional. DO/BO = OC/OA ⇒ OA/OC = OB/OD Hence, proved.
Q4. In the figure, QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR. Sol. In ΔPQR, ∠PQR = ∠PRQ ∴ PQ = PR ……(i) Given, QR/QS = QT/PR Using equation (i), we get QR/QS = QT/QP……….(ii) In ΔPQS and ΔTQR, by equation (ii), QR/QS = QT/QP ∠Q = ∠Q ∴ ΔPQS ~ ΔTQR [By SAS similarity criterion]
Q5. S and T are points on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS. Sol. Given, S and T are point on sides PR and QR of ΔPQR
And ∠P = ∠RTS. In ΔRPQ and ΔRTS, ∠RTS = ∠QPS (Given) ∠R = ∠R (Common angle) ∴ ΔRPQ ~ ΔRTS (AA similarity criterion)
Q6. In the figure, if ΔABE ≌ ΔACD, show that ΔADE ~ ΔABC. Sol. Given, ΔABE ≅ ΔACD. ∴ AB = AC [By CPCT] …………….(i) And, AD = AE [By CPCT] ………(ii) In ΔADE and ΔABC, dividing eq.(ii) by eq(i), AD/AB = AE/AC ∠A = ∠A [Common angle] ∴ ΔADE ~ ΔABC [SAS similarity criterion]
Q7. In the figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:
(i) ΔAEP ~ ΔCDP (ii) ΔABD ~ ΔCBE (iii) ΔAEP ~ ΔADB (iv) ΔPDC ~ ΔBEC Sol. Given, altitudes AD and CE of ΔABC intersect each other at the point P. (i) In ΔAEP and ΔCDP, ∠AEP = ∠CDP (90° each) ∠APE = ∠CPD (Vertically opposite angles) Hence, by AA similarity criterion, ΔAEP ~ ΔCDP (ii) In ΔABD and ΔCBE, ∠ADB = ∠CEB ( 90° each) ∠ABD = ∠CBE (Common Angles) Hence, by AA similarity criterion, ΔABD ~ ΔCBE (iii) In ΔAEP and ΔADB, ∠AEP = ∠ADB (90° each) ∠PAE = ∠DAB (Common Angles) Hence, by AA similarity criterion, ΔAEP ~ ΔADB (iv) In ΔPDC and ΔBEC, ∠PDC = ∠BEC (90° each) ∠PCD = ∠BCE (Common angles) Hence, by AA similarity criterion, ΔPDC ~ ΔBEC
Q8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that Δ ABE ~ ΔCFB. Sol. Given, E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Consider the figure below,
In ΔABE and ΔCFB, ∠A = ∠C (Opposite angles of a parallelogram) ∠AEB = ∠CBF (Alternate interior angles as AE || BC) ∴ ΔABE ~ ΔCFB (AA similarity criterion)
Q9. In the figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: (i) Δ ABC ~ Δ AMP (ii) CA/PA = BC/MP Sol. Given, ABC and AMP are two right triangles, right angled at B and M respectively. (i) In ΔABC and ΔAMP, we have, ∠CAB = ∠MAP (common angles) ∠ABC = ∠AMP = 90° (each 90°) ∴ ΔABC ~ ΔAMP (AA similarity criterion) (ii) As, ΔABC ~ ΔAMP (AA similarity criterion) If two triangles are similar then the corresponding sides are always equal, Hence, CA/PA = BC/MP
Q10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of Δ ABC and Δ EFG respectively. If ΔABC ~ ΔFEG, show that: (i) CD/GH = AC/FG (ii) ΔDCB ~ ΔHGE (iii) ΔDCA ~ ΔHGF Sol. Given, CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively.
(i) From the given condition, ΔABC ~ ΔFEG. ∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE Since, ∠ACB = ∠FGE ∴ ∠ACD = ∠FGH (Angle bisector) And, ∠DCB = ∠HGE (Angle bisector) In ΔACD and ΔFGH, ∠A = ∠F ∠ACD = ∠FGH ∴ ΔACD ~ ΔFGH (AA similarity criterion) ⇒ CD/GH = AC/FG (ii) In ΔDCB and ΔHGE, ∠DCB = ∠HGE (Already proved) ∠B = ∠E (Already proved) ∴ ΔDCB ~ ΔHGE (AA similarity criterion)
Q11. In the figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥BC and EF ⊥AC, prove that ΔABD ~ ΔECF.
Sol. Given, ABC is an isosceles triangle. ∴ AB = AC ⇒ ∠ABD = ∠ECF In ΔABD and ΔECF, ∠ADB = ∠EFC (Each 90°) ∠BAD = ∠CEF (Already proved) ∴ ΔABD ~ ΔECF (using AA similarity criterion)
Q12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of Δ PQR (see figure). Show that Δ ABC ~ ΔPQR. Sol. Given, ΔABC and ΔPQR, AB, BC and median AD of ΔABC are proportional to sides PQ, QR and median PM of ΔPQR i.e. AB/PQ = BC/QR = AD/PM We have to prove: ΔABC ~ ΔPQR As we know here, AB/PQ = BC/QR = AD/PM
⇒ AB/PQ = BC/QR = AD/PM (D is the midpoint of BC. M is the midpoint of QR) ⇒ ΔABD ~ ΔPQM [SSS similarity criterion] ∴ ∠ABD = ∠PQM [Corresponding angles of two similar triangles are equal] ⇒ ∠ABC = ∠PQR In ΔABC and ΔPQR AB/PQ = BC/QR ………………………….(i) ∠ABC = ∠PQR ……………………………(ii) From equation (i) and (ii), we get, ΔABC ~ ΔPQR [SAS similarity criterion]
Q13. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA2 = CB.CD. Sol. Given, D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC.
In ΔADC and ΔBAC, ∠ADC = ∠BAC (Already given) ∠ACD = ∠BCA (Common angles) ∴ ΔADC ~ ΔBAC (AA similarity criterion) We know that corresponding sides of similar triangles are in proportion. ∴ CA/CB = CD/CA ⇒ CA2 = CB.CD. Hence, proved.
Q14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ~ ΔPQR. Sol.
Given: Two triangles ΔABC and ΔPQR in which AD and PM are medians such that; AB/PQ = AC/PR = AD/PM We have to prove, ΔABC ~ ΔPQR Let us construct first: Produce AD to E so that AD = DE. Join CE, Similarly produce PM to N such that PM = MN, also Join RN. In ΔABD and ΔCDE, we have AD = DE [By Construction.] BD = DC [Since, AP is the median] and, ∠ADB = ∠CDE [Vertically opposite angles] ∴ ΔABD ≅ ΔCDE [SAS criterion of congruence] ⇒ AB = CE [By CPCT] …………………………..(i) Also, in ΔPQM and ΔMNR, PM = MN [By Construction.] QM = MR [Since, PM is the median] and, ∠PMQ = ∠NMR [Vertically opposite angles] ∴ ΔPQM = ΔMNR [SAS criterion of congruence] ⇒ PQ = RN [CPCT] ……………(ii) Now, AB/PQ = AC/PR = AD/PM From equation (i) and (ii), ⇒ CE/RN = AC/PR = AD/PM ⇒ CE/RN = AC/PR = 2AD/2PM ⇒ CE/RN = AC/PR = AE/PN [Since 2AD = AE and 2PM = PN] ∴ ΔACE ~ ΔPRN [SSS similarity criterion] Therefore, ∠2 = ∠4 Similarly, ∠1 = ∠3 ∴ ∠1 + ∠2 = ∠3 + ∠4 ⇒ ∠A = ∠P …………….(iii) Now, in ΔABC and ΔPQR, we have AB/PQ = AC/PR (Already given) From equation (iii), ∠A = ∠P ∴ ΔABC ~ ΔPQR [ SAS similarity criterion]
Q15. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower. Sol.
Given, Length of the vertical pole = 6m
Shadow of the pole = 4 m
Let Height of tower = h m
Length of shadow of the tower = 28 m
In ΔABC and ΔDEF, ∠C = ∠E (angular elevation of sum) ∠B = ∠F = 90° ∴ ΔABC ~ ΔDEF (AA similarity criterion) ∴ AB/DF = BC/EF (If two triangles are similar corresponding sides are proportional) ∴ 6/h = 4/28 ⇒ h = (6×28)/4 ⇒ h = 6 × 7 ⇒ h = 42 m Hence, the height of the tower is 42 m.
Q16. If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR, prove that AB/PQ = AD/PM.. Sol. Given, ΔABC ~ ΔPQR We know that the corresponding sides of similar triangles are in proportion. ∴ AB/PQ = AC/PR = BC/QR……………………………(i) Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R ………….…..(ii) Since AD and PM are medians, they will divide their opposite sides. ∴ BD = BC/2 and QM = QR/2 ……………..………….(iii) From equations (i) and (iii), we get AB/PQ = BD/QM ……………………….(iv) In ΔABD and ΔPQM, From equation (ii), we have ∠B = ∠Q From equation (iv), we have, AB/PQ = BD/QM ∴ ΔABD ~ ΔPQM (SAS similarity criterion) ⇒ AB/PQ = BD/QM = AD/PM
Q1: Find the sum of the following APs: (i) 2, 7, 12, …, to 10 terms.
Here, a = 2 d = 7 – 2 = 5 n = 10 Since, Sn = n/2 [2a + (n – 1) d] ∴ S10 = [2 × 2 + (10 – 1) × 5] ⇒ S10 = 5 [4 + 9 × 5] ⇒ S10 = 5 [49] = 245 Thus, the sum of first 10 terms is 245.
(ii) – 37, – 33, – 29, …, to 12 terms.
We have: a = – 37 d = – 33 – (- 37) = 4 n = 12 ∴ Sn = n/2 [2a + (n – 1) d] ⇒ S12 = 12/2 [2 (- 37) + (12 – 1) × 4] = 6 [- 74 + 11 × 4] = 6 [- 74 + 44] = 6 × [- 30] = – 180 Thus, sum of first 12 terms = -180.
(iii) 0.6, 1.7, 2.8, …, to 100 terms.
Here, a = 0.6 d = 1.7 – 0.6 = 1.1 n = 100 ∴ Sn = n/2 [2a + (n – 1) d] S100 = 100/2 [2 (0.6) + (100 – 1) × 1.1] = 50 [1.2 + 99 × 1.1] = 50 [1.2 + 108.9] = 50 [110.1] = 5505 Thus, the required sum of first 100 terms is 5505.
(iv) Solve the following , …, to 11 terms.
Here, a = 1/15 d = n = 11 ∴ Sn = n/2 [2a + (n – 1) d] S11 =
Thus, the required sum of first 11 terms = 33/20.
Q2: Find the sums given below: (i)
Here, a = 7
l = 84 Let n be the number of terms ∴ Tn = a + (n – 1) d ⇒ ⇒ ⇒ n = 22 + 1 = 23 Now, ⇒
Thus, the required sum =
(ii) 34 + 32 + 30 + … + 10
Here, a = 34 d = 32 – 34 = – 2 l = 10 Let the number of terms be n ∴ Tn = 10 Now Tn = a + (n – 1) d ⇒ 10 = 34 + (n – 1) × (- 2) ⇒ (n – 1) × (- 2) = 10 – 34 = – 24 ⇒ ⇒ n = 13
⇒ Now, ⇒ = = = = 13 × 22 = 286 OR S13 = n/2 (a + l)
Thus, the required sum is 286.
(iii) – 5 + (- 8) + (- 11) + … + (- 230)
Here, a = – 5 d = – 8 – (- 5) = – 3 l = – 230 Let n be the number of terms. ∴ Tn = – 230 ⇒ – 230 = – 5 + (n – 1) × (- 3) ⇒ (n – 1) × (- 3) = – 230 + 5 = – 225 ⇒ n – 1 = -225/-3 = 75 ⇒ n = 75 + 1 = 76 Now, = 38 × (- 235) = – 8930 ∴ The required sum = – 8930.
Q3: In an AP: (i) given a = 5, d = 3, an = 50, find n and Sn.
Here, a = 5, d = 3 and an = 50 = l ∵ an = a + (n – 1) d ∴ 50 = 5 + (n – 1) × 3 ⇒ 50 – 5 = (n – 1) × 3 ⇒ (n – 1) × 3 = 45 ⇒ (n – 1) = 45/3 =15 ⇒ n = 15 + 1 = 16 Now Sn = n/2 (a + l) = 16/2 (5 + 50) = 8 (55) = 440 Thus, n = 16 and Sn = 440
(ii) given a = 7, a13 = 35, find d and S13.
(ii) Here, a = 7 and a13 = 35 = l
∴ an = a + (n – 1) d ⇒ 35 = 7 + (13 – 1) d ⇒ 35 – 7 = 12d ⇒ 28 = 12d ⇒ d = 28/12 = 7/3 Now, using Sn = n/2 (a + l) S13 = 13/2 (7 + 35)
Sn = 273 and d = 7/3
(iii) given a12 = 37, d = 3, find a and S12.
Here, a12 = 37 = l and d = 3 Let the first term of the AP be ‘a’. Now a12 = a + (12 – 1) d ⇒ 37 = a + 11d ⇒ 37 = a + 11 × 3 ⇒ 37 = a + 33 ⇒ a = 37 – 33 = 4 Now, Sn = n/2 (a + l) ⇒ S12 = 12/2 (4 + 37) ⇒ S12 = 6 × (41) = 246 Thus, a = 4 and S12 = 246
(iv) given a3 = 15, S10 = 125, find d and a10.
Here, a3 = 15 = l S10 = 125 Let first term of the AP be ‘a’ and the common difference = d ∴ a3 = a + 2d ⇒ a + 2d = 15 …(1) Again Sn = n/2 [2a + (n – 1) d] ⇒ S10 = 10/2 [2a + (10 – 1) d] ⇒ 125 = 5 [2a + 9d] ⇒ 2a + 9d = 125/5 = 25 ⇒ 2a + 9d = 25 …(2) Multiplying (1) by 2 and subtracting (2) from it, we get 2 [a + 2d = 15] – [2a + 9d = 25] ⇒ 2a + 4d – 2a – 9d = 30 – 25 ⇒ – 5d = 5 ⇒ d =5/-5 = -1 ∴ From (1), a + 2 (- 1) = 15 ⇒ a = 15 + 2 ⇒ a = 17 Now, a10 = a + (10 – 1) d = 17 + 9 × (- 1) = 17 – 9 = 8 Thus, d = – 1 and a10 = 8
(v) given d = 5, S9 = 75, find a and a9.
Here, d = 5, S9 = 75 Let the first term of the AP is ‘a’. ∴ S9 = 9/2 [2a + (9 – 1) × 5] ⇒ 75 = 9/2 [2a + 40] ⇒ ⇒ 50/3 = 2a + 40 ⇒ ⇒ Now, a9 = a + (9 – 1) d = =
Thus,
(vi) given a = 2, d = 8, Sn = 90, find n and an.
Here, a = 2, d = 8 and Sn = 90 ∵ Sn = n/2 [2a + (n – 1) d] ∴ 90 = n/2 [2 × 2 + (n – 1) × 8] ⇒ 90 × 2 = 4n + n (n – 1) × 8 ⇒ 180 = 4n + 8n2 – 8n ⇒ 180 = 8n2 – 4n ⇒ 45 = 2n2 – n ⇒ 2n2 – n – 45 = 0 ⇒ 2n2 – 10n + 9n – 45 = 0 ⇒ 2n (n – 5) + 9 (n – 5) = 0 ⇒ (2n + 9) (n – 5) = 0 ∴ Either or n – 5 = 0 ⇒ n = 5 But is not required. ∴n = 5 Now, an = a + (n – 1) d ⇒ a5 = 2 + (5 – 1) × 8 = 2 + 32 = 34 Thus, n = 5 and a5 = 34.
(vii) given a = 8, an = 62, Sn = 210, find n and d.
Here, a = 8, an = 62 = l and Sn = 210 Let the common difference = d Now, Sn = 210 ⇒ 210 = n/2 (a + l) ⇒ 210 = n/2 (8 + 62) = ∴ n = 210/35 = 6 Again an = a + (n – 1) d ⇒ 62 = 8 + (6 – 1) × d ⇒ 62 – 8 = 5d ⇒ 54 = 5d ⇒ d = 54/5 Thus, n = 6 and d = 54/5 .
(viii) given an = 4, d = 2, Sn = – 14, find n and a.
Here, an = 4, d = 2 and Sn = – 14 Let the first term be ‘a’. ∵ an = 4 ∴ a + (n – 1) 2 = 4 ⇒ a + 2n – 2 = 4 ⇒ a = 4 – 2n + 2 ⇒ a = 6 – 2n …(1) Also Sn = – 14 ⇒ n/2 (a + l) = – 14 ⇒ n/2 (a + 4) = – 14 ⇒ n (a + 4) = – 28 …(2) Substituting the value of a from (1) into (2), n [6 – 2n + 4] = – 28 ⇒ n [10 – 2n] = – 28 ⇒ 2n [5 – n] = – 28 ⇒ n (5 – n) = – 14 [Dividing throughout by 2] ⇒ 5n – n2 + 14 = 0 ⇒ n2 – 5n – 14 = 0 ⇒ n2 – 7n + 2n – 14 = 0 ⇒ n (n – 7) + 2 (n – 7) = 0 ⇒ (n – 7) (n + 2) = 0 ∴ Either n – 7 = 0 ⇒ n = 7 or n + 2 = 0 ⇒ n = – 2 But n cannot be negative, ∴ n = 7 Now, from (1), we have a = 6 – 2 × 7 ⇒ a = – 8 Thus, a = – 8 and n = 7
(ix) given a = 3, n = 8, S = 192, find d.
Here, a = 3, n = 8 and Sn = 192 Let the common difference = d. ∵ Sn = n/2 [2a + (n – 1) d] ∴ 192 = 8/2 [2 (3) + (8 – 1) d] ⇒ 192 = 4 [6 + 7d] ⇒ 192 = 24 + 28d ⇒ 28d = 192 – 24 = 168 ⇒ d = 168/28 =6 Thus, d = 6.
(x) given l = 28, S = 144, and there are total 9 terms. Find a.
(x) Here, l = 28 and S9 = 144 Let the first term be ‘a’. Then Sn = n/2 (a + l) ⇒ S9 = 9/2 (a + 28) ⇒ 144 = 9/2 (a + 28) ⇒ ⇒ a = 32 – 28 = 4 Thus, a = 4.
Q4: How many terms of the AP: 9, 17, 25, … must be taken to give a sum of 636?
Here, a = 9 d = 17 – 9 = 8 Sn = 636 ∵ Sn = n/2 [2a + (n – 1) d] = 636 ∴ n/2 [(2 × 9) + (n – 1) × 8] = 636 ⇒ n [18 + (n – 1) × 8] = 1272 ⇒ n (8n + 10) = 1272 ⇒ 8n2 + 10n – 1272 = 0 ⇒ 4n2 + 5n – 636 = 0 ⇒ 4n2 + 53n – 48n – 636 = 0 ⇒ n (4n + 53) – 12 (4n + 53) = 0 ⇒ (n – 12) (4n + 53) = 0 ⇒ n = 12 and Rejecting , we have n = 12.
Q5: The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Here, a = 5 l = 45 = Tn Sn = 400 ∵ Tn = a + (n – 1) d ∴ 45 = 5 + (n – 1) d ⇒ (n – 1) d = 45 – 5 ⇒ (n – 1) d = 40 …(1) Also Sn = n/2 (a + l) ⇒ 400 = n/2 (5 + 45) ⇒ 400 × 2 = n × 50 ⇒ From (1), we get (16 – 1) d = 40 ⇒ 15d = 40 ⇒ d = 40/15 = 8/3
Q6: The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum
We have, First term a = 17 Last term l = 350 = Tn Common difference d = 9 Let the number of terms be ‘n’ ∵ Tn = a + (n – 1) d ∴ 350 = 17 + (n – 1) × 9 ⇒ (n – 1) × 9 = 350 – 17 = 333 ⇒ n – 1 = 333/9 = 37 ⇒ n = 37 + 1 = 38 Since, Sn = n/2 (a + l) ∴ S38 = 38/2 (17 + 350) = 19 (367) = 6973 Thus, n = 38 and Sn = 6973
Q7: Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Here, n = 22, T22 = 149 = l d = 7 Let the first term of the A.P. be ‘a’. ∴ Tn = a + (n – 1) d ⇒ Tn = a + (22 – 1) × 7 ⇒ a + 21 × 7 = 149 ⇒ a + 147 = 149 ⇒ a = 149 – 147 = 2 Now, S22 = n/2 [a + l] ⇒ S22 = 22/2 [2 + 149] = 11 [151] = 1661 Thus S22 = 1661
Q8: Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Here, n = 51, T2 = 14 and T3 = 18 Let the first term of the A.P. be ‘a’ and the common difference is d. ∴We have: T2 = a + d ⇒ a + d = 14 …(1) T3 = a + 2d ⇒ a + 2d = 18 …(2) Subtracting (1) from 2, we get a + 2d – a – d = 18 – 14 ⇒ d = 14 From (1), we get a + d = 14 ⇒ a + 4 = 14 ⇒ a = 14 – 4 = 10 Now, Sn = n/2 [2a + (n – 1) d] ⇒ S51 = 51/2 [(2 × 10) + (51 – 1) × 4] = 51/2 [20 + 200] = 51/2 [220] = 51 × 110 = 5610 Thus, the sum of 51 terms is 5610.
Q9: If the sum of first 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of first n terms.
Here, we have: S7 = 49 and S17 = 289 Let the first term of the A.P. be ‘a’ and ‘d’ be the common difference, then Sn = n/2 [2a + (n – 1) d] ⇒ S7 = 7/2 [2a + (7 – 1) d] = 49 ⇒ 7 (2a + 6d) = 2 × 49 = 98 ⇒ 2a + 6d = 98/7 = 14 ⇒ 2 [a + 3d] = 14 ⇒ a + 3d = 14/2 = 7 ⇒ a + 3d = 7 …(1) Also, S17 = 17/2[2a + (17 – 1) d] = 289 ⇒ 17/2 (2a + 16d) = 289 ⇒ a + 8d = 289/17 = 17 ⇒ a + 8d = 17 …(2) Subtracting (1) from (2), we have: a + 8d – a – 3d = 17 – 7 ⇒ 5d = 10 ⇒ d =10/5 = 2 Now, from (1), we have a + 3 (2) = 7 ⇒ a = 7 – 6 = 1 Now, Sn = n/2 [2a + (n – 1) d] = n/2 [2 × 1 + (n – 1) × 2] = n/2 [2 + 2n – 2] = n/2 [2n] = n × n = n2 Thus, the required sum of n terms = n2.
Q10: Show that a1, a2, …, an, … form an A.P. where an is defined as below: (i) an = 3 + 4n (ii) an = 9 – 5n Also find the sum of the first 15 terms in each case.
(i) Here, an = 3 + 4n Putting n = 1, 2, 3, 4, ….. n, we get: a1 = 3 + 4 (1) = 7 a2 = 3 + 4 (2) = 11 a3 = 3 + 4 (3) = 15 a4 = 3 + 4 (4) = 19 ….. ….. ….. an = 3 + 4n ∴ The A.P. in which a = 7 and d = 11 – 7 = 4 is: 7, 11, 15, 19, ….., (3 + 4n). Now S15 = 15/2 [(2 × 7) + (15 – 1) × 4] = 15/2 [14 + (14 × 4)] = 15/2 [14 + 56] = 15/2[70] = 15 × 35 = 525 (ii) Here, an = 9 – 5n Putting n = 1, 2, 3, 4, ….., n, we get a1 = 9 – 5 (1) = 4 a2 = 9 – 5 (2) = – 1 a3 = 9 – 5 (3) = – 6 a4 = 9 – 5 (4) = – 11 ….. ….. ∴ The A.P. is: 4, – 1, – 6, – 11, ….. 9 – 5 (n) [having first term as 4 and d = – 1 – 4 = – 5] ∴ S15 = 15/2 [(2 × 4) + (15 – 1) × ( – 5)] = 15/2[8 + 14 × (- 5)] = 15/2[8 – 70]
Q11: If the sum of the first n terms of an AP is 4n – n2, what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.
We have: Sn = 4n – n2 ∴ S1 = 4 (1) – (1)2 = 4 – 1 = 3 ⇒ First term = 3 S2 = 4 (2) – (2)2 = 8 – 4 = 4 ⇒ Sum of first two terms = 4 ∴ Second term (S2 – S1) = 4 – 3 = 1 S3 = 4 (3) – (3)2 = 12 – 9 = 3 ⇒ Sum of first 3 terms = 3 ∴ Third term (S3 – S2) = 3 – 4 = – 1 S9 = 4 (9) – (9)2 = 36 – 81 = – 45 S10 = 4 (10) – (10)2 = 40 – 100 = – 60 ∴ Tenth term = S10 – S9 = [- 60] – [- 45] = – 15 Now, Sn = 4 (n) – (n)2 = 4n – n2 Also Sn– 1 = 4 (n – 1) – (n – 1)2 = 4n – 4 – [n2 – 2n + 1] = 4n – 4 – n2 + 2n – 1 = 6n – n2 – 5 ∴ nth term = Sn – Sn – 1 = [4n – n2] – [6n – n2 – 5] = 4n – n2 – 6n + n2 + 5 = 5 – 2n Thus, S1 = 3 and a1 = 3 S2 = 4 and a2 = 1 S3 = 3 and a3 = – 1 a10 = – 15 and an = 5 – 2n
Q12: Find the sum of the first 40 positive integers divisible by 6.
The first 40 positive integers divisible by 6 are: 6, 12, 18, ….., (6 × 40). And, these numbers are in A.P. such that a = 6 d = 12 – 6 = 6 and an = 6 × 40 = 240 = l ∴ S40 = 40/2 [(2 × 6) + (40 – 1) × 6] = 20 [12 + 39 × 6] = 20 [12 + 234] = 20 × 246 = 4920 OR Sn = n/2 [a + l] S40 = 40/2[6 + 240] = 20 × 246 = 4920 Thus, the sum of first 40 multiples of 6 is 4920.
Q13: Find the sum of the first 15 multiples of 8.
The first 15 multiples of 8 are: 8, (8 × 2), (8 × 3), (8 × 4), ….., (8 × 15) or 8, 16, 24, 32, ….., 120. These numbers are in A.P., where a = 8 and l = 120 ∴ S15 = 15/2 [a + l] = 15/2 [8 + 120]
Thus, the sum of first positive 15 multiples of 8 is 960.
Q14: Find the sum of the odd numbers between 0 and 50.
Odd numbers between 0 and 50 are: 1, 3, 5, 7, ….., 49 These numbers are in A.P. such that a = 1 and l = 49 Here, d = 3 – 1 = 2 ∴ Tn = a + (n – 1) d ⇒ 49 = 1 + (n – 1) 2 ⇒ 49 – 1 = (n – 1) 2 ⇒ (n – 1) = 48/2 = 24 ∴ n = 24 + 1 = 25 Now, S25 = 25/2[1+49]
Thus, the sum of odd numbers between 0 and 50 is 625.
Q15: A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc., the penalty for each succeeding day being ₹ 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
Here, penalty for delay on 1st day = ₹ 200 2nd day = ₹ 250 3rd day = ₹ 300 …………… …………… Now, 200, 250, 300, ….. are in A.P. such that a = 200, d = 250 – 200 = 50 ∴ S30 is given by S30 = 30/2 [2 (200) + (30 – 1) × 50]
= 15 [400 + 29 × 50] = 15 [400 + 1450] = 15 × 1850 = 27,750 Thus, penalty for the delay for 30 days is ₹ 27,750.
Q16: A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performace. If each prize is ₹ 20 less than its preceding prize, find the value of each of the prizes.
Sum of all the prizes = ₹ 700 Let the first prize = a ∴ 2nd prize = (a – 20) 3rd prize = (a – 40) 4th prize = (a – 60) …………………………………. Thus, we have, first term = a Common difference = – 20 Number of prizes, n = 7 Sum of 7 terms Sn = 700 Since, Sn = n/2 [2a + (n – 1) d] ⇒ 700 = 7/2 [2 (a) + (7 – 1) × (- 20)] ⇒ 700 = 7/2 [2a + (6 × – 20)] ⇒ ⇒ 200 = 2a – 120 ⇒ 2a = 200 + 120 = 320 ⇒ a = 320/2 = 160 Thus, the values of the seven prizes are: ₹ 160, ₹ (160 – 20), ₹ (160 – 40), ₹ (160 – 60), ₹ (160 – 80), ₹ (160 – 100) and ₹ (160 – 120) ⇒ ₹ 160, ₹ 140, ₹ 120, ₹ 100, ₹ 80, ₹ 60 and ₹ 40.
Q17: In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
Number of classes = 12 ∵ Each class has 3 sections. ∴ Number of plants planted by class I = 1 × 3 = 3 Number of plants planted by class II = 2 × 3 = 6 Number of plants planted by class III = 3 × 3 = 9 Number of plants planted by class IV = 4 × 3 = 12 …………………………………………………………………………………………. Number of plants planted by class XII = 12 × 3 = 36 The numbers 3, 6, 9, 12, ……….., 36 are in A.P. Here, a = 3 and d = 6 – 3 = 3
∵ Number of classes = 12 i.e., n = 12 ∴ Sum of the n terms of the above A.P., is given by S12 = 12/2 [2 (3) + (12 – 1) 3]
= 6 [6 + 11 × 3] = 6 [6 + 33] = 6 × 39 = 234 Thus, the total number of trees = 234.
Q18: A spiral is made up of successive semi-circles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, ….. as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semi-circles? (Take π =22/7)
[Hint: Length of successive semi-circles is l1, l2, l3, l4, … with centres at A, B, A, B, …, respectively.]
Length of a semi-circle = semi-circumference = 1/2 (2πr) = πr ∴ l1 = π r1 = 0.5 π cm = 1 × 0.5 π cm l2 = π r2 = 1.0 π cm = 2 × 0.5 π cm l3 = π r3 = 1.5 π cm = 3 × 0.5 π cm l4 = π r4 = 2.0 π cm = 4 × 0.5 π cm …… …………… …………………. l13 = π r13 cm = 6.5 π cm = 13 × 0.5 π cm Now, length of the spiral = l1 + l2 + l3 + l4 + ….. + l13 = 0.5 π [1 + 2 + 3 + 4 + ….. + 13] cm …(1) ∵ 1, 2, 3, 4, ….., 13 are in A.P. such that a = 1 and l = 13 ∴ S13 = 13/2 [1+13]
∴ From (1), we have: Total length of the spiral = 1.5 π [91] cm
Q19: 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see figure). In how many rows are the 200 logs placed and how many logs are in the top row?
We have: The number of logs: 1st row = 20 2nd row = 19 3rd row = 18 obviously, the numbers 20, 19, 18, ….., are in A.P. such that a = 20 d = 19 – 20 = – 1 Let the numbers of rows be n. ∴ Sn = 200 Now, using, Sn = [2a + (n – 1) d], we get Sn = n/2 [2 (20) + (n – 1) × (- 1)] ⇒ 200 = n/2 [40 – (n – 1)] ⇒ 2 × 200 = n × 40 – n (n – 1) ⇒ 400 = 40n – n2 + n ⇒ n2 – 41n + 400 = 0 ⇒ n2 – 16n – 25n + 400 = 0 ⇒ n (n – 16) – 25 (n – 16) = 0 ⇒ (n – 16) (n – 25) = 0 Either ⇒ n – 16 = 0 ⇒ n = 16 or n – 25 = 0 ⇒ n = 25 Tn = 0 ⇒ a + (n – 1) d = 0 ⇒ 20 + (n – 1) × (- 1) = 0 ⇒ n – 1 = 20 ⇒ n = 21 i.e., 21st term becomes 0 ∴ n = 25 is not required. Thus, n = 16 ∴ Number of rows = 16 Now, T16 = a + (16 – 1) d = 20 + 15 × (- 1) = 20 – 15 = 5 ∴ Number of logs in the 16th (top) row is 5.
Q20: In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.
A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run? [Hint: To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3)]
Here, number of potatoes = 10 The up-down distance of the bucket: From the 1st potato = [5 m] × 2 = 10 m From the 2nd potato = [(5 + 3) m] × 2 = 16 m From the 3rd potato = [(5 + 3 + 3) m] × 2 = 22 m From the 4th potato = [(5 + 3 + 3 + 3) m] × 2 = 28 m ……………………………. ……………………… ∵ 10, 16, 22, 28, ….. are in A.P. such that a = 10 and d = 16 – 10 = 6 ∴ Using Sn = n/2 [2a + (n – 1) d], we have: S10 = 10/2[2 (10) + (10 – 1) × 6] = 5 [20 + 9 × 6] = 5 [20 + 54] = 5 [74] = 5 × 74 = 370 Thus, the sum of above distances = 370 m. ⇒ The competitor has to run a total distance of 370 m.
Exercise 5.4
Q1: Which term of the AP: 121, 117, 113, . . ., is its first negative term? [Hint: Find n for an < 0]
Given the AP series is 121, 117, 113, . . .,
Thus, first term, a = 121
Common difference, d = 117-121= -4
By the nth term formula,
an = a+(n −1)d
Therefore,
an = 121+(n−1)(-4)
= 121-4n+4
=125-4n
To find the first negative term of the series, an < 0
Therefore,
125-4n < 0
125 < 4n
n>125/4
n>31.25
Therefore, the first negative term of the series is 32nd term.
Q2: The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.
From the given statements, we can write,
a3 + a7 = 6 …………………………….(i)
And
a3 ×a7 = 8 ……………………………..(ii)
By the nth term formula,
an = a+(n−1)d
Third term, a3 = a+(3 -1)d
a3 = a + 2d………………………………(iii)
And Seventh term, a7= a+(7-1)d
a7 = a + 6d ………………………………..(iv)
From equation (iii) and (iv), putting in equation(i), we get,
a+2d +a+6d = 6
2a+8d = 6
a+4d=3
or
a = 3–4d …………………………………(v)
Again putting the eq.(iii) and (iv), in eq. (ii), we get,
(a+2d)×(a+6d) = 8
Putting the value of a from equation (v), we get,
(3–4d +2d)×(3–4d+6d) = 8
(3 –2d)×(3+2d) = 8
32 – 2d2 = 8
9 – 4d2 = 8
4d2 = 1
d = 1/2 or -1/2
Now, by putting both the values of d, we get,
a = 3 – 4d = 3 – 4(1/2) = 3 – 2 = 1, when d = 1/2
a = 3 – 4d = 3 – 4(-1/2) = 3+2 = 5, when d = -1/2
We know, the sum of nth term of AP is;
Sn = n/2 [2a +(n – 1)d]
So, when a = 1 and d=1/2
Then, the sum of first 16 terms are;
S16 = 16/2 [2 +(16-1)1/2] = 8(2+15/2) = 76
And when a = 5 and d= -1/2
Then, the sum of first 16 terms are;
S16 = 16/2 [2.5+(16-1)(-1/2)] = 8(5/2)=20
Q3: A ladder has rungs 25 cm apart. (see Fig. 5.7). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are apart, what is the length of the wood required for the rungs? [Hint: Number of rungs = -250/25 ].
Given,
Distance between the rungs of the ladder is 25cm.
Distance between the top rung and bottom rung of the ladder is =
= 5/2 ×100cm
= 250cm
Therefore, total number of rungs = 250/25 + 1 = 11
As we can see from the figure, the ladder has rungs in decreasing order from top to bottom. Thus, we can conclude now, that the rungs are decreasing in an order of AP.
And the length of the wood required for the rungs will be equal to the sum of the terms of AP series formed.
So,
First term, a = 45
Last term, l = 25
Number of terms, n = 11
Now, as we know, sum of nth terms is equal to,
Sn= n/2(a+ l)
Sn= 11/2(45+25) = 11/2(70) = 385 cm
Hence, the length of the wood required for the rungs is 385cm.
Q.4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x. [Hint :Sx – 1 = S49 – Sx ]
Given,
Row houses are numbers from 1,2,3,4,5…….49.
Thus we can see the houses numbered in a row are in the form of AP.
So,
First term, a = 1
Common difference, d=1
Let us say the number of xth houses can be represented as;
Sum of nth term of AP = n/2[2a+(n-1)d]
Sum of number of houses beyond x house = Sx-1
= (x-1)/2[2.1+(x-1-1)1]
= (x-1)/2 [2+x-2]
= x(x-1)/2 ………………………………………(i)
By the given condition, we can write,
S49 – Sx = {49/2[2.1+(49-1)1]}–{x/2[2.1+(x-1)1]}
= 25(49) – x(x + 1)/2 ………………………………….(ii)
As per the given condition, eq.(i) and eq(ii) are equal to each other;
Therefore,
x(x-1)/2 = 25(49) – x(x-1)/2
x = ±35
As we know, the number of houses cannot be a negative number. Hence, the value of x is 35.
Q5: A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of 1 4 m and a tread of 1 2 m. (see Fig. 5.8). Calculate the total volume of concrete required to build the terrace. [Hint : Volume of concrete required to build the first step = ¼ ×1/2 ×50 m3.]
As we can see from the given figure, the first step is ½ m wide, 2nd step is 1m wide and 3rd step is 3/2m wide. Thus we can understand that the width of step by ½ m each time when height is ¼ m. And also, given length of the steps is 50m all the time. So, the width of steps forms a series AP in such a way that;
½ , 1, 3/2, 2, ……..
Volume of steps = Volume of Cuboid = Length × Breadth Height
Now,
Volume of concrete required to build the first step = ¼ ×1/2 ×50 = 25/4 Volume of concrete required to build the second step =¼ ×1/×50 = 25/2 Volume of concrete required to build the second step = ¼ ×3/2 ×50 = 75/2
Now, we can see the volumes of concrete required to build the steps, are in AP series; 25/4 , 25/2 , 75/2 …..
Thus, applying the AP series concept, First term, a = 25/4
Common difference, d = 25/2 – 25/4 = 25/4
As we know, the sum of n terms is;
Sn = n/2[2a+(n-1)d] = 15/2(2×(25/4 )+(15/2 -1)25/4)
Upon solving, we get,
Sn = 15/2 (100)
Sn750
Hence, the total volume of concrete required to build the terrace is 750 m3.
Q1. In which of the following situations, does the list of numbers involved make an arithmetic progression and why? (i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.
We can write the given condition as:
Taxi fare for 1 km = 15
Taxi fare for first 2 kms = 15+8 = 23
Taxi fare for first 3 kms = 23+8 = 31
Taxi fare for first 4 kms = 31+8 = 39 and so on……
Thus, 15, 23, 31, 39 … forms an A.P. because every next term is 8 more than the preceding term.
(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.
Let the volume of air in a cylinder, initially, be V litres.
In each stroke, the vacuum pump removes 1/4th of air remaining in the cylinder at a time.
Or we can say, after every stroke, 1-1/4 = 3/4th part of air will remain.
Therefore, volumes will be V, 3V/4 , (3V/4)2, (3V/4)3…and so on
Clearly, we can see here, the adjacent terms of this series do not have a common difference between them. Therefore, this series is not an A.P.
(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.
We can write the given condition as:
Cost of digging a well for first metre = Rs.150
Cost of digging a well for first 2 metres = Rs.150+50 = Rs.200
Cost of digging a well for first 3 metres = Rs.200+50 = Rs.250
Cost of digging a well for first 4 metres =Rs.250+50 = Rs.300 and so on.
Clearly, 150, 200, 250, 300 … forms an A.P. with a common difference of 50 between each term.
(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.
We know that if Rs. P is deposited at r% compound interest per annum for n years, the amount of money will be:
P(1+r/100)n Therefore, after each year, the amount of money will be; 10000(1+8/100), 10000(1+8/100)2, 10000(1+8/100)3……
Clearly, the terms of this series do not have a common difference between them. Therefore, this series is not an A.P.
Q2. Write first four terms of the A.P. when the first term a and the common difference are given as follows: (i) a = 10, d = 10
Let us consider, the Arithmetic Progression series be a1, a2, a3, a4, a5 … a1 = a = 10 a2 = a1+d = 10+10 = 20 a3 = a2+d = 20+10 = 30 a4 = a3+d = 30+10 = 40 a5 = a4+d = 40+10 = 50 And so on… Therefore, the A.P. series will be 10, 20, 30, 40, 50 … And First four terms of this A.P. will be 10, 20, 30, and 40.
(ii) a = – 2, d = 0
Let us consider, the Arithmetic Progression series be a1, a2, a3, a4, a5 … a1 = a = -2 a2 = a1+d = – 2+0 = – 2 a3 = a2+d = – 2+0 = – 2 a4 = a3+d = – 2+0 = – 2 Therefore, the A.P. series will be – 2, – 2, – 2, – 2 … And, First four terms of this A.P. will be – 2, – 2, – 2 and – 2.
(iii) a = 4, d = – 3
Let us consider, the Arithmetic Progression series be a1, a2, a3, a4, a5 … a1 = a = 4 a2 = a1+d = 4-3 = 1 a3 = a2+d = 1-3 = – 2 a4 = a3+d = -2-3 = – 5 Therefore, the A.P. series will be 4, 1, – 2 – 5 … And, first four terms of this A.P. will be 4, 1, – 2 and – 5.
(iv) a = – 1, d = 1/2
Let us consider, the Arithmetic Progression series be a1, a2, a3, a4, a5 … a2 = a1+d = -1+1/2 = -1/2 a3 = a2+d = -1/2+1/2 = 0 a4 = a3+d = 0+1/2 = 1/2 Thus, the A.P. series will be-1, -1/2, 0, 1/2 And First four terms of this A.P. will be -1, -1/2, 0 and 1/2.
(v) a = – 1.25, d = – 0.25
Let us consider, the Arithmetic Progression series be a1, a2, a3, a4, a5 … a1 = a = – 1.25 a2 = a1 + d = – 1.25-0.25 = – 1.50 a3 = a2 + d = – 1.50-0.25 = – 1.75 a4 = a3 + d = – 1.75-0.25 = – 2.00 Therefore, the A.P series will be 1.25, – 1.50, – 1.75, – 2.00 …….. And first four terms of this A.P. will be – 1.25, – 1.50, – 1.75 and – 2.00.
Q3. For the following A.P.s, write the first term and the common difference. (i) 3, 1, – 1, – 3 … (ii) -5, – 1, 3, 7 … (iii) 1/3, 5/3, 9/3, 13/3 …. (iv) 0.6, 1.7, 2.8, 3.9 …
Ans.
(i) Given series, 3, 1, – 1, – 3 …
First term, a = 3 Common difference, d = Second term – First term ⇒ 1 – 3 = -2 ⇒ d = -2
(ii) Given series, – 5, – 1, 3, 7 …
First term, a = -5 Common difference, d = Second term – First term ⇒ ( – 1)-( – 5) = – 1+5 = 4 =d
(iii) Given series, 1/3, 5/3, 9/3, 13/3 ….
First term, a = 1/3 Common difference, d = Second term – First term ⇒ 5/3 – 1/3 = 4/3 = d
(iv) Given series, 0.6, 1.7, 2.8, 3.9 …
First term, a = 0.6 Common difference, d = Second term – First term ⇒ 1.7 – 0.6 ⇒ 1.1 = d
Q4. Which of the following are APs? If they form an A.P. find the common difference d and write three more terms. (i) 2, 4, 8, 16 … (ii) 2, 5/2, 3, 7/2 …. (iii) -1.2, -3.2, -5.2, -7.2 … (iv) -10, – 6, – 2, 2 … (v) 3, 3 + √2, 3 + 2√2, 3 + 3√2 (vi) 0.2, 0.22, 0.222, 0.2222 …. (vii) 0, – 4, – 8, – 12 … (viii) -1/2, -1/2, -1/2, -1/2 …. (ix) 1, 3, 9, 27 … (x) a, 2a, 3a, 4a … (xi) a, a2, a3, a4 … (xii) √2, √8, √18, √32 … (xiii) √3, √6, √9, √12 … (xiv) 12, 32, 52, 72 … (xv) 12, 52, 72, 73 … Ans. (i) Given, 2, 4, 8, 16 …
Here, the common difference is: a2 – a1 = 4 – 2 = 2 a3 – a2 = 8 – 4 = 4 a4 – a3 = 16 – 8 = 8 Since, an+1 – an or the common difference is not the same every time. Therefore, the given series are not forming an A.P.
(ii) Given, 2, 5/2, 3, 7/2 ….
Here, a2 – a1 = 5/2-2 = 1/2 a3 – a2 = 3-5/2 = 1/2 a4 – a3 = 7/2-3 = 1/2 Since, an+1 – an or the common difference is same every time. Therefore, d = 1/2 and the given series are in A.P. The next three terms are; a5 = 7/2+1/2 = 4 a6 = 4 +1/2 = 9/2 a7 = 9/2 +1/2 = 5
(iii) Given, -1.2, – 3.2, -5.2, -7.2 …
Here, a2 – a1 = (-3.2)-(-1.2) = -2 a3 – a2 = (-5.2)-(-3.2) = -2 a4 – a3 = (-7.2)-(-5.2) = -2 Since, an+1 – an or common difference is same every time. Therefore, d = -2 and the given series are in A.P. Hence, next three terms are; a5 = – 7.2-2 = -9.2 a6 = – 9.2-2 = – 11.2 a7 = – 11.2-2 = – 13.2
(iv) Given, -10, – 6, – 2, 2 …
Here, the terms and their difference are; a2 – a1 = (-6)-(-10) = 4 a3 – a2 = (-2)-(-6) = 4 a4 – a3 = (2 -(-2) = 4 Since, an+1 – an or the common difference is same every time. Therefore, d = 4 and the given numbers are in A.P. Hence, next three terms are; a5 = 2+4 = 6 a6 = 6+4 = 10 a7 = 10+4 = 14
(v) Given, 3, 3+√2, 3+2√2, 3+3√2
Here, a2 – a1 = 3+√2-3 = √2 a3 – a2 = (3+2√2)-(3+√2) = √2 a4 – a3 = (3+3√2) – (3+2√2) = √2 Since, an+1 – an or the common difference is same every time. Therefore, d = √2 and the given series forms a A.P. Hence, next three terms are; a5 = (3+√2) +√2 = 3+4√2 a6 = (3+4√2)+√2 = 3+5√2 a7 = (3+5√2)+√2 = 3+6√2
(vi) 0.2, 0.22, 0.222, 0.2222 ….
Here, a2 – a1 = 0.22-0.2 = 0.02 a3 – a2 = 0.222-0.22 = 0.002 a4 – a3 = 0.2222-0.222 = 0.0002 Since, an+1 – an or the common difference is not same every time. Therefore, and the given series doesn’t forms a A.P.
(vii) 0, -4, -8, -12 …
Here, a2 – a1 = (-4)-0 = -4 a3 – a2 = (-8)-(-4) = -4 a4 – a3 = (-12)-(-8) = -4 Since, an+1 – an or the common difference is same every time. Therefore, d = -4 and the given series forms a A.P. Hence, next three terms are; a5 = -12-4 = -16 a6 = -16-4 = -20 a7 = -20-4 = -24
(viii) -1/2, -1/2, -1/2, -1/2 ….
Here, a2 – a1 = (-1/2) – (-1/2) = 0 a3 – a2 = (-1/2) – (-1/2) = 0 a4 – a3 = (-1/2) – (-1/2) = 0 Since, an+1 – an or the common difference is same every time. Therefore, d = 0 and the given series forms a A.P. Hence, next three terms are; a5 = (-1/2)-0 = -1/2 a6 = (-1/2)-0 = -1/2 a7 = (-1/2)-0 = -1/2
(ix) 1, 3, 9, 27 …
Here, a2 – a1 = 3-1 = 2 a3 – a2 = 9-3 = 6 a4 – a3 = 27-9 = 18 Since, an+1 – an or the common difference is not same every time. Therefore, and the given series doesn’t form a A.P.
(x) a, 2a, 3a, 4a …
Here, a2 – a1 = 2a–a = a a3 – a2 = 3a-2a = a a4 – a3 = 4a-3a = a Since, an+1 – an or the common difference is same every time. Therefore, d = a and the given series forms a A.P. Hence, next three terms are; a5 = 4a+a = 5a a6 = 5a+a = 6a a7 = 6a+a = 7a
(xi) a, a2, a3, a4 …
Here, a2 – a1 = a2–a = a(a-1) a3 – a2 = a3 –a2 = a2(a-1) a4 – a3 = a4 – a3 = a3(a-1) Since, an+1 – an or the common difference is not same every time. Therefore, the given series doesn’t forms a A.P.
(xii) √2, √8, √18, √32 …
Here, a2 – a1 = √8-√2 = 2√2-√2 = √2 a3 – a2 = √18-√8 = 3√2-2√2 = √2 a4 – a3 = 4√2-3√2 = √2 Since, an+1 – an or the common difference is same every time. Therefore, d = √2 and the given series forms a A.P. Hence, next three terms are; a5 = √32+√2 = 4√2+√2 = 5√2 = √50 a6 = 5√2+√2 = 6√2 = √72 a7 = 6√2+√2 = 7√2 = √98
(xiii) √3, √6, √9, √12 …
Here, a2 – a1 = √6-√3 = √3×√2-√3 = √3(√2-1) a3 – a2 = √9-√6 = 3-√6 = √3(√3-√2) a4 – a3 = √12 – √9 = 2√3 – √3×√3 = √3(2-√3) Since, an+1 – an or the common difference is not same every time. Therefore, the given series doesn’t form a A.P.
(xiv) 12, 32, 52, 72 …
Or, 1, 9, 25, 49 ….. Here, a2 − a1 = 9−1 = 8 a3 − a2 = 25−9 = 16 a4 − a3 = 49−25 = 24 Since, an+1 – an or the common difference is not same every time. Therefore, the given series doesn’t form an A.P.
(xv) 12, 52, 72, 73 …
Or 1, 25, 49, 73 … Here, a2 − a1 = 25−1 = 24 a3 − a2 = 49−25 = 24 a4 − a3 = 73−49 = 24 Since, an+1 – an or the common difference is same every time. Therefore, d = 24 and the given series forms a A.P. Hence, next three terms are: a5 = 73+24 = 97 a6 = 97+24 = 121 a7 = 121+24 = 145
Exercise 5.2
Q1: Fill in the blanks in the following table, given that a is the first term, d is the common difference and an the nth term of the A.P.
Sol: (i) Given:
First term, a = 7 Common difference, d = 3 Number of terms, n = 8, We have to find the nth term, an = ? As we know, for an A.P., an = a+(n−1)d Putting the values: ⇒ 7+(8 −1) 3 ⇒ 7+(7) 3 ⇒ 7+21 = 28 Hence, an = 28
(ii) Given:
First term, a = -18 Common difference, d = ? Number of terms, n = 10 Nth term, an = 0 As we know, for an A.P., an = a+(n−1)d Putting the values, ⇒ 0 = − 18 +(10−1)d ⇒ 18 = 9d ⇒ d = 18/9 = 2 Hence, common difference, d = 2
(iii) Given:
First term, a = ? Common difference, d = -3 Number of terms, n = 18 Nth term, an = -5 As we know, for an A.P., an = a+(n−1)d Putting the values, ⇒ −5 = a+(18−1) (−3) ⇒ −5 = a+(17) (−3) ⇒ −5 = a−51 ⇒ a = 51−5 = 46 Hence, a = 46
(iv) Given:
First term, a = -18.9 Common difference, d = 2.5 Number of terms, n = ? Nth term, an = 3.6 As we know, for an A.P., ⇒ an = a +(n −1)d Putting the values, ⇒ 3.6 = − 18.9+(n −1)2.5 ⇒ 3.6 + 18.9 = (n−1)2.5 ⇒ 22.5 = (n−1)2.5 ⇒ (n – 1) = 22.5/2.5 ⇒ n – 1 = 9 ⇒ n = 10 Hence, n = 10
(v) Given:
First term, a = 3.5 Common difference, d = 0 Number of terms, n = 105 Nth term, an = ? As we know, for an A.P., an = a+(n −1)d Putting the values, ⇒ an = 3.5+(105−1) 0 ⇒ an = 3.5+104×0 ⇒ an = 3.5 Hence, an = 3.5
Q2: Choose the correct choice in the following and justify: (i) 30th term of the A.P: 10,7, 4, …, is (a) 97 (b) 77 (c) −77 (d) −87
Sol: Given here:
A.P. = 10, 7, 4, … Therefore, we can find, First term, a = 10 Common difference, d = a2 − a1 = 7−10 = −3 As we know, for an A.P., an = a +(n−1)d Putting the values; ⇒ a30 = 10+(30−1)(−3) ⇒ a30 = 10+(29)(−3) ⇒ a30 = 10−87 = −77 Hence, the correct answer is option c.
(ii) 11th term of the A.P. -3, -1/2, ,2 …. is (a) 28 (b) 22 (c) – 38 (d)
Given here:
A.P. = -3, -1/2, ,2 … Therefore, we can find, First term a = – 3 Common difference, d = a2 − a1 = (-1/2) -(-3) ⇒ (-1/2) + 3 = 5/2 As we know, for an A.P., an = a+(n−1)d Putting the values; ⇒ a11 = 3+(11-1)(5/2) ⇒ a11 = 3+(10)(5/2) ⇒ a11 = -3+25 ⇒ a11 = 22 Hence, the answer is option B.
Q3: In the following APs find the missing term in the boxes.
Sol: (i) For the given A.P., 2, 2, 26
The first and third term are; a = 2 a3 = 26 As we know, for an A.P., an = a+(n −1)d Therefore, putting the values here, ⇒ a3 = 2+(3-1)d ⇒ 26 = 2+2d ⇒ 24 = 2d ⇒ d = 12 ⇒ a2 = 2+(2-1)12 ⇒ a2 = 14 Therefore, 14 is the missing term.
(ii) For the given A.P., , 13, ,3
a2 = 13 and a4 = 3 As we know, for an A.P., an = a+(n−1) d Therefore, putting the values here, ⇒ a2 = a +(2-1)d ⇒ 13 = a+d ………………. (i) ⇒ a4 = a+(4-1)d ⇒ 3 = a+3d ………….. (ii) On subtracting equation (i) from (ii), we get, ⇒ – 10 = 2d ⇒ d = – 5 From equation (i), putting the value of d,we get ⇒ 13 = a+(-5) ⇒ a = 18 ⇒ a3 = 18+(3-1)(-5) ⇒ 18+2(-5) = 18-10 = 8 Therefore, the missing terms are 18 and 8 respectively.
(iii) For the given A.P.,
a = 5 and a4 = 19/2 As we know, for an A.P., an = a+(n−1)d Therefore, putting the values here, ⇒ a4 = a+(4-1)d ⇒ 19/2 = 5+3d ⇒ (19/2) – 5 = 3d ⇒ 3d = 9/2 ⇒ d = 3/2 ⇒ a2 = a+(2-1)d ⇒ a2 = 5+3/2 ⇒ a2 = 13/2 ⇒ a3 = a+(3-1)d ⇒ a3 = 5+2×3/2 ⇒ a3 = 8 Therefore, the missing terms are 13/2 and 8 respectively.
(iv) For the given A.P.,
a = −4 and a6 = 6 As we know, for an A.P., an = a +(n−1) d Therefore, putting the values here, ⇒ a6 = a+(6−1)d ⇒ 6 = − 4+5d ⇒ 10 = 5d ⇒ d = 2 ⇒ a2 = a+d = − 4+2 = −2 ⇒ a3 = a+2d = − 4+2(2) = 0 ⇒ a4 = a+3d = − 4+ 3(2) = 2 ⇒ a5 = a+4d = − 4+4(2) = 4 Therefore, the missing terms are −2, 0, 2, and 4 respectively.
(v) For the given A.P.,
a2 = 38 a6 = −22 As we know, for an A.P., an = a+(n −1)d Therefore, putting the values here, ⇒ a2 = a+(2−1)d ⇒ 38 = a+d ……………………. (i) ⇒ a6 = a+(6−1)d ⇒ −22 = a+5d …………………. (ii) On subtracting equation (i) from (ii), we get ⇒ − 22 − 38 = 4d ⇒ −60 = 4d ⇒ d = −15 ⇒ a = a2 − d = 38 − (−15) = 53 ⇒ a3 = a + 2d = 53 + 2 (−15) = 23 ⇒ a4 = a + 3d = 53 + 3 (−15) = 8 ⇒ a5 = a + 4d = 53 + 4 (−15) = −7 Therefore, the missing terms are 53, 23, 8, and −7 respectively.
Q4: Which term of the A.P. 3, 8, 13, 18, … is 78?
Sol: Given the A.P. series as3, 8, 13, 18, … First term, a = 3 Common difference, d = a2 − a1 = 8 − 3 = 5 Let the nth term of given A.P. be 78. Now as we know, an = a+(n−1)d Therefore, ⇒ 78 = 3+(n −1)5 ⇒ 75 = (n−1)5 ⇒ (n−1) = 15 ⇒ n = 16 Hence, 16th term of this A.P. is 78.
Q5: Find the number of terms in each of the following A.P. (i) 7, 13, 19, …, 205
Given, 7, 13, 19, …, 205 is the A.P
Therefore, The first term, a = 7 Common difference, d = a2 − a1 = 13 − 7 = 6 Let there are n terms in this A.P. an = 205 As we know, for an A.P., ⇒ an = a + (n − 1) d Therefore, 205 = 7 + (n − 1) 6 ⇒ 198 = (n − 1) 6 ⇒ 33 = (n − 1) ⇒ n = 34 Therefore, this given series has 34 terms in it.
First term, a = 18 Common difference, d = a2-a1 =
⇒ d = (31-36)/2 = -5/2 Let there are n terms in this A.P. an = 205 As we know, for an A.P., an = a+(n−1)d ⇒ -47 = 18+(n-1)(-5/2) ⇒ -47-18 = (n-1)(-5/2) ⇒ -65 = (n-1)(-5/2) ⇒ (n-1) = -130/-5 ⇒ (n-1) = 26 ⇒ n = 27 Therefore, this given A.P. has 27 terms in it.
Q6. Check whether -150 is a term of the A.P. 11, 8, 5, 2, …
For the given series, A.P. 11, 8, 5, 2.. First term, a = 11 Common difference, d = a2−a1 = 8−11 = −3 Let −150 be the nth term of this A.P. As we know, for an A.P., an = a+(n−1)d ⇒ -150 = 11+(n -1)(-3) ⇒ -150 = 11-3n +3 ⇒ -164 = -3n ⇒ n = 164/3 Clearly, n is not an integer but a fraction. Therefore, – 150 is not a term of this A.P.
Q7. Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73.
Sol: Given that, 11th term, a11 = 38 and 16th term, a16 = 73 We know that, an = a+(n−1)d ⇒ a11 = a+(11−1)d ⇒ 38 = a+10d ………………………………. (i) In the same way, ⇒ a16 = a +(16−1)d ⇒ 73 = a+15d ………………………………………… (ii) On subtracting equation (i) from (ii), we get ⇒ 35 = 5d ⇒ d = 7 From equation (i), we can write, ⇒ 38 = a+10×(7) ⇒ 38 − 70 = a ⇒ a = −32 a31 = a +(31−1) d ⇒ − 32 + 30 (7) ⇒ − 32 + 210 ⇒ 178 Hence, 31st term is 178.
Q8: An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Sol: Given that, 3rd term, a3 = 12 50th term, a50 = 106 We know that, an = a+(n−1)d a3 = a+(3−1)d 12 = a+2d ……………………………. (i) In the same way, a50 = a+(50−1)d 106 = a+49d …………………………. (ii) On subtracting equation (i) from (ii), we get 94 = 47d d = 2 = common difference From equation (i), we can write now, 12 = a+2(2) a = 12−4 = 8 a29 = a+(29−1) d a29 = 8+(28)2 a29 = 8+56 = 64 Therefore, 29th term is 64.
Q9: If the 3rd and the 9th terms of an A.P. are 4 and − 8 respectively. Which term of this A.P. is zero?
Sol: Given that, 3rd term, a3 = 4 and 9th term, a9 = −8 We know that, an = a+(n−1)d Therefore, a3 = a+(3−1)d 4 = a+2d ……………………………………… (i) a9 = a+(9−1)d −8 = a+8d ………………………………………………… (ii) On subtracting equation (i) from (ii), we will get here, −12 = 6d d = −2 From equation (i), we can write, 4 = a+2(−2) 4 = a−4 a = 8 Let nth term of this A.P. be zero. an = a+(n−1)d 0 = 8+(n−1)(−2) 0 = 8−2n+2 2n = 10 n = 5 Hence, 5th term of this A.P. is 0.
Q10: The 17th term of an A.P. exceeds its 10th term by 7. Find the common difference.
Sol:
We know that, for an A.P series: an = a+(n−1)d a17 = a+(17−1)d a17 = a +16d In the same way, a10 = a+9d As it is given in the question, a17 − a10 = 7 Therefore, (a +16d) − (a+9d) = 7 7d = 7 d = 1 Therefore, the common difference is 1.
Q11: Which term of the A.P. 3, 15, 27, 39,.. will be 132 more than its 54th term?
Sol: Given A.P. is 3, 15, 27, 39, … first term, a = 3 common difference, d = a2 − a1 = 15 − 3 = 12 We know that, an = a+(n−1)d Therefore, a54 = a+(54−1)d ⇒ 3+(53)(12) ⇒ 3+636 = 639 a54 = 639 We have to find the term of this A.P. which is 132 more than a54, i.e.771. Let nth term be 771. an = a+(n−1)d 771 = 3+(n −1)12 768 = (n−1)12 (n −1) = 64 n = 65 Therefore, 65th term was 132 more than 54th term. Or another method is; Let nth term be 132 more than 54th term. n = 54 + 132/2 = 54 + 11 = 65th term
Q12: Two APs have the same common difference. The difference between their 100th term is 100, what is the difference between their 1000th terms?
Sol: Let, the first term of two APs be a1 and a2 respectively And the common difference of these APs be d. For the first A.P.,we know, an = a+(n−1)d Therefore, a100 = a1+(100−1)d = a1 + 99d a1000 = a1+(1000−1)d a1000 = a1+999d For second A.P., we know, an = a+(n−1)d Therefore, a100 = a2+(100−1)d = a2+99d a1000 = a2+(1000−1)d = a2+999d Given that, difference between 100th term of the two APs = 100 Therefore, (a1+99d) − (a2+99d) = 100 a1−a2 = 100……………………………………………………………….. (i) Difference between 1000th terms of the two APs (a1+999d) − (a2+999d) = a1−a2 From equation (i), This difference, a1−a2 = 100 Hence, the difference between 1000th terms of the two A.P. will be 100.
Q13: How many three-digit numbers are divisible by 7?
Sol: First three-digit number that is divisible by 7 are; First number = 105 Second number = 105+7 = 112 Third number = 112+7 =119 Therefore, 105, 112, 119, … All are three digit numbers are divisible by 7 and thus, all these are terms of an A.P. having first term as 105 and common difference as 7. As we know, the largest possible three-digit number is 999. When we divide 999 by 7, the remainder will be 5. Therefore, 999-5 = 994 is the maximum possible 3-digit number that is divisible by 7. Now the series is as follows. 105, 112, 119, …, 994 Let 994 be the nth term of this A.P. first term, a = 105 common difference, d = 7 an = 994 n = ? As we know, an = a+(n−1)d 994 = 105+(n−1)7 889 = (n−1)7 (n−1) = 127 n = 128 Therefore, 128 three-digit numbers are divisible by 7.
Q14: How many multiples of 4 lie between 10 and 250?
Sol: The first multiple of 4 that is greater than 10 is 12. Next multiple will be 16. Therefore, the series formed as; 12, 16, 20, 24, … All these are divisible by 4 and thus, all these are terms of an A.P. with first term as 12 and common difference as 4. When we divide 250 by 4, the remainder will be 2. Therefore, 250 − 2 = 248 is divisible by 4. The series is as follows, now; 12, 16, 20, 24, …, 248 Let 248 be the nth term of this A.P. first term, a = 12 common difference, d = 4 an = 248 As we know, an = a+(n−1)d 248 = 12+(n-1)×4 236/4 = n-1 59 = n-1 n = 60 Therefore, there are 60 multiples of 4 between 10 and 250.
Q15: For what value of n, are the nth terms of two APs 63, 65, 67, and 3, 10, 17, … equal?
Sol: Given two APs as; 63, 65, 67,… and 3, 10, 17,…. Taking first AP, 63, 65, 67, … First term, a = 63 Common difference, d = a2−a1 = 65−63 = 2 We know, nth term of this A.P. = an = a+(n−1)d an= 63+(n−1)2 = 63+2n−2 an = 61+2n ………………………………………. (i) Taking second AP, 3, 10, 17, … First term, a = 3 Common difference, d = a2 − a1 = 10 − 3 = 7 We know that, nth term of this A.P. = 3+(n−1)7 an = 3+7n−7 an = 7n−4 ……………………………………………………….. (ii) Given, nth term of these A.P.s are equal to each other. Equating both these equations, we get, 61+2n = 7n−4 61+4 = 5n 5n = 65 n = 13 Therefore, 13th terms of both these A.P.s are equal to each other.
Q16: Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12.
Sol: Given, Third term, a3 = 16 As we know, a +(3−1)d = 16 a+2d = 16 ………………………………………. (i) It is given that, 7th term exceeds the 5th term by 12. a7 − a5 = 12 [a+(7−1)d]−[a +(5−1)d]= 12 (a+6d)−(a+4d) = 12 2d = 12 d = 6 From equation (i), we get, a+2(6) = 16 a+12 = 16 a = 4 Therefore, A.P. will be 4, 10, 16, 22, …
Q17: Find the 20th term from the last term of the A.P. 3, 8, 13, …, 253.
Sol: Given A.P. is3, 8, 13, …, 253 Common difference, d= 5. Therefore, we can write the given AP in reverse order as; 253, 248, 243, …, 13, 8, 5 Now for the new AP, first term, a = 253 and common difference, d = 248 − 253 = −5 n = 20 Therefore, using nth term formula, we get, a20 = a+(20−1)d a20 = 253+(19)(−5) a20 = 253−95 a = 158 Therefore, 20th term from the last term of the AP 3, 8, 13, …, 253 is 158.
Q18: The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the A.P.
Sol: We know that, the nth term of the AP is; an = a+(n−1)d a4 = a+(4−1)d a4 = a+3d In the same way, we can write, a8 = a+7d a6 = a+5d a10 = a+9d Given that, a4+a8 = 24 a+3d+a+7d = 24 2a+10d = 24 a+5d = 12 …………………………………………………… (i) a6+a10 = 44 a +5d+a+9d = 44 2a+14d = 44 a+7d = 22 …………………………………………………… (ii) On subtracting equation (i) from (ii), we get, 2d = 22 − 12 2d = 10 d = 5 From equation (i), we get, a+5d = 12 a+5(5) = 12 a+25 = 12 a = −13 a2 = a+d = − 13+5 = −8 a3 = a2+d = − 8+5 = −3 Therefore, the first three terms of this A.P. are −13, −8, and −3.
Q19: Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?
Sol: It can be seen from the given question, that the incomes of Subba Rao increases every year by Rs.200 and hence, forms an AP. Therefore, after 1995, the salaries of each year are: 5000, 5200, 5400, … Here, first term, a = 5000 and common difference, d = 200 Let after nth year, his salary be Rs 7000. Therefore, by the nth term formula of AP, an = a+(n−1) d 7000 = 5000+(n−1)200 200(n−1)= 2000 (n−1) = 10 n = 11 Therefore, in 11th year, his salary will be Rs 7000.
Q20: Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the nth week, her weekly savings become Rs 20.75, find n.
Sol: Given that, Ramkali saved Rs.5 in first week and then started saving each week by Rs.1.75. Hence, First term, a = 5 and common difference, d = 1.75 Also given, an = 20.75 Find, n = ? As we know, by the nth term formula, an = a+(n−1)d Therefore, 20.75 = 5+(n -1)×1.75 15.75 = (n -1)×1.75 (n -1) = 15.75/1.75 = 1575/175 = 63/7 = 9 n -1 = 9 n = 10 Hence, n is 10.
On comparing x2 − 4x + 6 to ax+ bx+ c=0 we find x2 − 4x + 6 is a quadratic polynomial Since x2 − 4x + 6 is a quadratic polynomial ∴ x2 − 2x = (−2) (3 − x) is a quadratic equation.
On comparing −3x + 1 = 0 to ax+ bx+ c=0 we find −3x + 1 = 0 is a quadratic polynomial Since −3x + 1 is a linear polynomial ∴ (x − 2) (x + 1) = (x − 1) (x + 3) is not quadratic equation.
On comparing x2 + 10x − 3= 0 to ax+ bx+ c=0 we find x2 + 10x − 3= 0 is a quadratic polynomial Since x2 + 10x − 3 is a quadratic polynomial ∴ (x − 3) (2x + 1) = x(x + 5) is a quadratic equation.
On comparing 7x − 3 = 0 to ax+ bx+ c=0 we find 7x − 3 = 0 is a quadratic polynomial Since 7x − 3 is a linear polynomial ∴ x2 + 3x + 1 = (x − 2)2 is not a quadratic equation.
On comparing− x3 + 6x2 + 14x + 8 = 0 to ax+ bx+ c=0 we find − x3 + 6x2 + 14x + 8 = 0 is NOT a quadratic polynomial Since −x3 + 6x2 + 14x + 8 is a polynomial of degree 3 ∴ (x + 2)3 = 2x(x2 − 1) is not a quadratic equation.
On comparing 2x2 − 13x + 9 = 0 to ax+ bx+ c=0 we find 2x2 − 13x + 9 = 0 is a quadratic polynomial Since 2x2 − 13x + 9 is a quadratic polynomial ∴ x3 − 4x2 − x + 1 = (x − 2)3 is a quadratic equation.
Q2.Represent the following situations in the form of quadratic equations: (i) The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
Let the breadth = x metres ∵ Length = 2 (Breadth) + 1 ∴ Length = (2x + 1) metres
Since Length × Breadth = Area ∴ (2x + 1) × x = 528 ⇒ 2x2 + x = 528 ⇒ 2x2 + x − 528 = 0 Thus, the required quadratic equation is 2x2 + x− 528 = 0
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
Let the two consecutive numbers be x and (x + 1). ∵Product of the numbers = 306 ∴ x (x + 1) = 306 ⇒ x2 + x = 306 ⇒ x2 + x − 306 = 0 Thus, the required quadratic equation is x2 + x − 306 = 0
(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
Let the present age = x ∴ Mother’s age = (x + 26) years After 3 years His age = (x + 3) years Mother’s age = [(x + 26) + 3] years =(x + 29) years According to the condition,
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
In the first case, Let the speed of the train = u km/hr Distance covered = 480 km Time taken = Distance ÷ Speed
FactorisationQ1. Find the roots of the following quadratic equations by factorisation: Solutions: (i) x2 – 3x – 10 = 0 Taking LHS, ⇒x2 – 5x + 2x – 10 ⇒x(x – 5) + 2(x – 5) ⇒(x – 5)(x + 2) The roots of this equation, x2 – 3x – 10 = 0 are the values of x for which (x – 5)(x + 2) = 0 Therefore, x – 5 = 0 or x + 2 = 0 ⇒ x = 5 or x = -2
(ii) 2x2 + x – 6 = 0 Taking LHS, ⇒ 2x2 + 4x – 3x – 6 ⇒ 2x(x + 2) – 3(x + 2) ⇒ (x + 2)(2x – 3) The roots of this equation, 2x2 + x – 6=0 are the values of x for which (x – 5)(x + 2) = 0 Therefore, x + 2 = 0 or 2x – 3 = 0 ⇒ x = -2 or x = 3/2
(iii) √2 x2 + 7x + 5√2 = 0 Taking LHS, ⇒ √2 x2 + 5x + 2x + 5√2 ⇒ x (√2x + 5) + √2(√2x + 5)= (√2x + 5)(x + √2) The roots of this equation, √2 x2 + 7x + 5√2=0 are the values of x for which (x – 5)(x + 2) = 0 Therefore, √2x + 5 = 0 or x + √2 = 0 ⇒ x = -5/√2 or x = -√2
(iv) 2x2 – x +1/8 = 0 Taking LHS, ⇒1/8 (16x2 – 8x + 1) ⇒ 1/8 (16x2 – 4x -4x + 1) ⇒1/8 (4x(4x– 1) -1(4x – 1)) ⇒1/8 (4x – 1)2 The roots of this equation, 2x2 – x + 1/8 = 0, are the values of x for which (4x – 1)2= 0 Therefore, (4x – 1) = 0 or (4x – 1) = 0 ⇒ x = 1/4
(v) Given, 100x2 – 20x + 1=0 Taking LHS, ⇒100x2 – 10x – 10x + 1 ⇒10x(10x – 1) -1(10x – 1) ⇒(10x – 1)2 The roots of this equation, 100x2 – 20x + 1=0, are the values of x for which (10x – 1)2= 0 ∴ (10x – 1) = 0 or (10x – 1) = 0 ⇒x = 1/10
Q2. Solve the problems given in Example 1. Represent the following situations mathematically: (i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with. (ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs 750. We would like to find out the number of toys produced on that day. Solutions: (i) Let us say, the number of marbles John has = x.
Therefore, the number of marble Jivanti have = 45 – x After losing 5 marbles each, Number of marbles John have = x – 5 Number of marble Jivanti have = 45 – x – 5 = 40 – x Given that the product of their marbles is 124. ∴ (x – 5)(40 – x) = 124 ⇒ x2 – 45x + 324 = 0 ⇒ x2 – 36x – 9x + 324 = 0 ⇒ x(x – 36) -9(x – 36) = 0 ⇒ (x – 36)(x – 9) = 0 Thus, we can say, x – 36 = 0 or x – 9 = 0 ⇒ x = 36 or x = 9 Therefore, If, John’s marbles = 36, Then, Jivanti’s marbles = 45 – 36 = 9 And if John’s marbles = 9, Then, Jivanti’s marbles = 45 – 9 = 36
(ii) Let us say, the number of toys produced in a day is x. Therefore, cost of production of each toy = Rs(55 – x) Given, the total cost of production of the toys = Rs 750 ∴ x(55 – x) = 750 ⇒ x2 – 55x + 750 = 0 ⇒ x2 – 25x – 30x + 750 = 0 ⇒ x(x – 25) -30(x – 25) = 0 ⇒ (x – 25)(x – 30) = 0 Thus, either x -25 = 0 or x – 30 = 0 ⇒ x = 25 or x = 30 Hence, the number of toys produced in a day, will be either 25 or 30.
Q3. Find two numbers whose sum is 27 and product is 182. Solution:
Let us say, first number be x and the second number is 27 – x. Therefore, the product of two numbers x(27 – x) = 182 ⇒ x2 – 27x – 182 = 0 ⇒ x2 – 13x – 14x + 182 = 0 ⇒ x(x – 13) -14(x – 13) = 0 ⇒ (x – 13)(x -14) = 0 Thus, either, x = -13 = 0 or x – 14 = 0 ⇒ x = 13 or x = 14 Therefore, if first number = 13, then second number = 27 – 13 = 14 And if first number = 14, then second number = 27 – 14 = 13 Hence, the numbers are 13 and 14.
Q4. Find two consecutive positive integers, sum of whose squares is 365.
Solution: Let us say, the two consecutive positive integers be x and x + 1. Therefore, as per the given questions, x2 + (x + 1)2 = 365 ⇒ x2 + x2 + 1 + 2x = 365 ⇒ 2x2 + 2x – 364 = 0 ⇒ x2 + x – 182 = 0 ⇒ x2 + 14x – 13x – 182 = 0 ⇒ x(x + 14) -13(x + 14) = 0 ⇒ (x + 14)(x – 13) = 0 Thus, either, x + 14 = 0 or x – 13 = 0, ⇒ x = – 14 or x = 13 since, the integers are positive, x can be 13, only. ∴ x + 1 = 13 + 1 = 14 Therefore, two consecutive positive integers will be 13 and 14.
Q5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Solution: Let us say, the base of the right triangle be x cm. Given, the altitude of right triangle = (x – 7) cm From Pythagoras theorem, we know, Base2 + Altitude2 = Hypotenuse2 ∴ x2 + (x – 7)2 = 132 ⇒ x2 + x2 + 49 – 14x = 169 ⇒ 2x2 – 14x – 120 = 0 ⇒ x2 – 7x – 60 = 0 ⇒ x2 – 12x + 5x – 60 = 0 ⇒ x(x – 12) + 5(x – 12) = 0 ⇒ (x – 12)(x + 5) = 0 Thus, either x – 12 = 0 or x + 5 = 0, ⇒ x = 12 or x = – 5 Since sides cannot be negative, x can only be 12. Hence, the base of the given triangle is 12 cm and the altitude of this triangle will be (12 – 7) cm = 5 cm.
Q6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs.90, find the number of articles produced and the cost of each article.
Solution: Let us say, the number of articles produced is x. Therefore, cost of production of each article = Rs (2x + 3) Given, the total cost of production is Rs.90 ∴ x(2x + 3) = 90 ⇒ 2x2 + 3x – 90 = 0 ⇒ 2x2 + 15x -12x – 90 = 0 ⇒ x(2x + 15) -6(2x + 15) = 0 ⇒ (2x + 15)(x – 6) = 0 Thus, either 2x + 15 = 0 or x – 6 = 0 ⇒ x = -15/2 or x = 6 As the number of articles produced can only be a positive integer, therefore, x can only be 6. Hence, the number of articles produced = 6 Cost of each article = 2 × 6 + 3 = Rs 15.
Exercise 4.3
Q1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them; (i) 2x2 – 3x + 5 = 0 (ii) 3x2 – 4√3x + 4 = 0 (iii) 2x2 – 6x + 3 = 0 Solutions: (i) Given, 2x2 – 3x + 5 = 0 Comparing the equation with ax2 + bx + c = 0, we get a = 2, b = -3 and c = 5 We know, Discriminant = b2 – 4ac = ( – 3)2 – 4 (2) (5) = 9 – 40 = – 31 As you can see, b2 – 4ac < 0 Therefore, no real root is possible for the given equation, 2x2 – 3x + 5 = 0. (ii) 3x2 – 4√3x + 4 = 0 Comparing the equation with ax2 + bx + c = 0, we get a = 3, b = -4√3 and c = 4 We know, Discriminant = b2 – 4ac = (-4√3)2 – 4(3)(4) = 48 – 48 = 0 As b2 – 4ac = 0, Real roots exist for the given equation and they are equal to each other. Hence the roots will be –b/2a and –b/2a. –b/2a = -(-4√3)/2×3 = 4√3/6 = 2√3/3 = 2/√3 Therefore, the roots are 2/√3 and 2/√3. (iii) 2x2 – 6x + 3 = 0 Comparing the equation with ax2 + bx + c = 0, we get a = 2, b = -6, c = 3 As we know, Discriminant = b2 – 4ac = (-6)2 – 4 (2) (3) = 36 – 24 = 12 As b2 – 4ac > 0, Therefore, there are distinct real roots exist for this equation, 2x2 – 6x + 3 = 0. = (-(-6) ± √(-62-4(2)(3)) )/ 2(2) = (6±2√3 )/4 = (3±√3)/2 Therefore the roots for the given equation are (3+√3)/2 and (3-√3)/2
Q2. Find the values of k for each of the following quadratic equations, so that they have two equal roots. (i) 2x2 + kx + 3 = 0 (ii) kx (x – 2) + 6 = 0 Solutions: (i) 2x2+ kx + 3 = 0 Comparing the given equation with ax2 + bx + c = 0, we get, a = 2, b = k and c = 3 As we know, Discriminant = b2 – 4ac = (k)2 – 4(2) (3) = k2 – 24 For equal roots, we know, Discriminant = 0 k2 – 24 = 0 k2 = 24 k = ±√24 = ±2√6 (ii) kx(x – 2) + 6 = 0 or kx2 – 2kx + 6 = 0 Comparing the given equation with ax2 + bx + c = 0, we get a = k, b = – 2k and c = 6 We know, Discriminant = b2 – 4ac = ( – 2k)2 – 4 (k) (6) = 4k2 – 24k For equal roots, we know, b2 – 4ac = 0 4k2 – 24k = 0 4k (k – 6) = 0 Either 4k = 0 or k = 6 = 0 k = 0 or k = 6 However, if k = 0, then the equation will not have the terms ‘x2‘ and ‘x‘. Therefore, if this equation has two equal roots, k should be 6 only.
Q3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth. Solution: Let the breadth of mango grove be l. Length of mango grove will be 2l. Area of mango grove = (2l) (l)= 2l2 2l2 = 800 l2 = 800/2 = 400 l2 – 400 =0 Comparing the given equation with ax2 + bx + c = 0, we get a = 1, b = 0, c = 400 As we know, Discriminant = b2 – 4ac => (0)2 – 4 × (1) × ( – 400) = 1600 Here, b2 – 4ac > 0 Thus, the equation will have real roots. And hence, the desired rectangular mango grove can be designed. l = ±20 As we know, the value of length cannot be negative. Therefore, breadth of mango grove = 20 m Length of mango grove = 2 × 20 = 40 m
Q4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48. Solution: Let’s say, the age of one friend be x years. Then, the age of the other friend will be (20 – x) years. Four years ago, Age of First friend = (x – 4) years Age of Second friend = (20 – x – 4) = (16 – x) years As per the given question, we can write, (x – 4) (16 – x) = 48 16x – x2 – 64 + 4x = 48 – x2 + 20x – 112 = 0 x2 – 20x + 112 = 0 Comparing the equation with ax2 + bx + c = 0, we get a = 1, b = -20 and c = 112 Discriminant = b2 – 4ac = (-20)2 – 4 × 112 = 400 – 448 = -48 b2 – 4ac < 0 Therefore, there will be no real solution possible for the equations. Hence, condition doesn’t exist.
Q5. Is it possible to design a rectangular park of perimeter 80 and area 400 m2? If so find its length and breadth. Solution: Let the length and breadth of the park be l and b. Perimeter of the rectangular park = 2 (l + b) = 80 So, l + b = 40 Or, b = 40 – l Area of the rectangular park = l×b = l(40 – l) = 40l – l2 = 400 l2–40l + 400= 0, which is a quadratic equation. Comparing the equation with ax2 + bx + c = 0, we get a = 1, b = -40, c = 400 Since, Discriminant = b2 – 4ac =(-40)2 – 4 × 400 = 1600 – 1600 = 0 Thus, b2 – 4ac = 0 Therefore, this equation has equal real roots. Hence, the situation is possible. Root of the equation, l = –b/2a l = (40)/2(1) = 40/2 = 20 Therefore, length of rectangular park, l = 20 m And breadth of the park, b = 40 – l = 40 – 20 = 20 m.
Q1. Form the pair of linear equations in the following problems, and find their solutions graphically. (i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz. (ii) 5 pencils and 7 pens together cost 50, whereas 7 pencils and 5 pens together cost 46. Find the cost of one pencil and that of one pen. Ans: (i)
Let number of boys = x Number of girls = y Given that total number of student is 10. So, x + y = 10 Subtract y both side we get, x = 10 – y Putting y = 0 , 5, 10 we get, x = 10 – 0 = 10 x = 10 – 5 = 5 x = 10 – 10 = 0x105y05
Given that If the number of girls is 4 more than the number of boys. So, y = x + 4 Putting x = -4, 0, 4, and we get, y = – 4 + 4 = 0 y = 0 + 4 = 4 y = 4 + 4 = 8x-404y048
Graphical representation: Therefore, number of boys = 3 and number of girls = 7
(ii) Let 1 pencil costs Rs.x and 1 pen costs Rs.y. According to the question, the algebraic expression cab be represented as; 5x + 7y = 50 7x + 5y = 46 For, 5x + 7y = 50 or x = (50 – 7y) / 5, the solutions are;x30y57.14
For 7x + 5y = 46 or x = (46 – 5y) / 7, the solutions are;x6.573y05
Hence, the graphical representation is as follows:
From the graph, it can be seen that the given lines cross each other at point (3, 5).So, the cost of a pencil is Rs. 3 and cost of a pen is Rs. 5.
Q2. On comparing the ratios a1 / a2 , b1 / b2 , c1 / c2 find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident: (i) 5x – 4y + 8 = 0 7x + 6y – 9 = 0 (ii) 9x + 3y + 12 = 0 18x + 6y + 24 = 0 (iii) 6x – 3y + 10 = 0 2x – y + 9 = 0 Ans: (i) Given expressions; 5x − 4y + 8 = 0 7x + 6y − 9 = 0 Comparing these equations with a1x + b1y + c1 = 0 And a2x + b2y + c2 = 0 We get, a1 = 5, b1 = -4, c1 = 8 a2 = 7, b2 = 6, c2 = -9 (a1 / a2) = 5 / 7 (b1 / b2) = -4 / 6 = -2 / 3 (c1/ c2) = 8 / -9 Since, (a1 / a2) ≠ (b1 / b2) So, the pairs of equations given in the question have a unique solution and the lines cross each other at exactly one point.
(ii) Given expressions; 9x + 3y + 12 = 0 18x + 6y + 24 = 0 Comparing these equations with a1x + b1y + c1 = 0 And a2x + b2y + c2 = 0 We get, a1 = 9, b1 = 3, c1 = 12 a2 = 18, b2 = 6, c2= 24 (a1/ a2) = 9 / 18 = 1 / 2 (b1 / b2) = 3 / 6 = 1 / 2 (c1/ c2) = 12 / 24 = 1 / 2 Since (a1/ a2) = (b1 / b2) = (c1/ c2) So, the pairs of equations given in the question have infinite possible solutions and the lines are coincident.
(iii) Given Expressions; 6x – 3y + 10 = 0 2x – y + 9 = 0 Comparing these equations with a1x + b1y + c1= 0 And a2x + b2y + c2= 0 We get, a1 = 6, b1 = -3, c1= 10 a2 = 2, b2= -1, c2= 9 (a1 / a2) = 6 / 2 = 3 / 1 (b1 / b2) = -3 / -1 = 3 / 1 (c1 / c2) = 10 / 9 Since (a1 / a2) = (b1 / b2) ≠ (c1/ c2) So, the pairs of equations given in the question are parallel to each other and the lines never intersect each other at any point and there is no possible solution for the given pair of equations.
Q3. On comparing the ratio, (a1 / a2), (b1 / b2), (c1/ c2) find out whether the following pair of linear equations are consistent, or inconsistent. (i) 3x + 2y = 5 ; 2x – 3y = 7 (ii) 2x – 3y = 8 ; 4x – 6y = 9 (iii) (3 / 2)x + (5 / 3)y = 7; 9x – 10y = 14 (iv) 5x – 3y = 11 ; – 10x + 6y = –22 (v) (4 / 3)x + 2y = 8 ; 2x + 3y = 12 Ans: (i)Given: 3x + 2y = 5 or 3x + 2y -5 = 0 and 2x – 3y = 7 or 2x – 3y -7 = 0 Comparing these equations with a1x + b1y + c1= 0 And a2x + b2y + c2= 0 We get, a1 = 3, b1 = 2, c1= -5 a2 = 2, b2= -3, c2= -7 (a1 / a2) = 3 / 2 (b1 / b2) = 2 / -3 (c1/ c2) = -5 / -7 = 5 / 7 Since, (a1 / a2) ≠ (b1 / b2) So, the given equations intersect each other at one point and they have only one possible solution. The equations are consistent.
(ii) Given: 2x – 3y = 8 and 4x – 6y = 9 Therefore, a1 = 2, b1 = -3, c1= -8 a2= 4, b2= -6, c2 = -9 (a1 / a2) = 2 / 4 = 1 / 2 (b1 / b2) = -3 / -6 = 1 / 2 (c1/ c2) = -8 / -9 = 8 / 9 Since , (a1/ a2) = (b1/ b2) ≠ (c1/ c2) So, the equations are parallel to each other and they have no possible solution. Hence, the equations are inconsistent.
(iii)Given: (3 / 2)x + (5 / 3)y = 7 and 9x – 10y = 14 Therefore, a1= 3 / 2, b1= 5 / 3, c1= -7 a2= 9, b2= -10, c2= -14 (a1/ a2) = 3 / (2 × 9) = 1 / 6 (b1/ b2) = 5 / (3× – 10)= -1 / 6 (c1/ c2) = -7 / -14 = 1 / 2 Since, (a1/ a2) ≠ (b1/ b2) So, the equations are intersecting each other at one point and they have only one possible solution. Hence, the equations are consistent.
(iv)Given:5x – 3y = 11 and – 10x + 6y = –22 Therefore, a1= 5, b1 = -3, c1= -11 a2= -10, b2 = 6, c2= 22 (a1/ a2) = 5 / (-10) = -5 / 10 = -1 / 2 (b1/ b2) = -3 / 6 = -1 / 2 (c1/ c2) = -11 / 22 = -1 / 2 Since (a1/ a2) = (b1/ b2) = (c1/ c2) These linear equations are coincident lines and have infinite number of possible solutions. Hence, the equations are consistent.
(v) Given: (4 / 3)x + 2y = 8 and 2x + 3y = 12 a1= 4 / 3 , b1= 2 , c1= -8 a2= 2, b2 = 3 , c2= -12 (a1/ a2) = 4 / (3 × 2)= 4 / 6 = 2 / 3 (b1/ b2) = 2 / 3 (c1/ c2) = -8 / -12 = 2 / 3 Since (a1/ a2) = (b1/ b2) = (c1/ c2) These linear equations are coincident lines and have infinite number of possible solutions. Hence, the equations are consistent.
Q4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: (i) x + y = 5, 2x + 2y = 10 (ii) x – y = 8, 3x – 3y = 16 (iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0 (iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0 Ans: (i) Given, x + y = 5 and 2x + 2y = 10 (a1/ a2) = 1 / 2 (b1/ b2) = 1 / 2 (c1/ c2) = 1 / 2 Since (a1/ a2) = (b1/ b2) = (c1/ c2) ∴The equations are coincident and they have infinite number of possible solutions. So, the equations are consistent. For, x + y = 5 or x = 5 – yx05y50
For 2x + 2y = 10 or x = (10 – 2y) / 2x05y50
So, the equations are represented in graphs as follows:
From the figure, we can see, that the lines are overlapping each other. Therefore, the equations have infinite possible solutions.
(ii) Given, x – y = 8 and 3x – 3y = 16 (a1/ a2) = 1 / 3 (b1/ b2) = -1 / -3 = 1 / 3 (c1/ c2) = 8 / 16 = 1 / 2 Since, (a1/ a2) = (b1/ b2) ≠ (c1/ c2) The equations are parallel to each other and have no solutions. Hence, the pair of linear equations is inconsistent.
(iii) Given, 2x + y – 6 = 0 and 4x – 2y – 4 = 0 (a1/ a2) = 2 / 4 = 1 / 2 (b1/ b2) = 1 / -2 (c1/ c2) = -6 / -4 = 3 / 2 Since, (a1/ a2) ≠ (b1/ b2) The given linear equations are intersecting each other at one point and have only one solution. Hence, the pair of linear equations is consistent. Now, for 2x + y – 6 = 0 or y = 6 – 2xx03y60
And for 4x – 2y – 4 = 0 or y = (4x – 4) / 2x01y-20
So, the equations are represented in graphs as follows:
From the graph, it can be seen that these lines are intersecting each other at only one point (2,2).
(iv) Given, 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0 (a1/ a2) = 2 / 4 = 1 / 2 (b1/ b2) = -2 / -4 = 1 / 2 (c1/ c2) = 2 / 5 Since, a1/ a2= b1/ b2≠ c1/ c2 Thus, these linear equations have parallel and have no possible solutions. Hence, the pair of linear equations are inconsistent.
Q5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden. Ans: Let us consider. The width of the garden is y and length is x. Now, according to the question, we can express the given condition as; x – y = 4 and y + x = 36 Now, taking x – y = 4 or x = y + 4x40y0-4
For y + x = 36, y = 36 – xx1220y2416
The graphical representation of both the equation is as follows: From the graph you can see, the lines intersects each other at a point(20, 16). Hence, the width of the garden is 16 and length is 20.
Q6. Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is: (i) Intersecting lines (ii) Parallel lines (iii) Coincident lines Ans: (i) Given the linear equation 2x + 3y – 8 = 0. To find another linear equation in two variables such that the geometrical representation of the pair so formed is intersecting lines, it should satisfy below condition; (a1/ a2) ≠ (b1/ b2) Thus, another equation could be 2x – 7y + 9 = 0, such that; (a1/ a2) = 2 / 2 = 1 and (b1/ b2) = 3 / -7 Clearly, you can see another equation satisfies the condition. (ii) Given the linear equation 2x + 3y – 8 = 0. To find another linear equation in two variables such that the geometrical representation of the pair so formed is parallel lines, it should satisfy below condition; (a1/ a2) = (b1/ b2) ≠ (c1/ c2) Thus, another equation could be 6x + 9y + 9 = 0, such that; (a1/ a2) = 2 / 6 = 1 / 3 (b1/ b2) = 3 / 9= 1 / 3 (c1/ c2) = -8 / 9 Clearly, you can see another equation satisfies the condition. (iii) Given the linear equation 2x + 3y – 8 = 0. To find another linear equation in two variables such that the geometrical representation of the pair so formed is coincident lines, it should satisfy below condition; (a1/ a2) = (b1/ b2) = (c1/ c2) Thus, another equation could be 4x + 6y – 16 = 0, such that; (a1/ a2) = 2 / 4 = 1 / 2 ,(b1/ b2) = 3 / 6 = 1 / 2, (c1/ c2) = -8 / -16 = 1 / 2 Clearly, you can see another equation satisfies the condition.
Q7. Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region. Ans: Given, the equations for graphs are x – y + 1 = 0 and 3x + 2y – 12 = 0. For, x – y + 1 = 0 or y = 1 + xx0-1y10
For, 3x + 2y – 12 = 0 or x = (12 – 2y) / 3x04y60
Hence, the graphical representation of these equations is as follows:
From the figure, it can be seen that these lines are intersecting each other at point (2, 3) and x-axis at (−1, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), (−1, 0), and (4, 0).
Exercise 3.2
Q1: Solve the following pair of linear equations by the substitution method
(i) x + y = 14 x – y = 4 Sol: Given, x + y = 14 and x – y = 4 are the two equations. From 1st equation, we get, x = 14 – y Now, substitute the value of x in second equation to get, (14 – y) – y = 4 14 – 2y = 4 2y = 10 Or y = 5 By the value of y, we can now find the exact value of x; ∵ x = 14 – y ∴ x = 14 – 5 Or x = 9 Hence, x = 9 and y = 5.
(ii) s – t = 3 (s/3) + (t/2) = 6 Sol:Given, s – t = 3 and (s/3) + (t/2) = 6 are the two equations. From 1st equation, we get, s = 3 + t ________________(1) Now, substitute the value of s in second equation to get, (3+t)/3 + (t/2) = 6 ⇒ (2(3+t) + 3t )/6 = 6 ⇒ (6+2t+3t)/6 = 6 ⇒ (6+5t) = 36 ⇒5t = 30 ⇒t = 6 Now, substitute the value of t in equation (1) s = 3 + 6 = 9 Therefore, s = 9 and t = 6.
(iii) 3x – y = 3 9x – 3y = 9 Sol: 3x – y = 3 and 9x – 3y = 9 are the two equations. From 1st equation, we get, x = (3+y)/3 Now, substitute the value of x in the given second equation to get, 9(3+y)/3 – 3y = 9 ⇒9 +3y -3y = 9 ⇒ 9 = 9 Therefore, y has infinite values and since, x = (3+y) /3, so x also has infinite values.
(iv) 0.2x + 0.3y = 1.3 0.4x + 0.5y = 2.3 Sol: Given, 0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3are the two equations. From 1st equation, we get, x = (1.3- 0.3y)/0.2 _________________(1) Now, substitute the value of x in the given second equation to get, 0.4(1.3-0.3y)/0.2 + 0.5y = 2.3 ⇒ 2(1.3 – 0.3y) + 0.5y = 2.3 ⇒ 2.6 – 0.6y + 0.5y = 2.3 ⇒ 2.6 – 0.1 y = 2.3 ⇒ 0.1 y = 0.3 ⇒ y = 3 Now, substitute the value of y in equation (1), we get, x = (1.3-0.3(3))/0.2 = (1.3-0.9)/0.2 = 0.4/0.2 = 2 Therefore, x = 2 and y = 3.
(v) √2 x + √3 y = 0 √3 x – √8 y = 0 Sol:Given, √2 x + √3 y = 0 and √3 x – √8 y = 0 are the two equations. From 1st equation, we get, x = – (√3/√2)y __________________(1) Putting the value of x in the given second equation to get, √3(-√3/√2)y – √8y = 0 ⇒ (-3/√2)y- √8 y = 0 ⇒ y = 0 Now, substitute the value of y in equation (1), we get, x = 0 Therefore, x = 0 and y = 0.
(vi) (3x / 2) – (5y / 3) = -2 (x / 3) + (y / 2) = (13 / 6) Sol: Given, (3x/2)-(5y/3) = -2 and (x/3) + (y/2) = 13/6 are the two equations. From 1st equation, we get, (3/2)x = -2 + (5y/3) ⇒ x = 2(-6+5y)/9 = (-12+10y)/9 ………………………(1) Putting the value of x in the given second equation to get, ((-12+10y)/9)/3 + y/2 = 13/6 ⇒y/2 = 13/6 –( (-12+10y)/27 ) + y/2 = 13/6
Now, substitute the value of y in equation (1), we get, (3x/2) – 5(3)/3 = -2 ⇒ (3x/2) – 5 = -2 ⇒ x = 2 Therefore, x = 2 and y = 3.
Q2: Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3. Sol: 2x + 3y = 11…………………………..(I) 2x – 4y = -24………………………… (II) From equation (II), we get x = (11-3y)/2 ………………….(III) Substituting the value of x in equation (II), we get 2(11-3y)/2 – 4y = 24 11 – 3y – 4y = -24 -7y = -35 y = 5……………………………………..(IV) Putting the value of y in equation (III), we get x = (11-3×5)/2 = -4/2 = -2 Hence, x = -2, y = 5 Also, y = mx + 3 5 = -2m +3 -2m = 2 m = -1 Therefore the value of m is -1.
Q3: Form the pair of linear equations for the following problems and find their solution by substitution method. (i) The difference between two numbers is 26 and one number is three times the other. Find them. Sol: Let the two numbers be x and y respectively, such that y > x. According to the question, y = 3x ……………… (1) y – x = 26 …………..(2) Substituting the value of (1) into (2), we get 3x – x = 26 x = 13 ……………. (3) Substituting (3) in (1), we get y = 39 Hence, the numbers are 13 and 39.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them. Sol: Let the larger angle by xo and smaller angle be yo. We know that the sum of two supplementary pair of angles is always 180o. According to the question, x + y = 180o……………. (1) x – y = 18o ……………..(2) From (1), we get x = 180o – y …………. (3) Substituting (3) in (2), we get 180o – y – y =18o 162o = 2y y = 81o ………….. (4) Using the value of y in (3), we get x = 180o – 81o = 99o Hence, the angles are 99o and 81o.
(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs.3800. Later, she buys 3 bats and 5 balls for Rs.1750. Find the cost of each bat and each ball. Sol: Let the cost a bat be x and cost of a ball be y. According to the question, 7x + 6y = 3800 ………………. (I) 3x + 5y = 1750 ………………. (II) From (I), we get y = (3800-7x)/6………………..(III) Substituting (III) in (II). we get, 3x+5(3800-7x)/6 =1750 ⇒ 3x+ 9500/3 – 35x/6 = 1750 ⇒ 3x- 35x/6 = 1750 – 9500/3 ⇒ (18x-35x)/6 = (5250 – 9500)/3 ⇒-17x/6 = -4250/3 ⇒-17x = -8500 x = 500 ……………………….. (IV) Substituting the value of x in (III), we get y = (3800-7 ×500)/6 = 300/6 = 50 Hence, the cost of a bat is Rs 500 and cost of a ball is Rs 50.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km? Sol: Let the fixed charge be Rs x and per km charge be Rs y. According to the question, x + 10y = 105 …………….. (1) x + 15y = 155 …………….. (2) From (1), we get x = 105 – 10y ………………. (3) Substituting the value of x in (2), we get 105 – 10y + 15y = 155 5y = 50 y = 10 …………….. (4) Putting the value of y in (3), we get x = 105 – 10 × 10 = 5 Hence, fixed charge is Rs 5 and per km charge = Rs 10 Charge for 25 km = x + 25y = 5 + 250 = Rs 255
(v) A fraction becomes 9/11 , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction. Sol: Let the fraction be x/y. According to the question, (x+2) /(y+2) = 9/11 11x + 22 = 9y + 18 11x – 9y = -4 …………….. (1) (x+3) /(y+3) = 5/6 6x + 18 = 5y +15 6x – 5y = -3 ………………. (2) From (1), we get x = (-4+9y)/11 …………….. (3) Substituting the value of x in (2), we get 6(-4+9y)/11 -5y = -3 -24 + 54y – 55y = -33 -y = -9 y = 9 ………………… (4) Substituting the value of y in (3), we get x = (-4+9×9 )/11 = 7 Hence the fraction is 7/9.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages? Solutions: Let the age of Jacob and his son be x and y respectively. According to the question, (x + 5) = 3(y + 5) x – 3y = 10 …………………………………….. (1) (x – 5) = 7(y – 5) x – 7y = -30 ………………………………………. (2) From (1), we get x = 3y + 10 ……………………. (3) Substituting the value of x in (2), we get 3y + 10 – 7y = -30 -4y = -40 y = 10 ………………… (4) Substituting the value of y in (3), we get x = 3 x 10 + 10 = 40 Hence, the present age of Jacob’s and his son is 40 years and 10 years respectively.
Exercise 3.3
Q1. Solve the following pair of linear equations by the elimination method and the substitution method: (i) x + y = 5 and 2x – 3y = 4 (ii) 3x + 4y = 10 and 2x – 2y = 2 (iii) 3x – 5y – 4 = 0 and 9x = 2y + 7 (iv) x / 2 + 2y / 3 = -1 and x – y / 3 = 3 Sol: (i) x + y = 5 and 2x – 3y = 4 By the method of elimination: x + y = 5 ……………………(i) 2x – 3y = 4 …………………(ii) When the equation (i) is multiplied by 2, we get 2x + 2y = 10 ………………(iii) When the equation (ii) is subtracted from (iii) we get, 5y = 6 y = 6 / 5 ………………(iv) Substituting the value of y in eq. (i) we get, x = 5 − 6 / 5 = 19 / 5 ∴ x = 19 / 5, y = 6 / 5 By the method of substitution: From the equation (i), we get: x = 5 – y………………(v) When the value is put in equation (ii) we get, 2(5 – y) – 3y = 4 -5y = -6 y = 6 / 5 When the values are substituted in equation (v), we get: x = 5 − 6 / 5 = 19 / 5 ∴ x = 19 / 5 ,y = 6 / 5
(ii) 3x + 4y = 10 and 2x – 2y = 2 By the method of elimination: 3x + 4y = 10………………(i) 2x – 2y = 2 …………………(ii) When the equation (i) and (ii) is multiplied by 2, we get: 4x – 4y = 4 …………………(iii) When the Equation (i) and (iii) are added, we get: 7x = 14 x = 2 ………………….(iv) Substituting equation (iv) in (i) we get, 6 + 4y = 10 4y = 4 y = 1 Hence, x = 2 and y = 1 By the method of Substitution: From equation (ii) we get, x = 1 + y……………………(v) Substituting equation (v) in equation (i) we get, 3(1 + y) + 4y = 10 7y = 7 y = 1 When y = 1 is substituted in equation (v) we get, A = 1 + 1 = 2 Therefore, A = 2 and B = 1
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7 By the method of elimination: 3x – 5y – 4 = 0 ………………………(i) 9x = 2y + 7 9x – 2y – 7 = 0 ………………………(ii) When the equation (i) and (iii) is multiplied we get, 9x – 15y – 12 = 0 ……………………(iii) When the equation (iii) is subtracted from equation (ii) we get, 13y = -5 y = -5 / 13 …………………………(iv) When equation (iv) is substituted in equation (i) we get, 3x + 25 / 13 − 4 = 0 3x = 27 / 13 x = 9 / 13 ∴ x = 9 / 13 and y = -5 / 13 By the method of Substitution: From the equation (i) we get, x = (5y + 4) / 3 …………………………(v) Putting the value (v) in equation (ii) we get, 9(5y + 4) / 3 − 2y − 7 = 0 13y = -5 y = -5 / 13 Substituting this value in equation (v) we get, x = (5(-5 / 13) + 4) / 3 x = 9 / 13 ∴ x = 9 / 13, y = -5 / 13
(iv) x / 2 + 2y / 3 = -1 and x – y / 3 = 3 By the method of Elimination: 3x + 4y = -6 …………………………. (i) x – y / 3 = 3 3x – y = 9 ……………………………. (ii) When the equation (ii) is subtracted from equation (i) we get, 5y = -15 y = – 3 ……………………….(iii) When the equation (iii) is substituted in (i) we get, 3x – 12 = -6 3x = 6 x = 2 Hence, x = 2 , y = -3 By the method of Substitution: From the equation (ii) we get, x = (y + 9) / 3…………………………(v) Putting the value obtained from equation (v) in equation (i) we get, 3(y + 9) / 3 + 4y = −6 5y = -15 y = -3 When y = -3 is substituted in equation (v) we get, x = (-3 + 9) / 3 = 2 Therefore, x = 2 and y = -3
Q.2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method: (i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes if we only add 1 to the denominator. What is the fraction? Sol: Let the fraction be a / b According to the given information, (a + 1) / (b – 1) = 1 => a – b = -2 ……………..(i) a / (b + 1) = 1 / 2 => 2a – b = 1……………………(ii) When equation (i) is subtracted from equation (ii) we get, a = 3 ………………….(iii) When a = 3 is substituted in equation (i) we get, 3 – b = -2 -b = -5 b = 5 Hence, the fraction is 3 / 5.
(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu? Sol: Let us assume, present age of Nuri is x And present age of Sonu is y. According to the given condition, we can write as; x – 5 = 3(y – 5) x – 3y = -10………………………(1) Now, x + 10 = 2(y +10) x – 2y = 10…………………(2) Subtract eq. 1 from 2, to get, y = 20 …………………………(3) Substituting the value of y in eq.1, we get, x – 3.20 = -10 x – 60 = -10 x = 50 Therefore, Age of Nuri is 50 years Age of Sonu is 20 years.
(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number. Sol: Let the unit digit and tens digit of a number be x and y respectively. Then, Number (n) = 10B + A N after reversing order of the digits = 10A + B According to the given information, A + B = 9…………………….(i) 9(10B + A) = 2(10A + B) 88 B – 11 A = 0 -A + 8B = 0 ……………………(ii) Adding the equations (i) and (ii) we get, 9B = 9 B = 1……………………………(3) Substituting this value of B, in the equation (i) we get A= 8 Hence the number (N) is 10B + A = 10 x 1 + 8 = 18
(iv) Meena went to a bank to withdraw Rs.2000. She asked the cashier to give her Rs.50 and Rs.100 notes only. Meena got 25 notes in all. Find how many notes of Rs.50 and Rs.100 she received. Sol: Let the number of Rs.50 notes be A and the number of Rs.100 notes be B According to the given information, A + B = 25 …………………(i) 50A + 100B = 2000 ……………………(ii) When equation (i) is multiplied with (ii) we get, 50A + 50B = 1250 ……………………(iii) Subtracting the equation (iii) from the equation (ii) we get, 50B = 750 B = 15 Substituting in the equation (i) we get, A = 10 Hence, Meena has 10 notes of Rs.50 and 15 notes of Rs.100.
(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs.27 for a book kept for seven days, while Susy paid Rs.21 for the book she kept for five days. Find the fixed charge and the charge for each extra day. Sol: Let the fixed charge for the first three days be Rs.A and the charge for each day extra be Rs.B. According to the information given, A + 4B = 27 …………………… (i) A + 2B = 21 ………………………(ii) When equation (ii) is subtracted from equation (i) we get, 2B = 6 B = 3 …………………(iii) Substituting B = 3 in equation (i) we get, A + 12 = 27 A = 15 Hence, the fixed charge is Rs.15 And the Charge per day is Rs.3
Q1: The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.
Sol: Graphical method to find zeroes: Total number of zeroes in any polynomial equation = total number of times the curve intersects x-axis. (i) In the given graph, the number of zeroes of p(x) is 0 because the graph is parallel to x-axis does not cut it at any point. (ii) In the given graph, the number of zeroes of p(x) is 1 because the graph intersects the x-axis at only one point. (iii) In the given graph, the number of zeroes of p(x) is 3 because the graph intersects the x-axis at any three points. (iv) In the given graph, the number of zeroes of p(x) is 2 because the graph intersects the x-axis at two points. (v) In the given graph, the number of zeroes of p(x) is 4 because the graph intersects the x-axis at four points. (vi) In the given graph, the number of zeroes of p(x) is 3 because the graph intersects the x-axis at three points.
(Exercise 2.2)
Q1: Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. (i) x2–2x –8 Sol: ⇒x2– 4x+2x–8 = x(x–4)+2(x–4) = (x-4)(x+2) Therefore, zeroes of polynomial equation x2–2x–8 are (4, -2) Sum of zeroes = 4–2 = 2 = -(-2)/1 = -(Coefficient of x)/(Coefficient of x2) Product of zeroes = 4×(-2) = -8 =-(8)/1 = (Constant term)/(Coefficient of x2)
(ii) 4s2–4s+1 Sol: ⇒4s2–2s–2s+1 = 2s(2s–1)–1(2s-1) = (2s–1)(2s–1) Therefore, zeroes of polynomial equation 4s2–4s+1 are (1/2, 1/2) Sum of zeroes = (½)+(1/2) = 1 = -(-4)/4 = -(Coefficient of s)/(Coefficient of s2) Product of zeros = (1/2)×(1/2) = 1/4 = (Constant term)/(Coefficient of s2 )
(iii) 6x2–3–7x Sol: ⇒6x2–7x–3 = 6x2 – 9x + 2x – 3 = 3x(2x – 3) +1(2x – 3) = (3x+1)(2x-3) Therefore, zeroes of polynomial equation 6x2–3–7x are (-1/3, 3/2) Sum of zeroes = -(1/3)+(3/2) = (7/6) = -(Coefficient of x)/(Coefficient of x2) Product of zeroes = -(1/3)×(3/2) = -(1/2) = (Constant term) /(Coefficient of x2 )
(iv) 4u2+8u Sol: ⇒ 4u(u+2) Therefore, zeroes of polynomial equation 4u2 + 8u are (0, -2). Sum of zeroes = 0+(-2) = -2 = -(8/4) = = -(Coefficient of u)/(Coefficient of u2) Product of zeroes = 0×-2 = 0 = 0/4 = (Constant term)/(Coefficient of u2 )
(v) t2–15 ⇒ t2 = 15 or t = ± √15 Therefore, zeroes of polynomial equation t2 –15 are (√15, -√15) Sum of zeroes =√15 + (-√15) = 0 = -(0/1)= -(Coefficient of t) / (Coefficient of t2) Product of zeroes = √15 × (-√15) = -15 = -15/1 = (Constant term) / (Coefficient of t2 )
(vi) 3x2–x–4 ⇒ 3x2–4x+3x–4 = x(3x-4)+1(3x-4) = (3x – 4)(x + 1) Therefore, zeroes of polynomial equation3x2 – x – 4 are (4/3, -1) Sum of zeroes = (4/3)+(-1) = (1/3)= -(-1/3) = -(Coefficient of x) / (Coefficient of x2) Product of zeroes=(4/3)×(-1) = (-4/3) = (Constant term) /(Coefficient of x2 )
Q2: Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes, respectively. (i) 1/4 , -1 Sol: From the formulas of sum and product of zeroes, we know, Sum of zeroes = α+β Product of zeroes = α β Sum of zeroes = α+β = 1/4 Product of zeroes = α β = -1 ∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:- x2–(α+β)x +αβ = 0 x2–(1/4)x +(-1) = 0 4x2–x-4 = 0 Thus, 4x2–x–4 is the quadratic polynomial.
(ii) √2, 1/3 Sol: Sum of zeroes = α + β =√2 Product of zeroes = α β = 1/3 ∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:- x2–(α+β)x +αβ = 0 x2 –(√2)x + (1/3) = 0 3x2-3√2x+1 = 0 Thus, 3x2-3√2x+1 is the quadratic polynomial.
(iii) 0, √5 Sol: Given, Sum of zeroes = α+β = 0 Product of zeroes = α β = √5 ∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:- x2–(α+β)x +αβ = 0 x2–(0)x +√5= 0 Thus, x2+√5 is the quadratic polynomial.
(iv) 1, 1 Sol: Given, Sum of zeroes = α+β = 1 Product of zeroes = α β = 1 ∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:- x2–(α+β)x +αβ = 0 x2–x+1 = 0 Thus, x2–x+1 is the quadratic polynomial.
(v) -1/4, 1/4 Sol: Given, Sum of zeroes = α+β = -1/4 Product of zeroes = α β = 1/4 ∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:- x2–(α+β)x +αβ = 0 x2–(-1/4)x +(1/4) = 0 4x2+x+1 = 0 Thus, 4x2+x+1 is the quadratic polynomial.
(vi) 4, 1 Sol: Given, Sum of zeroes = α+β =4 Product of zeroes = αβ = 1 ∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:- x2–(α+β)x+αβ = 0 x2–4x+1 = 0 Thus, x2–4x+1 is the quadratic polynomial.
from the starting point of 2nd line. Therefore, the coordinates of this point P is (2, 25).
from the starting point of 8th line. Therefore, the coordinates of this point Q are (8, 20)
