Q1: Find the area of a triangle whose sides are 11 m, 60 m and 61 m.
Ans: Let a = 11 m, b = 60 m and c = 61 m : Firstly, to calculate semi perimeter (s) ∴ a + b + c2 = 11 + 60 + 612 = 1322 = 66 m Now, s − a = 66 − 11 = 55 m s − b = 66 − 60 = 6 m s − c = 66 − 61 = 5 m ∴ Area = √[s(s − a)(s − b)(s − c)] = √[66(55)(6)(5)] = √108900 = 330 m2
Q2: Suman has a piece of land, which is in the shape of a rhombus. She wants her two sons to work on the land and produce different crops. She divides the land in two equal parts by drawing a diagonal. If its perimeter is 400 m and one of the diagonals is of length 120 m, how much area each of them will get for his crops ?
Ans: Here, perimeter of the rhombus is 400 m. ∴ Side of the rhombus = 400/4 = 100 m Let diagonal BD = 120 m and this diagonal divides the rhombus ABCD into two equal parts.
∴ s = 100 + 120 + 1002 = 3202 = 160 ∴Area of ΔABD = √[s(s − a)(s − b)(s − c)] = √[160(160 − 100)(160 − 100)(160 − 120)] = √[160 × 60 × 60 × 40] = 80 × 60 = 4800 m2 Hence, area of land allotted to two sons for their crops is 4800 m2 each.
Q3: The perimeter of a triangular field is 144 m and its sides are in the ratio 3:4:5. Find the length of the perpendicular from the opposite vertex to the side whose length is 60 m.
Ans: Let the sides of the triangle be 3x, 4x and 5x ∴ The perimeter of the triangular field = 144 m ⇒ 3x + 4x + 5x = 144m ⇒ 12x = 144m
⇒ x = 14412 = 12 m Sides of the triangle are: (3 × 12 = 36) m, (4 × 12 = 48) m, (5 × 12 = 60) m ∴ s = a + b + c2 = 36 + 48 + 602 = 1442 = 72 m Area of ΔABC = √[s(s − a)(s − b)(s − c)] (using Heron’s formula) = √[72(72 − 36)(72 − 48)(72 − 60)] = √[72 × 36 × 24 × 12] = √746496 = 864 m2 ∴ Also, Area (ΔABC) = 12 × AD × BC (Using the formula for area) = 1 2 × AD × 60 = 30 x AD ∴ 30 x AD = 864 AD = 86430 = 28.8 m
Q4:Find the area of the triangle whose perimeter is 180 cm and two of its sides are of lengths 80 cm and 18 cm. Also, calculate the altitude of the triangle corresponding to the shortest side.
Ans:
Perimeter of given triangle = 180 cm Two sides are 18 cm and 80 cm ∴ Third side = 180 – 18 – 80 = 82 cm s = 1802 = 90 cm Area of triangle using Heron’s Formula: Area = √[s(s − a)(s − b)(s − c)] = √[90(90 − 18)(90 − 80)(90 − 82)] = √[90 × 72 × 10 × 8] = √518400 = 720 cm2 Use area formula to calculate height (h): 12 × base × height = Area 12 × 18 × h = 720 9 × h = 720 h = 7209 = 80 cm
Hence, area of triangle is 720 cm2and altitude of the triangle corresponding to the shortest side is 80 cm.
Q5: Calculate the area of the shaded region.
Ans:
Area of ΔAOB:
Area = 12 × OA × OB
= 12 × 12 × 5 = 30 cm2
Calculate AB:
AB2 = OA2 + OB2
= 122 + 52 = 144 + 25 = 169
AB = √169= 13 cm
Area of ΔABC: a = BC = 14 cm, b = CA = 15 cm, c = AB = 13 cm s = a + b + c2 = 14 + 15 + 132 = 422 = 21 cm
Using Heron’s Formula: Area = √[s(s − a)(s − b)(s − c)] = √[21(21 − 14)(21 − 15)(21 − 13)] = √[21 × 7 × 6 × 8] = √[3 × 7 × 7 × 2 × 3 × 2 × 2 × 2] = 2 × 2 × 3 × 7 = 84 cm2 Area of shaded region = Area of ∆ABC – Area of ∆AOB = 84 cm2– 30 cm2= 54 cm2
Q6: The sides of a triangular park are 8 m, 10 m and 6 m respectively. A small circular area of diameter 2 m is to be left out and the remaining area is to be used for growing roses. How much area is used for growing roses ? (use n = 3.14)
Ans: The sides of the triangular park are 8 m, 10 m and 6 m. ∴ s = a + b + c2 = 8 + 10 + 62 = 242 = 12 m Area of the park using Heron’s Formula: Area = √[s(s − a)(s − b)(s − c)] = √[12(12 − 8)(12 − 10)(12 − 6)] = √[12 × 4 × 2 × 6] = √[2 × 2 × 3 × 2 × 2 × 2 × 3] = 2 × 2 × 2 × 3 = 24 m2 Radius of the circle = 2/2 = 1 m Area of the circle = πr2 = 3.14 × 1 × 1 = 3.14 m2 ∴ Area to be used for growing roses = Area of the park – area of the circle ⇒ 24 – 3.14 = 20.86 m2
Q7: The Triangular garden has sides 40 m, 60 m, and 80 m. A landscaping company needs to install a pond in the center of the garden. How much area will the company need to clear for the pond?
Ans:
Given, The sides of the triangular garden are 40 m, 60 m, and 80 m.
Q8: The sides of a triangle are in the ratio of 12: 17: 25 and its perimeter is 540 cm. Find its area.
Ans:
Given, Ratio of the sides of the triangle is 12: 17: 25
Let the sides of triangle be 12x, 17x and 25x
Given, perimeter of the triangle = 540 cm
12x + 17x + 25x = 540 cm
⇒ 54x = 540cm
So, x = 10
Thus, the sides of the triangle are:
12 x 10 = 120 cm
17 x 10 = 170 cm
25 x 10 = 250 cm
Semi perimeter, s = 5402 = 270 cm
Using Heron’s formula,
Area of the triangle = √[s (s-a) (s-b) (s-c)]
= √[270 (270 – 120) (270 – 170) (270 – 250)]
= √(270 x 150 x 100 x 20) = 9000 cm2
Q9: Find the area of a triangle whose two sides are 14 cm and 12 cm, respectively, and the perimeter is 38 cm.
Ans: Given Two sides of the triangle are 14 cm and 12 cm, and the perimeter is 38 cm. Let the third side be c. The perimeter of the triangle is given as 38 cm: a + b + c = 38 14 + 12 + c = 38 c = 38 – 26 = 12 cm The semi-perimeter s is calculated as: s = 382 = 19 cm Using Heron’s formula: Area = √[s(s – a)(s – b)(s – c)] Substitute the values: Area = √[19(19 – 14)(19 – 12)(19 – 12)] Area = √[19 × 5 × 7 × 7] Area = √[4655] The area of the triangle is approximately 68.22 cm²
Q10: The sides of a quadrilateral, taken in order as 5, 12, 14, and 15 meters, respectively, and the angle contained by the first two sides is a right angle. Find its area.
Ans:
Here, AB = 5 m, BC = 12 m, CD =14 m and DA = 15 m
Join the diagonal AC.
Now, the area of △ABC = 1/2 ×AB×BC
= 1/2×5×12 = 30
The area of △ABC is 30 m2
In △ABC, (right triangle).
From Pythagoras theorem,
AC2 = AB2 + BC2
AC2 = 52 + 122
AC2 = 25 + 144 = 169
or AC = 13 m
Now in △ADC,
All sides are known, apply Heron’s Formula:
Area of triangle = √[s(s − a)(s − b)(s − c)] Semi-Perimeter, s = a + b + c2 Where: a, b, and c are the sides of the triangle. Perimeter of △ADC = 2s = AD + DC + AC 2s = 15 m +14 m +13 m s = 21 m Area of triangle ADC = √[21 × (21 − 13) × (21 − 14) × (21 − 15)] Area of triangle ADC = √[21 × 8 × 7 × 6] Area of △ADC = 84 m2 Area of quadrilateral ABCD = Area of △ABC + Area of △ADC = (30 + 84) m2 = 114 m2
Question 1. Look at the adjoining figure. If O is the center of the circle. PQ=12cm and ST = 3 cm, then find the radius of the circle when RS ⊥ PQ. Solution: Let us join O and P such that OP = r. ∵ RS ⊥ PQ
∴ T is the mid-point of PQ .⇒ PT = (1/2) PQ⇒ PT = (1/2) x 12 cm = 6 cm [∵ PQ = 12 cm (Given)] And ∠ OTP = 90° Also OS = r and TS = 3 cm ∴ OT = OS – TS = (r – 3) cm Now, in right ΔOTP, we have OP2 = PT2 + OT2 ⇒ r2 = 62 + (r – 3)2 ⇒ r2 = 36 + r2 + 9 – 6r ⇒ 6r = 45
⇒ r =(45/6) = (15/2) = 7.5 cm Thus, the radius of the circle is 7.5 cm.
Question 2. An equilateral triangle ABCABC is inscribed in a circle. Each side of the triangle is 99cm. Find the radius of the circle. Solution: Let ABC be an equilateral triangle such that AB = BC = AC = 9 cm (each) Let us draw a median AD corresponding to BC. ∴ BD =(1/2) BC ⇒ BD = (1/2) x 9 cm = (9/2)cm Also, AD ⊥ BC [∵ O is the centre of the circle] Now, in right ΔADB,
AD2 = AB2 – BD2
Since, in an equilateral triangle, the centroid and circumcentre coincide.
Q1: In the figure, P, Q and R are the mid-points of the sides BC, AC and AB of ΔABC. If BQ and PR intersect at X and CR and PQ intersect at Y, then show that XY = 1/4 BC.
Ans: R and Q are mid-points of AB and AC respectively. ∴ By the mid-point theorem, RQ || BC and RQ = 1/2 BC. Also P is the mid-point of BC, so BP = PC = 1/2 BC. Thus RQ ∥ BP and RQ = BP. Consider quadrilateral BPQR with vertices B-P-Q-R. Since one pair of opposite sides (BP and RQ) are equal and parallel, BPQR is a parallelogram. Therefore its diagonals PR and BQ bisect each other, so X (the intersection of PR and BQ) is the mid-point of PR. Similarly, in quadrilateral PCQR, the opposite sides PC and RQ are equal and parallel, so PCQR is a parallelogram. Hence its diagonals PQ and CR bisect each other, so Y (the intersection of PQ and CR) is the mid-point of PQ. Now, in ΔPQR, X and Y are mid-points of PR and PQ respectively. Therefore XY is parallel to RQ and XY = 1/2 · RQ (mid-point theorem applied in ΔPQR). But RQ = 1/2 BC, so XY = 1/2 · (1/2 BC) = 1/4 BC. Thus XY = 1/4 BC.
Q2: In ΔABC, AB = 8 cm, BC = 9 cm and AC = 10 cm. X, Y and Z are mid-points of AO, BO and CO respectively as shown in the figure. Find the lengths of the sides of ΔXYZ.
Ans: Given AB = 8 cm, BC = 9 cm and AC = 10 cm. In ΔAOB, X and Y are mid-points of AO and BO respectively. ∴ By the mid-point theorem, XY = 1/2 · AB = 1/2 × 8 cm = 4 cm. In ΔBOC, Y and Z are mid-points of BO and CO respectively. ∴ YZ = 1/2 · BC = 1/2 × 9 cm = 4.5 cm. In ΔCOA, Z and X are mid-points of CO and AO respectively. ∴ ZX = 1/2 · AC = 1/2 × 10 cm = 5 cm. Hence the sides of ΔXYZ are XY = 4 cm, YZ = 4.5 cm and ZX = 5 cm.
Q3: ABCD is a parallelogram. If the bisectors DP and CP of angles D and C meet at P on side AB, then show that P is the mid-point of side AB. Ans:
Ans: Let DP and CP bisect ∠D and ∠C respectively. So the two halves of ∠D are equal and the two halves of ∠C are equal. Since AB ∥ DC and CP is a transversal, an angle formed at P with AB is equal to the corresponding half of ∠C. Hence an angle at B in triangle BCP equals the corresponding half of ∠D. Therefore, in triangle BCP two angles are equal, which implies BC = BP. (angles opposite equal sides are equal and conversely) Similarly, since AB ∥ DC and DP is a transversal, an angle at A in triangle ADP equals the corresponding half of ∠D and so in triangle ADP two angles are equal; hence DA = AP. But in a parallelogram opposite sides are equal, so BC = AD. From BC = BP, DA = AP and BC = AD, we get BP = AP. Hence P is the mid-point of AB.
Q4: In the figure, ΔBCD is a trapezium in which AB || DC. E and F are the mid-points of AD and BC respectively. DF and AB are produced to meet at G. Also, AC and EF intersect at the point O. Show that : (i) EO || AB (ii) AO = CO
Ans: F is the mid-point of BC, so BF = FC. Lines DF and AB are produced to meet at G, so points D, F, G are collinear. Consider triangles BFG and CFD: BF = FC, ∠BFG = ∠CFD (vertical angles), and ∠BGF = ∠CDF (alternate interior angles, as AB ∥ DC). Thus ΔBFG ≅ ΔCFD by AAS, and so DF = FG. Since D, F, G are collinear and DF = FG, F is the mid-point of DG. Now, in ΔAGD, E is the mid-point of AD and F is the mid-point of DG. By the mid-point theorem, EF ∥ AG. But G lies on AB, so AG is a segment of AB; therefore EF ∥ AB and, in particular, EO (part of EF) ∥ AB. This proves (i). Because AB ∥ DC, EF ∥ AB implies EF ∥ DC as well. In ΔADC, E is the mid-point of AD and EO ∥ DC, so the line through the mid-point E parallel to DC bisects AC. Hence O is the mid-point of AC and AO = CO. This proves (ii).
Q5: PQRS is a square and ∠ABC = 90° as shown in the figure. If AP = BQ = CR, then prove that ∠BAC = 45°
Ans: In square PQRS, all sides are equal, so PQ = QR. (1) Given AP = BQ = CR. Subtracting the equal lengths BQ and CR from PQ and QR respectively gives: PQ – BQ = QR – CR ⇒ PB = QC. (2) In ΔAPB and ΔBQC: AP = BQ (given), ∠APB = ∠BQC = 90° (angles in a square), and PB = QC (from (2)). So ΔAPB ≅ ΔBQC by SAS congruence. Hence AB = BC. In ΔABC, if AB = BC, the base angles opposite those equal sides are equal; let each be x°. Then ∠B + ∠ACB + ∠BAC = 180° ⇒ 90° + x + x = 180° ⇒ 2x = 90° ⇒ x = 45°. Therefore ∠BAC = 45°.
Q6: In the given figure, AE = DE and BC || AD. Prove that the points A, B, C and D are concyclic. Also, prove that the diagonals of the quadrilateral ABCD are equal.
Ans: Since AE = DE ∠D = ∠A …. (i) [∵ ∠s opp. to equal sides of a Δ] Again, BC || AD ∠EBC = ∠A …. (ii) (corresponding ∠s] From (i) and (ii), we have ∠D = ∠EBC …. (iii) But ∠EBC + ∠ABC = 180° (a linear pair] ∠D + ∠ABC = 180° (using (iii)] Now, a pair of opposite angles of quadrilateral ABCD is supplementary Thus, ABCD is a cyclic quadrilateral i.e., A, B, C and D’are concyclic. In ΔABD and ΔDCA ∠ABD = ∠ACD [∠s in the same segment for cyclic quad. ABCD] ∠BAD = ∠CDA [using (i)] AD = AD (common] So, by using AAS congruence axiom, we have ΔABD ≅ ΔDCA Hence, BD = CA [c.p.c.t.]
Q1: In the given figure, AP and DP are bisectors of two adjacent angles A and D of quadrilateral ABCD. Prove that 2 ∠APD = ∠B + 2C. Ans: Here, AP and DP are angle bisectors of ∠A and ∠D ∴ ∠DAP = 1/2∠DAB and ∠ADP = 1/2∠ADC ……(i) In ∆APD, ∠APD + ∠DAP + ∠ADP = 180° ⇒ ∠APD + 1/2 ∠DAB + 1/2∠ADC = 180° ⇒ ∠APD = 180° – 1/2(∠DAB + ∠ADC) ⇒ 2∠APD = 360° – (∠DAB + ∠ADC) ……(ii) Also, ∠A + ∠B + C + ∠D = 360° ∠B + ∠C = 360° – (∠A + ∠D) ∠B + ∠C = 360° – (∠DAB + ∠ADC) ……(iii) From (ii) and (iii), we obtain 2∠APD = ∠B + ∠C
Q2: In right triangle ABC, right-angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see fig.). Show that : (i) ∆AMC ≅ ∆BMD (ii) ∠DBC = 90° (ii) ∆DBC ≅ ∆ACB (iv) CM = 1/2AB Ans: Given : ∆ACB in which 4C = 90° and M is the mid-point of AB. To Prove : (i) ∆AMC ≅ ∆BMD (ii) ∠DBC = 90° (iii) ∆DBC ≅ ∆ACB (iv) CM = 1/2AB Proof : Consider ∆AMC and ∆BMD, we have AM = BM [given] CM = DM [given – construction] ∠AMC = ∠BMD [vertically opposite angles] ∴ ∆AMC ≅ ∆BMD [by SAS congruence axiom] ⇒ AC = DB …(i) [by c.p.c.t.] and ∠1 = ∠2 [by c.p.c.t.] But ∠1 and ∠2 are alternate angles.
⇒ BD || CA Now, BD || CA and BC is transversal. ∴ ∠ACB + ∠CBD = 180° ⇒ 90° + CBD = 180° ⇒ ∠CBD = 90° In ∆DBC and ∆ACB, we have CB = BC [common] DB = AC [using (i)] ∠CBD = ∠BCA ∴ ∆DBC ≅ ∆ACB ⇒ DC = AB ⇒ 1/2AB = 1/2DC ⇒ 1/2AB = CM or CM = 1/2AB (∵ CM = 1/2DC)
Q3: Prove that two triangles are congruent if any two angles and the included side of one triangle is equal to any two angles and the included side of the other triangle. Ans:
Given : Two As ABC and DEF in which ∠B = ∠E, ∠C = ∠F and BC = EF To Prove : ∆ABC = ∆DEF Proof : We have three possibilities Case I. If AB = DE, we have AB = DE, ∠B = ∠E and BC = EF. So, by SAS congruence axiom, we have ∆ABC ≅ ∆DEF
Case II. If AB < ED, then take a point Mon ED such that EM = AB. Join MF. Now, in ∆ABC and ∆MEF, we have AB = ME, ∠B = ∠E and BC = EF. So, by SAS congruence axiom, we have ΔΑΒC ≅ ΔΜEF ⇒ ∠ACB = ∠MFE But ∠ACB = ∠DFE ∴ ∠MFE = ∠DFE
Which is possible only when FM coincides with B FD i.e., M coincides with D. Thus, AB = DE ∴ In ∆ABC and ∆DEF, we have AB = DE, ∠B = ∠E and BC = EF So, by SAS congruence axiom, we have ∆ABC ≅ ∆DEF Case III. When AB > ED Take a point M on ED produced such that EM = AB. Join MF Proceeding as in Case II, we can prove that ∆ABC = ∆DEF Hence, in all cases, we have ∆ABC = ∆DEF.
Q4: In the given figure, side QR is produced to the point S. If the bisectors of ∠PQR and ∠PRS meet at T, prove that ∠QTR = 1/2 ∠QPR.
In triangle QTR, ∠TRS is an exterior angle.
:. ∠QTR + ∠TQR = ∠TRS
∠QTR = ∠TRS – ∠TQR ——(1)
For triangle PQR, ∠PRS is an external angle.
:. ∠QPR + ∠PQR = ∠PRS
∠QPR + 2∠TQR = 2∠TRS (As QT and RT are angle bisectors)
∠QPR = 2(∠TRS – ∠TQR)
∠QPR = 2∠QTR [By using equation (1)]
LQTR= 1/2 ∠QPR
Q5: In figure, ABC is an isosceles triangle with AB = AC. D is a point in the interior of ∆ABC such that ∠BCD = ∠CBD. Prove that AD bisects ∠BAC of ∆ABC.
Ans: In ∆BDC, we have ∠DBC = ∠DCB (given). ⇒ CD = BD (sides opp. to equal ∠s of ∆DBC) Now, in ∆ABD and ∆ACD, we have AB = AC [given] BD = CD [proved above] AD = AD [common] ∴ By using SSS congruence axiom, we obtain ∆ABD ≅ ∆ACD ⇒ ∠BAD = ∠CAD [c.p.ç.t.] Hence, AD bisects ∠BAC of ∆ABC.
Q6: In figure, ABCD is a square and EF is parallel to diagonal BD and EM = FM. Prove that (i) DF = BE (ii) AM bisects ∠BAD.
Ans: (i) EF || BD = ∠1 = ∠2 and ∠3 = ∠4 [corresponding ∠s] Also, ∠2 = ∠4 ⇒ ∠1 = ∠3 ⇒ CE = CF (sides opp. to equals ∠s of a ∆] ∴ DF = BE [∵ BC – CE = CD – CF)
(ii) In ∆ADF and ∆ABE, we have AD = AB [sides of a square] DF = BE [proved above] ∠D = ∠B = 90° ⇒ ∆ADF ≅ ∆ABE [by SAS congruence axiom] ⇒ AF = AE and ∠5 = ∠6 … (i) [c.p.c.t.] In ∆AMF and ∆AME AF = AE [proved above] AM = AM [common] FM = EM (given) ∴ ∆AMF ≅ ∆AME [by SSS congruence axiom] ∴ ∠7 = ∠8 …(ii) [c.p.c.t.] Adding (i) and (ii), we have ∠5 + ∠7 = ∠6 + ∠8 ∠DAM = ∠BAM ∴ AM bisects ∠BAD.
Q1: In the adjoining figure, AB || CD. If ∠ APQ = 54° and ∠ PRD = 126°, then find x and y. Solution: ∵ AB || CD and PQ is a transversal, then interior alternate angles are equal.
⇒ ∠ APQ = ∠ PQR [alt. interior angles] ⇒ 54° = x [∵ ∠ APQ = 54° (Given)] Again, AB || CD and PR is a transversal, then ∠ APR = ∠ PRD [Interior alternate angles] But ∠ PRD = 126° [Given] ∴ ∠ APR = 126° Now, exterior∠ PRD + ∠ PRQ = 180° ⇒ 126°+∠ PRQ = 180° ⇒∠ PRQ= 180°-126° = 54°
Q2: In the adjoining figure AB || CD || EG, find the value of x. Solution: Let us draw FEG || AB || CD through E. Now, since FE || AB and BE are transversals, ∴ ∠ ABE + ∠ BEF = 180° [Interior opposite angles]
⇒ 127° + ∠ BEF = 180° [co int. angles] ⇒ ∠ BEF = 180° – 127° = 53° Again, EG || CD and CE is a transversal. ∴ ∠ DCE + ∠ CEG = 180° [Interior opposite angles] ⇒ 108° + ∠ CEG = 180° ⇒ ∠ CEG = 180° – 108° = 72° Since FEG is a straight line, then ⇒ ∠BEF + ∠BEC + ∠CEG = 180° [Sum of angles at a point on the same side of a line = 180°] ⇒ 53° + x + 72° = 180° ⇒ x = 180° – 53° – 72° = 55° Thus, the required measure of x = 55°.
Q3: AB and CD are parallel, and EF is a transversal. If ∠BEF = 70°, find ∠EFD and ∠CFE
Ans: Since AB || CD and EF is a transversal, by the Corresponding Angles Theorem:
∠BEF = ∠EFD
Thus, ∠EFD = 70°.
By the Linear Pair Axiom:
∠EFD + ∠CFE = 180°
70° + ∠CFE = 180°
∠CFE = 110°.
So, ∠EFD = 70° and ∠CFE = 110°.
Q4: If two parallel lines are intersected by a transversal, prove that the bisectors of the two pairs of interior angles enclose a rectangle.
Ans:
Two parallel lines AB and CD are intersected by a transversal L at P and R
respectively
PQ, RQ, RS and PS are bisectors of ∠APR, ∠PRC, ∠PRD and ∠BPR respectively.
Since AB || CD and L is a transversal
∠APR = ∠PRD ( alt. interior angles)
∠APQ = 1/2(∠PRD) = ∠QPR = ∠PRS
these are alternate interior angles.
QP || RS. Similarly QR || PS.
PQRS is a parallelogram.
Also ray PR stands on AB
∠APR + ∠BPR = 180° ( linear pair)
∠QPR + (1/2) ∠BPR = 90°
∠QPR + ∠SPR = 90°
∠QPS = 90°
Therefore PQRS is a parallelograrn, one of whose angle is 90°.
Hence PQRS satisfies all the properties of being a rectangle.
Long Question Answer: Introduction To Euclid’s Geometry
Q1: Which of the following statements are true and which are false? Give reasons for your answers. (i) Only one line can pass through a single point. Ans: False Correct statement: Infinite many lines can pass through a single point. This is self-evident and can be seen visually by the student given below:
(ii) There are infinite number of lines which pass through two distinct points. Ans: False The given statement contradicts the postulate I of the Euclid that assures that there is a unique line that passes through two distinct points.
Through two points P and Q a unique line can be drawn. (iii) A terminated line can be produced indefinitely on both the sides. Ans: True
We need to consider Euclid’s Postulate 2: “A terminated line can be produced indefinitely. (iv) If two circles are equal, then their radii are equal. Ans: True Let us consider two circles with same radii. We can conclude that, when we make the two circles overlap with each other, we will get a superimposed figure of the two circles. Therefore, we can conclude that the radii of both the circles will also coincide and still be same.
(v) In Fig. if AB=PQ and PQ=XY, then AB=XY
Ans: True We are given that AB=PQ and PQ=XY. We need to consider the axiom: “Given two distinct points, there is a unique line that passes through them.” Therefore, we can conclude that AB,PQ and XY are the lines with same dimensions, and hence if AB=PQ and PQ=XY, then AB=XY.
Q2: Give a definition for each of the following terms. Are there other terms that need to be defined first? What are they, and how might you define them? (i) Parallel lines Ans: Two lines are said to be parallel, when the perpendicular distance between these lines is always constant or we can say that the lines that never intersect each other are called as parallel lines.
We need to define line first, in order to define parallel lines.
(ii) Perpendicular lines Ans: Two lines are said to be perpendicular lines, when angle between these two lines is 90∘.
We need to define line and angle, in order to define perpendicular lines.
(iii) Line segment Ans: A line of a fixed dimension between two given points is called as a line segment.
We need to define line and point, in order to define a line segment.
(iv) Radius of a circle Ans: The distance of any point lying on the boundary of a circle from the center of the circle is called as radius of a circle.
We need to define circle and center of a circle, in order to define radius of a circle.
(v) Square Ans: A quadrilateral with all four sides equal and all four angles of 90∘ is called as a square.
We need to define quadrilateral and angle, in order to define a square.
Q3: If a point C lies between two points A and B such that AC = BC, then prove that AC = 1/2AB⋅ Explain by drawing the figure. Ans: We are given that a point C lies between two points B and C, such that AC = BC. We need to prove that AC = 1/2AB⋅ Let us consider the given below figure.
We are given that AC = BC−…(i) An axiom of the Euclid says that “If equals are added to equals, the wholes are equal.” Let us add AC to both sides of equation (i). AC + AC = BC + AC. An axiom of the Euclid says that “Things which coincide with one another are equal to one another.” “ We can conclude that BC+AC coincide with AB, or AB = BC + AC.…(ii) An axiom of the Euclid says that “Things which are equal to the same thing are equal to one another.” From equations (i) and (ii), we can conclude that AC + AC = AB, or 2AC = AB An axiom of the Euclid says that “Things which are halves of the same things are equal to one another.” Therefore, we can conclude that AC = 1/2AB
Q4: In the following figure, if AC = BD, then prove that AB = CD.
Ans: We are given that AC = BD We need to prove that AB = CD in the figure given below.
From the figure, we can conclude that AC = AB + BC, and BD = CD + BC. An axiom of the Euclid says that “Things which are equal to the same thing are equal to one another.” AB + BC = CD + BC An axiom of the Euclid says that “when equals are subtracted from equals, the remainders are also equal.” We need to subtract BC from equation(i) to get AB + BC − BC = CD + BC− BC
AB = CD
Therefore, we can conclude that the desired result is proved.
Q5: Why is axiom 5, in the list of Euclid’s axioms, considered as a ‘universal truth’? (Note that the question if not about fifth postulate) Ans: We need to prove that Euclid’s fifth axiom is considered as a universal truth. Euclid’s fifth axiom states that “the whole is greater than the part.” The above given axiom is a universal truth. We can apply the fifth axiom not only mathematically but also universally in daily life. Mathematical proof: Let us consider a quantity z, which has different parts as a,b,x and y So, z = a + b + x + y. Therefore, we can conclude that z will always be greater than its corresponding parts a,b,x and y. Universal proof: We know that Mumbai is located in Maharashtra and Maharashtra is located in India. In other words, we can conclude that Mumbai is a part of Maharashtra and Maharashtra is a part of India. Therefore, we can conclude that whole India will be greater than Mumbai or Maharashtra or both. Therefore, we can conclude that Euclid’s fifth axiom is considered as a ‘Universal truth’. Q6: How would you rewrite Euclid’s fifth postulate so that it would be easier to understand? Ans: We need to rewrite Euclid’s fifth postulate so that it is easier to understand. We know that Euclid’s fifth postulate states that “No intersection of lines will take place when the sum of the measures of the interior angles on the same side of the falling line is exactly 180∘. We know that Playfair’s axiom states that “For every line l and for every point P not lying on 1, there exists a unique line m passing through P and parallel to Γ. The above mentioned Playfair’s axiom is easier to understand in comparison to the Euclid’s fifth postulate. Let us consider a line l that passes through a point p and another line m. Let these lines be at a same plane. Let us consider the perpendicular CD on l and FE on m.
From the above figure, we can conclude that CD = EF. Therefore, we can conclude that the perpendicular distance between lines m and l will b. constant throughout, and the lines m and l will never meet each other or in other words, w can say that the lines m and I are equidistant from each other.
Q7: Does Euclid’s fifth postulate imply the existence of parallel lines? Explain. Ans: We need to verify whether Euclid’s fifth postulate imply the existence of parallel lines or not. The answer to the above statement is Yes. Let us consider two lines m and l In the figure given below, we can conclude that the lines m and l will intersect further.
From the figure, we can conclude that ∠1 + ∠2 < 180∘ and ∠3 +∠4 > 180∘ We know that Euclid’s fifth postulate states that “No intersection of lines will take place when the sum of the measures of the interior angles on the same side of the falling line is exactly 180∘. Let us consider lines l and m.
From the above figure, we can conclude that lines I and m will never intersect from either side. Therefore, we can conclude that the lines l and m are parallel.
Q8: Consider the two ‘postulates’ given below: (i) Given any two distinct points A and B, there exists a third point C, which is between A and B. (ii) There exists at least three points that are not on the same line. Do these postulates contain any undefined terms? Are these postulates consistent? Do they follow from Euclid’s postulates? Explain. Ans: Given any two distinct points A and B, there exists a third point C, which is between A and B. There exists at least three points that are not on the same line. The undefined terms in the given postulates are point and line. The two given postulates are consistent, as they do not refer to similar situations and they refer to two different situations. We can also conclude that, it is impossible to derive at any conclusion or any statement that contradicts any well-known axiom and postulate. The two given postulates do not follow from the postulates given by Euclid. The two given postulates can be observed following from the axiom, “Given two distinct points, there is a unique line that passes through them”
Q9: In the above question, point C is called a mid-point of line segment AB, prove that every line segment has one and only one mid-point. Ans: We need to prove that every line segment has one and only one mid-point. Let us consider the given below line segment AB and assume that C and D are the mid-points of the line segment AB.
If C is the mid-point of line segment AB, then AC = CB An axiom of the Euclid says that “If equals are added to equals, the wholes are equal.” AC + AC = CB + AC.(i) From the figure, we can conclude that CB + AC will coincide with AB. An axiom of the Euclid says that “Things which coincide with one another are equal to one another.” “ AC+AC = AB.(ii) An axiom of the Euclid says that “Things which are equal to the same thing are equal to one another.” “ Let us compare equations (i) and (ii), to get AC+AC = AB, or 2AC = AB.(iii) If D is the mid-point of line segment AB, then AD = DB An axiom of the Euclid says that “If equals are added to equals, the wholes are equal.” AD + AD = DB + AD. (iv) From the figure, we can conclude that DB+AD will coincide with AB. An axiom of the Euclid says that “Things which coincide with one another are equal to one another.” AD + AD = AB.(v) An axiom of the Euclid says that “Things which are equal to the same thing are equal to one another:” Let us compare equations (iv) and (v), to get AD + AD = AB, or 2AD = AB.(vi) An axiom of the Euclid says that “Things which are equal to the same thing are equal to one another.” Let us compare equations (iii) and (vi), to get 2AC = 2AD An axiom of the Euclid says that “Things which are halves of the same things are equal to one another.” AC = AD Therefore, we can conclude that the assumption that we made previously is false and a line segment has one and only one mid-point.
If a point C lies between two points A and B such that AB = BC, then prove that AC = ½ AB
Find the number of dimensions a solid, surface, and point have.
Let x + y = 10, and x = z. Show that y + z = 10
There are two sales employees who received equal incentives during the month of July. In August, each sales employee received double incentives in comparison to the month of July. Compare the employees’ incentives for the month of August.
Prove that an equilateral triangle can be formed on any of the given line segments.
Q1: Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k. Ans: The given equation is: 2x + 3y = k According to the question, x = 2 and y = 1. Now, Substituting the values of x and y in the equation 2x + 3y = k, We get, ⇒ (2 x 2)+ (3 × 1) = k ⇒ 4+3 = k ⇒ 7 = k ⇒ k = 7 The value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k, is 7.
Q2: Write two solutions for each of the following equations: (i) 2x + y = 7 Sol: To find the four solutions of 2x + y = 7 we substitute different values for x and y 1. Let x = 0 Then, 2x + y = 7 (2×0)+y = 7 y = 7 Therefore, one solution is = (0,7) 2. Let y = 1 Then, 2x + y = 7 2x+ 1 = 7 2x = 7 – 1 2x = 6 x = 3 Therefore, one solution is = (3, 1) (ii) πx + y = 9 Sol: To find the four solutions of πx + y = 9 we substitute different values for x and y 1. Let x = 0 Then, πx + y = 9 (π × 0)+y = 9 y = 9 Therefore, one solution is = (0,9) 2. Let y = 0 Then, πx + y = 9 πx +0 = 9 πx = 9 x =9/π Therefore, one solution is = (9/π,0)
Q3: The price of a notebook is twice the cost of a pen. Note a linear equation in two variables to illustrate this statement. (Taking the price of a notebook to be ₹ x and that of a pen to be ₹ y) Sol: Let the price of one notebook be = ₹ x Let the price of one pen be = ₹ y As per the question, The price of one notebook is twice the cost of one pen. i.e., the price of one notebook = 2×price of a pen x = 2×y x = 2y x-2y = 0 x-2y = 0 is the required linear equation in two variables to illustrate the statement, ‘The price of one given notebook is twice the cost of a pen.
Q4: Find out the value of k, if x = 2, y = 1 is a given solution of the equation 2x + 3y = k. Sol: The provided equation is 2x + 3y = k As per the given question, x = 2 and y = 1. Then, Replacing the values of x and y in the equation 2x + 3y = k, We get, ⇒ (2 x 2)+ (3 × 1) = k ⇒ 4+3 = k ⇒ 7 = k ⇒ k = 7 The required value of k, if x = 2, y = 1 is a given solution of the equation 2x + 3y = k, is 7.
Q5: Establish that the required points A (1, 2), B ( – 1, – 16) and C (0, – 7) lie on the graph of the required linear equation y = 9x – 7. Sol: We include the equation, y = 9x – 7 For A (1, 2), Replacing (x,y) = (1, 2), We obtain, 2 = 9(1) – 7 2 = 9 – 7 2 = 2 For B (–1, –16), Replacing (x,y) = (–1, –16), We get, –16 = 9(–1) – 7 -16 = – 9 – 7 -16 = – 16 For C (0, –7), Replacing (x, y) = (0, –7), We obtain, – 7 = 9(0) – 7 -7 = 0 – 7 -7 = – 7 Therefore, the points A (1, 2), B (–1, –16) and C (0, –7) satisfy the line y = 9x – 7. Therefore, A (1, 2), B (–1, –16) and C (0, –7) are answers to the linear equation y = 9x – 7 Thus, points A (1, 2), B (–1, –16), and C (0, –7) lie on the graph of the linear equation y = 9x – 7.
Q6: Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case: (i) x – y/5 – 10 = 0 (ii) -2x + 3y = 6 (iii) y – 2 = 0 Sol: (i) The equation x – y/5 – 10 = 0 can be written as: (1)x + (-1/5) y + (-10) = 0 Now compare the above equation with ax + by + c = 0 Thus, we get; a = 1 b = -1/5 c = -10 (ii) –2x + 3y = 6 Re-arranging the given equation, we get, –2x + 3y – 6 = 0 The equation –2x + 3y – 6 = 0 can be written as, (–2)x + 3y +(– 6) = 0 Now comparing (–2)x + 3y +(– 6) = 0 with ax + by + c = 0 We get, a = –2 b = 3 c = -6 (iii) y – 2 = 0 y – 2 = 0 The equation y – 2 = 0 can be written as, 0x + 1y + (–2) = 0 Now comparing 0x + 1y + (–2) = 0with ax + by + c = 0 We get, a = 0 b = 1 c = –2
Q7: Give the geometric representations of 2x + 9 = 0 as an equation (i) in one variable (ii) in two variables Sol: (i) 2x + 9 = 0 We have, 2x + 9 = 0 2x = – 9 x = -9/2 which is the required linear equation in one variable, that is, x only. Therefore, x= -9/2 is a unique solution on the number line as shown below: (ii) 2x +9=0 We can write 2x + 9 = 0 in the two variables as 2x + 0, y + 9 = 0 or x = −9−0.y/2 ∴ When y = 1, x = −9−0.(1)/2 = -9/2 y=2 , x = −9−0.(2)/2 = -9/2 y = 3, x = −9−0.(3)/2= -9/2 Therefore, we obtain the following table:
Now, plotting the ordered pairs (−9/2,3), (−9/2,3) and (−9/2,3) on graph paper and connecting them, we get a line PQ as the solution of 2x + 9 = 0.
Q8: Find the value of k for which x = 0, y = 8 is a solution of 3x – 6y = k. Sol: Since x = 0 and y = 8 is a solution of given equation 3x – 6y = k 3(0) – 6(8) = k ⇒ k = – 48
Q9: The cost of a table is 100 more than half the cost of a chair. Write this statement as a linear equation in two variables. Sol: Let the cost price of a table be ₹ x and that of a chair be ₹ y. Since the cost price of a table is 100 more than half the cost price of a chair. ∴ x = 1/2y + 100 ⇒ 2x = y + 200 or 2x – y – 200 = 0.
Q10: Give equation of two lines on same plane which are intersecting at the point (2, 3). Sol: Since there are infinite lines passing through the point (2, 3). Let, first equation is x + y = 5 and second equation is 2x + 3y = 13. Clearly, the lines represented by both equations intersect at the point (2, 3).
Q1: If the points A(x, y), B(3, 6) and C(-3, 4) are collinear, show that x – 3y + 15 = 0. Ans: Pts. A(x, y), B(3, 6), C(-3, 4) are collinear. ∴ Area of ∆ = 0 As area of ∆ = 12[ x 1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] ∴Area of ∆ABC = x(6 – 4) + 3(4 – y) + (-3) (y – 6) = 0 = 2x + 12 – 3y – 3y + 18 = 0 = 2x – 6y + 30 = 0 ∴ x – 3y + 15 = 0
Q2: If two vertices of an equilateral triangle are (3, 0) and (6, 0), find the third vertex. (2011D) Ans: Let A (3,0), B (6, 0), C (x, y). ∴ ∆ABC is an equilateral ∴ AB = BC = AC AB2 = BC2 = AC2 …[Squaring throughout (6 – 3)2 + (0 – 0)2 = (x – 6)2 + (1 – 0)2 = (x – 3)2 + (y – 0)2 9 = x2 – 12x + 36 + y2 = x2 – 6x + 9 + y2
Q3: If the vertices of a triangle are (1, -3), (4, p) and (-9, 7) and its area is 15 sq. units, find the value(s) of p. Ans: Area of ∆ = 15 sq. units 12 [1(p – 7) + 4(7 + 3)) + (-9)(-3 – p)] = ±15 p – 7 + 40 + 27 + 9p = ±30 10p + 60 = ± 30 10p = -60 ± 30 p = −60±30 / 10 ∴ Taking +ve sign, p = −60 + 30 / 10 = −30 / 10 = -3 Taking -ve sign, p = −60 − 30 / 10 = −90 / 10 = -9
Q4: If A(4, 2), B(7,6) and C(1, 4) are the vertices of a AABC and AD is its median, prove that the median AD divides AABC into two triangles of equal areas. Ans: Area of ∆ABD = 1/2 (x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 1/2[4(6 – 5) + 7(5 – 2) + 4(2 – 6)) = 1/2(4 + 21 – 16) = 9/2 sq.units …(i) Area of ∆ADC = 1/2 [4(5 – 4) + 4(4 – 2) + 162 – 5)] = 1/2(4 + 8 – 3) = 9/2sq.units From (i) and (ii), Area of ∆ABD = Area of ∆ADC ∴ Median AD divides ∆ABC into two triangles of equal area.
Q5: Prove that the area of a triangle with vertices (t, t – 2), (t + 2, t + 2) and (t + 3, t) is independent of t. Ans: Let A(t, t – 2), B(t + 2, + + 2), C(t + 3, t). Area of ∆ABC = 1/2[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 1/2 [t(t + 2 – t) + (t + 2)(t – (t – 2)) +(t + 3)((t – 2) – (t + 2))] = 1/2 [t(2) + (t + 2)(2) + (t + 3)(-4)] = 1/2 (2+ + 2+ + 4 – 46 – 12] = 1/2 [-8] = -4 Area of ∆ is always positive. ∴ Area of ∆ = 4 sq. units, which is independent of t.
Q6: If the area of triangle ABC formed by A(x, y), B(1, 2) and C(2, 1) is 6 square units, then prove that x + y = 15 or x + y = -9. Ans: Let A(x, y), B(1, 2), C(2, 1). Area of ∆ABC = 6 sq. units …[Given As 1/2 [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 6 ∴ x(2 – 1) + 1(1 – y) + 2(y – 2) = ±12 x + 1 – y + 2y – 4 = ±12 Taking +ve sign x + y = 12 + 4 – 1 ∴ x + y = 15 Taking -ve sign x + y = -12 + 4 – 1 ∴ x + y = -9
Q7: Point P(x, 4) lies on the line segment joining the points A(-5, 8) and B(4, -10). Find the ratio in which point P divides the line segment AB. Also find the value of x. Ans:
Q8: Find the ratio in which the point P(x, 2) divides the line segment joining the points A(12, 5) and B(4, -3). Also, find the value of x. Ans:
Q9: Find the area of the quadrilateral ABCD whose vertices are A(-3, -1), B(-2, -4), C(4, -1) and D(3, 4). Ans: Area of ∆ = 12[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] ∴ Area of ∆ABC = 1/2[(-3)(4 + 1) + (-2){-1 – (-1)} +4{-1 – (- 4}}] = 1/2[9+ 0 + 12] = 21 sq. units …(i) ∴ Area of ∆ACD = 1/2[-3(-1 – 4) + 4{(4 – (-1)} + 34{-1 – (-1)}] = 1/2 [15 + 20 + 0] = 1/2[15 + 20] = 35/2 sq. units ∴ ar(quad. ABCD) = ar (∆ABC) + ar(∆ACD) = 21/2+35/2 …[From (i) & (ii) = 56/2 = 28 sq. units
Q10: If the points A(1, -2), B(2, 3), C(-3, 2) and D(-4, -3) are the vertices of parallelogram ABCD, then taking AB as the base, find the height of this parallelogram. Ans: 1st method. A(1, -2), B(2, 3), D(-4, -3) .. Area of ∆ABD = 1/2 [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 1/2 [1(3 + 3) + 2(-3 + 2) + (-4)(-2 – 3)] = 1/2 [6 – 2 + 20) = 242 sq.units ar(∆ABD) = 12 sq.units = 1/2 × AB × DM = 12 sq.units …[Area of ∆ = 1/2 × Base × Altitude
Q11: If P(2, 4) is equidistant from Q(7, 0) and R(x, 9), find the values of x. Also find the distance P. Ans: PQ = PR …[GivenPQ2 = PR2 … [Squaring both sides ∴ (7 – 2)2 + (0 – 4)2 = (x – 2)2 + (9 – 4)2 ⇒ 25 + 16 = (x – 2)2 + 25 ⇒ 16 = (x – 2)2 ⇒ ±4 = x – 2 …[Taking sq. root of both sides ⇒ 2 ± 4 = x ⇒ x = 2 + 4 = 6 or x = 2 – 4 = -2
Q12: Find the value of k, if the points P(5, 4), Q(7, k) and R(9, – 2) are collinear. Ans: Given points are P(5, 4), Q(7, k) and R(9, -2). x1 (y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 0 …[∵ Points are collinear ∴ 5 (k + 2) + 7 (- 2 – 4) + 9 (4 – k) = 0 5k + 10 – 14 – 28 + 36 – 9k = 0 4 = 4k ∴ k=1
Q13: If (3, 3), (6, y), (x, 7) and (5, 6) are the vertices of a parallelogram taken in order, find the values of x and y. Ans: Let A (3, 3), B (6, y), C (x, 7) and D (5, 6).
Q14: A point P divides the line segment joining the points A(3, -5) and B(-4, 8) such that AP/PB=K/1. If P lies on the line x + y = 0, then find the value of K. Ans:
Q15: If the centroid of ∆ABC, in which A(a, b), B(b, c), C(c, a) is at the origin, then calculate the value of (a3 + b3 + c). Ans: Centroid = a + b + c = 0 If a + b + c = 0 then, as we know a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ac) ∴ a3 + b3 + c3 – 3abc = 0 … [Since a + b + c = 0 ∴ a3 + b3 + c3 = 3abc …(Hence proved)
Q16: Find the values of k so that the area of the triangle with vertices (k + 1, 1), (4, -3) and (7, -k) is 6 sq. units. Ans: Let A(k + 1, 1), B(4, -3) and C(7, -k). We have, Area of ∆ABC = 6 … [Given 6 = 1/2 [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] 6 = 1/2[(k + 1)(-3 + k) + 4(-k – 1) + 7(1 + 3)] 12 = (-3k + k2 – 3 + k – 4k – 4 + 28] 12 = [k2 – 6k + 21] ⇒ k2 – 6k + 21 – 12 = 0 ⇒ k2 – 6k + 9 = 0 ⇒ k2 – 3k – 3k + 9 = 0 ⇒ k(k – 3) – 3(k – 3) = 0 = ⇒ (k – 3) (k – 3) = 0 ⇒ k – 3 = 0 or k – 3 = 0 ⇒ k = 3 or k = 3 Solving to get k = 3.
Q17: For the triangle ABC formed by the points A(4, -6), B(3,-2) and C(5, 2), verify that median divides the triangle into two triangles of equal area. Ans: Let A(4, -6), B(3, -2) and C(5, 2) be the vertices of ∆ABC. Since AD is the median ∴ D is the mid-point of BC. ⇒ D(3 + 5 / 2,−2 + 2 / 2) ⇒ D(4,0) Area of ∆ABD = 1/2 [4(-2 – 0) + 3(0 + 6) + 4(-6 + 2)] = 1/2 [-8 + 18 – 16) = 1/2 [-6] = -3 But area of A cannot be negative. ∴ ar(∆ABD) = 3 sq.units …(i) Area of ∆ADC = 1/2 [4(0 – 2) + 4(2 + 6) + 5(-6 – 0)] = 1/2(-8 + 32 – 30] = 12 [-6] = -3 But area of ∆ cannot be negative. ∴ ar(∆ADC) = 3 sq.units From (i) and (ii), ∴ Median AD of AABC divides it into two ∆s of equal area.
Q18: Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are A(2, 1), B(4, 3) and C(2, 5). Ans:
Q19: Find the values of k for which the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear. Ans: A (k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k). When points are collinear, area of ∆ is 0. ∴ Area of triangle = 0 ⇒ [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 0 ⇒ [(k + 1) (2k + 3 – 5k) + 3k (5k – 2k) + (5k – 1) (2k – 2k – 3)] = 0 ⇒ [(k + 1) (3 – 3k) + 3k(3k) + (5k – 1)(-3)] = 0 ⇒ [3k – 3k2 + 3 – 3k + 9k2 – 15k + 3) = 0 ⇒ 6k2 – 15k + 6 = 0 ⇒ 2k2 – 5k + 2 = 0 ⇒ 2k2 – 4k – 1k + 2 = 0 ⇒ 2k(k – 2) – 1(k – 2) = 0 ⇒ (k – 2)(2k – 1) = 0 ⇒ k – 2 = 0 or 2k – 1 = 0 ⇒ k = 2 or k = 1/2 We get, k = 2, 1/2
Question 3. If the polynomials 2x3 + 3x2 – a and ax3 – 5x + 2 leave the same remainder when each is divided by x – 2, find the value of ‘a’ Solution: Let p(x) = 2x3 + 3x2 – a and f(x) = ax3 – 5x + 2
when p(x) is divided by x – 2 then remainder = p(2) since p(2) = 2(2)3 + 3(2)2 – a = 2(8) + 3(4) – a = 16 + 12 – a ∴ Remainder = 28 – a
when f(x) is divided by x – 2, then remainder = f(2) since, f(2) = a(2)3 – 5(2) + 2 = a(8) – 10 + 2 = 8a – 8 ∴ Remainder = 8a – 8
28 – a = 8a – 8 ⇒ 8a + a = 28 + 8 ⇒ 9a = 36 Thus , a = 4
Question 4. Find the values of ‘p’ and ‘q’, so that (x – 1) and (x + 2) are the factors of x3 + 10x2 + px + q. Solution: Here f(x) = x3 + 10x2 + px + q Since, x + 2 = 0 [∵ x + 2 is a factor of f(x)] ⇒ x= –2 If x + 2 is a factor f(x), then f(–2) = 0 i.e. (–2)3 + 10(–2)2 + p(–2) + q = 0 [∵Factor theorem] ⇒ –8 + 40 + (–2p) + q = 0 ⇒ 32 – 2p + q = 0 …(1) ⇒ 2p – q = 32 Also x – 1 = 0 ⇒ x = 1 If (x – 1) is a factor of f(x), then f(1) must be equal to 0. [∵Factor theorem] i.e. (1)3 + 10(1)2 + p(1) + q = 0 ⇒ 1 + 10 + p + q = 0 ⇒ 11 + p + q = 0 ⇒ p + q = –11 …(2) Now, by adding (1) and (2), we get
Now we put p = 7 in (2), we have 7 + q = –11 ⇒ q = –11 – 7 = –18 Thus, the required value of p and q are 7 and –18 respectively.
Question 5. If (x2 – 1) is a factor of the polynomial px4 + qx3 + rx2 + sx + t, then prove that p + r + t = q + s = 0. Solution: We have f(x) = px4 + qx3 + rx2 + sx + t Since, (x2 – 1) is a factor of f(x), [∵ x2 – 1 = (x + 1)(x – 1)] then (x + 1) and (x – 1) are also factors of f(x). ∴ By factor theorem, we have f(1) = 0 and f(–1) = 0 For f(1) = 0, p(1)4 + q(1)3 + r(1)2 + s(1) + t = 0 ⇒ p + q + r + s + t = 0 …(1) For f(–1) = 0, p(–1)4 + q(–1)3 + r(–1)2 + s(–1) + t = 0 ⇒ p – q + r – s + t = 0 …(2)
From (4) and (3), we get p + r + t = q + s = 0
Question 6. If x + y = 12 and xy = 27, find the value of x3 + y3. Solution: Since, (x + y)3 = x3 + y3 + 3xy (x + y) ∴ Substituting x + y = 12 and xy = 27, we have: (12)3 =x3 + y3 + 3 (27) (12) ⇒ x3 + y3 = 81(12) – 123 = [92 – 122] × 12 = [(9 + 12) (9 – 12)] × 12 = 21 × 3 × 12 = 756
Question 7. If a + b + c = 5 and ab + bc + ca = 10, Then prove that a3 + b3 + c3 – 3abc = –25. Solution: Since, a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca) ∴ a3 + b3 + c3 – 3 abc = (a + b + c) [(a2 + b2 + c2 + 2ab+ 2bc+ 2cb) − 3ab − 3bc − 3ca] = (a + b + c) [(a + b + c)2 − 3 (ab + bc + ca)] = 5 [ 52 − 3(10)] = 5[25 – 30] = 5[–5] = –25
Question 8. If a, b, c are all non-zero and a + b + c = 0, prove that Solution: Since, a + b + c = 0 ∴ a3 + b3 + c3 = 3abc ….. (1) Now, in = 3, we have
∴ T is the mid-point of PQ
Area of ∆ = 12[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
1st method. A(1, -2), B(2, 3), D(-4, -3)
PQ = PR …[GivenPQ2 = PR2 … [Squaring both sides
Centroid = 
= 3, we have
….. (2)
,