Q1. Write the numerical co-efficient and degree of each term of: x2 − 3x² + 52 x³ − 5x⁴

Sol:

Q2. Write the coefficient of x² in each of the following:
(i) 9 – 12x + x³
(ii) ∏/6 x² – 3x + 4
(iii) √3x – 7

Sol:
(i) 9 – 12x + x³
Coefficient of x² =0
(ii) ∏/6 x² – 3x + 4
Coefficient of x² = ∏/6
(iii) √3x – 7
Coefficient of x² = 0

Q3. Factorise z²  – 4z –12

Sol: To factorize this expression, we need to find two numbers α and β such that α + β = -4 and αβ = -12

z² – 6z + 2z – 12

z(z – 6) + 2(z – 6)

(z + 2)(z – 6)

Q4. (a) For what value of k, the polynomial x² + (4 – k)x + 2 is divisible by x – 2?
(b) For what value of ‘m’ is x³ – 2mx² + 16 is divisible by (x + 2)?

Sol: (a) Here p(x) = x² + 4x – kx + 2
If p(x) is exactly divisible by x – 2, then p(2) = 0
i.e. (2)² + 4(2) – k(2) + 2 = 0
⇒ 4 + 8 – 2k + 2 = 0
⇒ 14 – 2k = 0
⇒ 2k = 14
⇒ k= 14/2 = 7
Thus, the required value of k is 7.

(b) Here, p(x) = x³ – 2mx² + 16
∴ p(–2) = (–2)³ –2(–2)²m + 16
⇒ –8 –8m + 16
⇒ –8m + 8
Since, p(x) is divisible by x + 2
∴ p(–2) = 0
or –8m + 8 = 0
⇒ m = 1

Q5. Factorize x² – x – 12.

Sol: We have  x² – x – 12
⇒ x² – 4x + 3x – 12
⇒ x(x – 4) + 3(x – 4)
⇒ (x – 4)(x + 3)
Thus, x² – x – 12 = (x – 4)(x + 3)

Q6.Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following cases:
(i) f(x) = 3x + 1, x = −1/3
(ii) f(x) = x² – 1, x = 1,−1

Sol: (i) f(x) = 3x + 1, x = −1/3

f(x) = 3x + 1
Substitute x = −1/3 in f(x)
f( −1/3) = 3(−1/3) + 1
= -1 + 1
= 0
Since, the result is 0, so x = −1/3 is the root of 3x + 1

(ii) f(x) = x² – 1, x = 1,−1

f(x) = x² – 1
Given that x = (1 , -1)
Substitute x = 1 in f(x)
f(1) = 1² – 1
= 1 – 1
= 0
Now, substitute x = (-1) in f(x)
f(-1) = (−1)² – 1
= 1 – 1
= 0
Since , the results when x = 1 and x = -1 are 0, so (1 , -1) are the roots of the polynomial f(x) = x² – 1

Q7. Check whether (x – 1) is a factor of the polynomial x³ – 27x² + 8x + 18.

Sol: Here, p(x) = x³ – 27x² + 8x + 18, (x – 1) will be a factor of p(x) only if (x – 1) divides p(x) leaving a remainder 0.
For  x – 1 = 0
⇒ x = 1
∴ p(1) = (1)³ – 27(1)² + 8(1) + 18
⇒ 1 – 27 + 8 + 18
⇒ 27 – 27
⇒ 0
Since, p(1) = 0
∴ (x – 1) is a factor of p(x).
Thus, (x – 1) is a factor of x³ – 27x² + 8x + 18.

Q8.Evaluate each of the following using identities:
(i) (2x – 1/x)²
(ii) (2x + y) (2x – y)

Sol:
(i) (2x – 1/x)²
[Use identity: (a – b)² = a² + b² – 2ab ]
(2x – 1/x)² = (2x)² + (1/x)² – 2 (2x)(1/x)
= 4x² + 1/x² – 4
(ii) (2x + y) (2x – y)
[Use identity: (a – b)(a + b) = a² – b² ]
(2x + y) (2x – y) = (2x )² – (y)²
= 4x² – y²

Q9. Find the value of k, if (x – k) is a factor of x⁶ – kx⁵ + x⁴ – kx³ + 3x – k + 4.

Sol: Here, p(x) = x⁶ – kx⁵ + x⁴ – kx³ + 3x – k + 4
If (x – k) is a factor of p(x), then p(k) = 0
i.e (k)⁶ – k(k⁵) + k⁴ – k(k³) + 3k – k + 4 = 0
⇒ k⁶ – k⁶ + k⁴ – k⁴ + 3k – k + 4 = 0
⇒ 2k + 4 = 0
⇒ 2k = – 4
⇒ k = (-4/2) = –2
Thus, the required value of k is –2.

Q10. Factorize: 9a² – 9b² + 6a + 1

Sol: 9a² – 9b² + 6a + 1
⇒ [9a² + 6a + 1] – 9b² 
⇒ [(3a)² + 2(3a)(1) + (1)2] – (3b)² 
⇒ (3a + 1)² – (3b)²
⇒ [(3a + 1) + 3b][(3a + 1) – 3b] {using x² – y² = (x – y)(x + y)}
⇒ (3a + 1 + 3b)(3a + 1 – 3b)

Q1. Find a rational number between 1 and 2.

Sol: Express 1 and 2 as rational numbers with the same denominator:
Now, the rational numbers between  are:Out of these, select any five:

Answer: The five rational numbers between 1 and 2 are:

Q2. Find two rational numbers between 0.1 and 0.3.

Sol:  Express $0.1$ and $0.3$ as rational numbers with the same denominator:$0.1=\frac{1}{10}, 0.3=\frac{3}{10}$

Now, the rational numbers between $\frac{1}{10}$ and $\frac{3}{10}$ can be written with a larger denominator to find numbers in between.
Let us express them with a denominator of $100$100:$0.1=\frac{10}{\mathrm{100 }},0.3=\frac{30}{\mathrm{100 }}\mathrm{ .}$

The rational numbers between $\frac{10}{100}$ and $\frac{30}{100}$ are:$\frac{11}{100},\frac{12}{100},\frac{13}{100},\dots ,\frac{29}{100}.$

any two:$\frac{12}{100},\frac{25}{100}.$

Q3. Express   in the form of a decimal.

We have, Now, dividing 25 by 8,
 

Since, the remainder is 0.
∴ The process of division terminates.

Q4. Express   as a rational number.

Multiplying (1) by 100, we have 100x = 100 x 0.3333…
⇒ 100x = 33.3333              …(2)
Subtracting (2) from (1), we have
100x – x = 33.3333… – 0.3333…
⇒ 99x = 33

Q5. Simplify: (4+ √3) (4 −√3)

∵ (a + b)(a – b) = a² – b²
(4 +√3) (4 −√3) = (4)² – ( √3)² = 16 – 3 = 13
Thus, (4 +√3) (4 −√3) = 13

Q6. Simplify: (√3 +√2)²

∵ (a + b)² = a² + 2ab + b²
(√3 +√2)² = (√3)² + √2 ( √3 ×2) + (√2)² = 3 + 2 √6 + 2 = 5 + 2 √6
Thus, (√3 +√2)² = 5 + 2√6

Q7. Rationalise the denominator of  

Multiply and divide the given number by √6 + √5

Q8. Find (64)-1/3

Q9. Find a rational number lying between 1/5  and  1/2.

Rational numbers between   1/2 and  1/5 are infinite. Some of them are  3/10 ,   4/10 ,   45/100 ,  35/100 .
Step-by-step explanation:
As per the question, We need to find drational numbers lying between   1/5  and  1/2  As we know,

• Rational Numbers are numbers that can be expressed in the form of p/q where q is not equal to zero.
• Now, we know that  1/5 = 0.2 and 1/2  = 0.5
• So, numbers between 0.2 and 0.5 are infinite. Some of them are 0.3,0.4,0.45,0.35 etc.
• And these may be written as  310 ,   410 ,   45100 ,  35100  etc.

Hence, Rational numbers between 15  and  12 are infinite. Some of them are 3/10 ,   4/10 ,   45/100 ,  35/100

Q10. Express  0.245  as a fraction in the simplest form.

We know that

$0.245=\frac{245}{1000}$

because 0.245 means 245 thousandths.

Now we simplify the fraction.

First, divide both numerator and denominator by 5:

Next, check if 49 and 200 have any common factor.
49 = 7 × 7
200 = 2 × 2 × 2 × 5 × 5

There is no common factor, so the fraction is already in simplest form.

∴ The simplest form of 0.245 is

$\frac{49}{200}$

Question 1. The marks obtained by 17 students in a mathematics test (out of 100) are given below:
48, 66, 68, 49, 91, 72, 64, 46, 90, 79, 76, 82, 65, 96, 100, 82, 100
Find the range of the data.
Solution: Lowest observation = 46  
Highest observation = 100
∴ Range = 100 –46 = 54

Question 2. The class-mark of the class 130 – 150 is:
Solution: Class-mark =  ((130+150)/2)
 = (280/2)
  = 140
 

Question 3. The median of the following numbers arranged in descending order is 25. Find the value of
x: 40, 38, 35, 2x + 10, 2x + 1, 15, 11, 8, 5
Solution: Number of observations = 9

∴ The median is the  ((n+1)/2)th  term i.e. ((9+1)/2) th or the 5^th term.
⇒ 2x + 1 = 25
⇒ x= (25-1)/2= 12

Question 4. If the mean of 6, x, 4, and 12 is 8, then find the value of x.
 Solution: 

 ∴        = 8 or 22 + x = 32
or  x = 32 – 22 = 10

Question 5. Find the range of the data 9, 7, 5, 7, 9, 9, 6, 18, 9 and 8.
Solution:  Highest data = 18
Lowest data = 5
⇒ Range = 18 – 5 = 13

Question 6. Calculate the median of : 152, 155, 160, 144, 145, 148, 147, 149, 150
Solution:  Writing the given data in ascending order, we get: 144, 145, 147, 148, 149,  150, 152, 155, 160
Here, n = 9

∴ Median =   term =  term = 5^th term

⇒ Median = 149

Question 7. What is the median of 70, 40, 50, 100, 75, 75, 65 and 95?
Solution:  In ascending order, the given data is:
40, 50, 65, 70, 75, 75, 95, 100
Here, n = 8 (Even number)

⇒ Median =   

Q1. What is the longest pole that can be put in a room of dimensions l = 10 cm, b = 10 cm and h = 5 cm?
Ans: The longest diagonal of a cuboid =

∴ The length  of the required pole (diagonal) = 

Q2. The total surface area of a cube is 96 cm². What is its volume?
Ans: Total surface area of the cube = 6l²

∴   

Thus, the volume of the cube = l³ = 4³ = 64 cm³ 

Q3. The radius of a sphere doubled. What percent of its volume is increased?
Ans: Original volume = (4/3)πr³

Increased volume = (4/3)π(2r)³ = 32/3πr³

Increase in volume =  32/3πr³ –  4/3πr³ = 28/3πr³

∴ Percent increase in volume = 

Q4. Write ‘True or False’ for the following statements: (i) A right circular cylinder just encloses a sphere of radius r as shown in the figure. The area of the sphere is equal to the curved surface area of the cylinder.
Ans: True.

∵ [Radius of the sphere] = [Radius of the cylinder] = r

∴ The diameter of the sphere = 2r
⇒ Height of the cylinder (h) = 2r
Now, the surface area of the sphere = 4πr² 
And curved surface area of the cylinder = 2πrh = 2πr (2r) = 4πr²

(ii) An edge of a cube measures ‘r’ cm. If the largest possible right circular cone is cut out of this cube, then the volume of the cone (in cm³) is 1/6πr³
Ans: False.
∵ Height of the cone = r cm
∴ The diameter of the base of the cone = r cm

⇒ Radius of the base of the cone = (r/2) cm

Now volume of the cone 

Q5. If the total surface area of a sphere is 154 cm². Find its total volume.
Ans: Let ‘r’ be the radius of the sphere
∴ Total S.A. = 4 π r² = 154 cm²

or        

Now,                 

Q6. If the radius of a sphere is 3r then what is its volume?
Ans: 

Q1.Find the area of a triangle whose sides are 5 cm, 6 cm, and 7 cm.

Sol:

s = 5 + 6 + 72 = 9 cm.

Heron’s formula for the area of the triangle:

Area = √s(s − 5)(s − 6)(s − 7) cm²

Area = √(9)(9 − 5)(9 − 6)(9 − 7) cm²

Area = √(9 × 4 × 3 × 2) cm²

Area = √216 cm²

Area = 14.7 cm²

Q2. Find the area of a triangle whose sides are 3 cm, 4 cm and 5 cm.

Sol:

s = 3 + 4 + 52 = 6 cm

Heron’s formula for the area of the triangle:

Area = √s(s − 3)(s − 4)(s − 5) cm²

Area = √(6)(6 − 3)(6 − 4)(6 − 5) = √(6 × 3 × 2 × 1)

Area = √36 cm²

The area of the triangle is 6 cm².

Q3.  Find the area of a triangle whose two sides are 18 cm and 10 cm and the perimeter is 42cm.

Sol: Assume that the third side of the triangle to be “x”.

Now, the three sides of the triangle are 18 cm, 10 cm, and “x” cm

It is given that the perimeter of the triangle = 42cm

So, x = 42 – (18 + 10) cm = 14 cm

∴ The semi perimeter of triangle = 42/2 = 21 cm

Using Heron’s formula,

Area of the triangle,

= √[s (s-a) (s-b) (s-c)]

= √[21(21 – 18) (21 – 10) (21 – 14)] cm²

= √[21 × 3 × 11 × 7] m²

= 21√11 cm²

Q4.If the perimeter of an equilateral triangle is 180 cm. Then its area will be ?

Solution. Given, Perimeter = 180 cm

3a = 180 (Equilateral triangle)

a = 60 cm

Semi-perimeter = 180/2 = 90 cm

Now as per Heron’s formula,

= √s (s-a)(s-b)(s-c)

In the case of an equilateral triangle, a = b = c = 60 cm

Substituting these values in the Heron’s formula, we get the area of the triangle as:

A = √[90(90 – 60)(90 – 60)(90 – 60)]

= √(90× 30 × 30 × 30)

A = 900√3 cm²

Q5.The sides of a triangle are 122 m, 22 m and 120 m respectively. The area of the triangle is ?

Sol: Given,

a = 122 m

b = 22 m

c = 120 m

Semi-perimeter, s = (122 + 22 + 120)/2 = 132 m

Using heron’s formula:

= √s (s-a)(s-b)(s-c)

= √[132(132 – 122)(132 – 22)(132 – 120)]

= √(132 × 10 × 110 × 12)

= 1320 sq.m

Q6. The sides of a triangle are in the ratio 12: 17: 25 and its perimeter is 540 cm. The area is:

Sol: The ratio of the sides is 12: 17: 25

Perimeter = 540 cm

Let the sides of the triangle be 12x, 17x and 25x.

Hence,

12x + 17x + 25x = 540 cm

54x = 540 cm

x = 10

Therefore,

a = 12x = 12 × 10 = 120

b = 17x = 17 × 10 = 170

c = 25x = 25 × 10 = 250

Semi-perimeter, s = 540/2 = 270 cm

Using Heron’s formula:$𝐴=\; \surd s\; (s-a)(s-b)(s-c)$

= √[270(270 – 120)(270 – 170)(270 – 250)]

= √(270 × 150 × 100 × 20)

= 9000 sq.cm

Q7. Find the area of an equilateral triangle having side length equal to √3/4 cm (using Heron’s formula)

Sol: Here, a = b = c = √3/4

Semiperimeter = (a + b + c)/2 = 3a/2 = 3√3/8 cm

Using Heron’s formula,$𝐴=\; \surd s\; (s-a)(s-b)(s-c)$

= √[(3√3/8) (3√3/8 – √3/4)(3√3/8 – √3/4)(3√3/8 – √3/4)]

= 3√3/64 sq.cm

Q8. A rhombus-shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?

Sol: Draw a rhombus-shaped field first with the vertices as ABCD. The diagonal AC divides the rhombus into two congruent triangles which are having equal areas. The diagram is as follows.

Consider the triangle BCD,

Its semi-perimeter = (48 + 30 + 30)/2 m = 54 m

Using Heron’s formula,

Area of the ΔBCD =

Area = √[s(s − a)(s − b)(s − c)]

= √[54(54 − 48)(54 − 30)(54 − 30)] m²

= √[54 × 6 × 24 × 24] m²

= 432 m²

∴ Area of field = 2 × area of the ΔBCD = (2 × 432) m² = 864 m²

Thus, the area of the grass field that each cow will be getting = (864/18) m² = 48 m²

Q9. The area of an isosceles triangle having a base 2 cm and the length of one of the equal sides 4 cm, is

Solution: Given that a = 2 cm, b= c = 4 cm 

s = (2 + 4 + 4)/2 = 10/2 = 5 cm

By using Heron’s formula, we get:

A =√[5(5 – 2)(5 – 4)(5 – 4)] = √[(5)(3)(1)(1)] = √15 cm².

Q10. The perimeter of an equilateral triangle is 60 m. The area is

Solution:

Given: Perimeter of an equilateral triangle = 60 m

3a = 60 m (As the perimeter of an equilateral triangle is 3a units)

a = 20 cm.

We know that area of equilateral triangle = (√3/4)a²square units

A = (√3/4)20²

A = (√3/4)(400) = 100√3 m².

Q1. In the figure, if ∠ ACB = 35°, then find the measure of ∠ OAB.

Ans: ∠AOB = 180° –  (∠1 + ∠2)
= 180° – (35° + 35°)                     [∵ AO = OB ∴ ∠1 = ∠2]
= 110°
∠ACB = (1/2)(∠AOB) = (1/2) (110°) = 55°

Q2. If points A, B and C are such that AB ⊥ BC and AB = 12 cm, BC = 16 cm. Find the radius of the circle passing through the points A, B and C.
Ans: ∵ AB ⊥ BC
∴ ∠B = 90°                  [Angle in a semicircle]
⇒ AC is a diameter 

AC = 20

Radius = 10cm

Q3. The angles subtended by a chord at any two points of a circle are equal. Write true or false for the above statement and justify your answer.
Ans: False. If two points lie in the same segment only, then the angles will be equal otherwise they are not equal.

Q4. Two chords of a circle of length 10 cm and 8 cm are at the distance 8.0 cm and 3.5 cm, respectively from the centre, state true and false for the above statement.
Ans: False. As the larger chord is at a smaller distance from the centre.

Q5. If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is 

Ans: 

Assuming that AB = 12 cm and BC = 16 cm

If BC and AB are in a circle, then AC will be the circle’s diameter.

[circular’s diameter forms a right angle to the circle]

Apply the Pythagorean theorem to the ABC right angle.

AC²=AB²+BC²⇒

AC²=(12)²+(16)²⇒

AC²=144+256⇒ AC²=400

⇒ AC²=√400=20 cm

[using the square root of a positive number since the diameter is always positive]

∴ Circle radius = 1/2(AC)=1/220=10 cm

As a result, the circle’s radius is 10 cm.

Q6. In Figure, if ∠ABC = 20º, then ∠AOC is equal to:  

Ans: Assumed: ABC = 20°…(1)

We are aware that an arc’s angle at the circle’s center is twice as large as its angle at the rest of the circle.

= ∠AOC = 2(20°) (From (1))

=  ∠AOC = 40°

∠AOC thus equals 40°

Q7. Two chords AB and AC of a circle with center O are on the opposite sides of OA. Then ∠OAB = ∠OAC. 

Write true or false for the above statement and justify your answer.

Ans:

Two chords are used: AB and AC.

Join us, OB and OC.

Specifically, triangle OAC and OAB

OA (common side) = OA

Triangles OAB and OAC are not congruent, therefore it is impossible to prove that any third side or angle is equivalent. OC = OB (radius of the circle).

<OAC and <OAB

The claim is untrue as a result.

Q8. ABCD is a cyclic quadrilateral such that ∠A = 90°, ∠B = 70°, ∠C = 95° and∠D = 105°.

Write true or false for the above statement and justify your answer.

Ans: Given that ABCD is a cyclic quadrilateral with angles of 90°, 70°, 95°, and 105°,

We are aware that the sum of the opposite angles in a cyclic quadrilateral is 180°.

It can be expressed mathematically as follows: ∠A +∠ C = 90° + 95° = 185° 180°

∠B +∠ D = 70° + 105° = 175° 

The sum of the opposing angles in this situation is not 180°.

It is therefore not a cyclic quadrilateral.

The claim is false as a result.

Q9. In Fig, if AOB is the diameter of the circle and AC = BC, then ∠CAB is equal to: 

Ans: We are aware that the circle’s diameter forms a right angle.

∠BCA = 90^∘ ……(i)

AC = BC, and (ii) <ABC=<CAB (the angle opposite to equal sides is equal)

In △ ABC, ∠CBA+∠ABC+∠BCA=180^∘ [Triangle’s angle-sum attribute]]
 ∠CAB+∠CAB+90^∘=180^∘
 2 ∠CAB=90^∘
 ∠CAB=45^∘. 

Q10.  Two chords AB and CD of a circle are each at distances 4 cm from the center. Then AB = CD.

Write true or false for the above statement and justify your answer. 

Ans: The right answer is True.

The above assertion is accurate since chords that are equally spaced from a circle’s center have equal lengths.

AB=CD 

Q11.  If A, B, C, and D are four points such that ∠BAC = 30° and ∠BDC = 60°, then D is the center of the circle through A, B and C. 

Write true or false for the above statement and justify your answer. 
Ans: False
Because there are numerous places D where ∠BDC=60°, none of which can be the center of the circle formed by points A, B, and C.

Question 1. In the adjoining figure, ABCD is a trapezium in which AB || DC. If ∠A = 35º and ∠B = 75º, then find ∠C and ∠D.
 Solution: AB || DC and AD is a transversal.
Therefore, ∠A + ∠D = 180° (co-interior angles on the same side of the transversal).
∴ ∠D = 180° – ∠A = 180° – 35° = 145°.
 

 Similarly, since AB || DC and BC is a transversal, ∠B + ∠C = 180° (co-interior angles).
∴ ∠C = 180° – ∠B = 180° – 75° = 105°.
Thus, ∠C = 105° and ∠D = 145°.

Question 2. Fill in the blanks:  
 (i) In a parallelogram, opposite angles are ____.(ii) A ____ of a parallelogram divides it into two congruent triangles. (iii) The sum of the angles of a quadrilateral is ____. (iv ) In a parallelogram, the opposite sides are parallel and _____.
 Solution: Ans: (i) equal
(ii) diagonal
(iii) 360°
(iv) equal

Question 3. One angle of a quadrilateral is 140° and other three angles are in the ratio of 3 : 3 : 2.
 Find the measure of the smallest angle of the quadrilateral.
 Solution: Remaining three angles = 360° – 140° = 220°.
Ratio of the three angles = 3 : 3 : 2.
Sum of ratio parts = 3 + 3 + 2 = 8.
∴ One part = 220° ÷ 8 = 27.5°.
∴ The smallest angle = 2 × 27.5° =

= 55°.
Hence, the smallest angle of the quadrilateral is 55°.

Q.1. In a ΔABC, BC = CA and ∠ A = 50°, which is longer BC or AB?
Sol. ∵ ∠ 1 = ∠ 2  [∵ AC = BC]
and ∠A = ∠1 = 50°

⇒ ∠ 2 = 50°

 ⇒ ∠1+ ∠2+ ∠3= 180 [ angle sum property ]
∴ ∠ C = 180º – [50° + 50°] = 80°
⇒ Since, ∠C > ∠A
⇒ AB > BC Thus, AB is greater. [The greatest side is opp. to the greatest angle.]

Q.2. In ΔABC, ∠ B = 60° and ∠ C = 63°. Name the greatest side.
Sol. In ΔABC, ∠ B = 60° and ∠ C = 63°

⇒ ∠A+ ∠B+ ∠C= 180 [ angle sum property ]

⇒  ∠ A = 180º – (60° + 63°) = 57°
⇒ The greatest side is opp. to the greatest angle, i.e., 63°
∴ Side AB is the greatest.

Q.3. In ΔABC, if BC = AB and ∠ B = 80°, then find the measure of∠ A.
Sol. BC = AB ⇒ ∠A = ∠C

⇒ ∠A+ ∠B+ ∠C= 180 [ angle sum property ]

∵ ∠B = 80°
∴ ∠A + ∠C = 180° – 80° = 100°
⇒ ∠A = ∠C = 50º

Q.4. Which of the following is not the criterion for congruence of triangles?

  (i) SAS (ii) SSA (iii) ASA (iv) RHS 

Ans: SSA is not the criterion for congruency.

Q.5. If two angles are (30 – a)º and (125 + 2a)º and they are supplement of each other. Find the value of ‘a’.
Sol. ∵ (30 – a)º and (125 + 2a)º are supplement to each other.
∴ (30 – a + 125 + 2a)º = 180º
⇒ a = 180º – 125º – 30º = 25º
⇒ Value of a = 25°

Q.6. Each of the equal angles of an isosceles triangle is 38°. What is the measure of the third angle?
Sol. Let the third angle = x
∴ x + 38° + 38° = 180°
⇒ x = 180° – 38° – 38° = 104º

Q.7. In an isosceles ΔABC, AB = AC and ∠ A= 80°. What is the measure of ∠ B?
Sol. We have AB = AC ⇒ ∠ B = ∠ C          [∵ Angles opposite to equal sides are equal]
∴ ∠A + ∠B + ∠C = 180°
⇒ ∠A + ∠B + ∠B = 180°
⇒ ∠A + 2∠B = 180°
⇒ 80° + 2∠B = 180°
⇒ 2∠B = 180°- 80°
= 100
⇒ ∠B= (100^o/2) = 50°

Q.8. Find the measure of each acute angle in a right angle isosceles triangle.
Sol. Let the measure of each of the equal acute angle of the Δ be x
∴ We have: x + x + 90° = 180°
⇒ x + x = 180° – 90° = 90°
⇒ x= (90^o/2)= 45°

Q.1. What is the measure of an angle whose measure is 32° less than its supplement?
Sol. Let the required angle be x
∴ x = (180°- x) – 32°
⇒ x = 74°

Q.2. If the supplement of an angle is 4 times its complement, find the angle.
Sol. Let the required angle be x
∴  (180°- x) = 4 (90° – x)
⇒ x = 60°

Q.3. An exterior angle of a Δ is 110° and its two opposite interior angles are equal. What is the measure of each angle?
Sol. Let each of the interior opposite angle be x
∴ x + x  = 110°
⇒ x = 55°

Q.4. In a rt. ΔABC, ∠A = 90° and AB = AC. What are the values of ∠B and ∠C?
Sol. ∵ AB = AC
⇒ ∠B = ∠C
Also, ∠A = 90°
⇒ ∠B + ∠C = 90°
⇒ ∠B = ∠C = (90^o/2) = 45°

Q.5. In the figure, what is the value of x?
Sol. ∵ ℓ || m and p is a transversal

∴ ∠1 + 70° = 180°       [co-interior angles]
⇒ ∠1 = 180° – 70° = 110°
Now, 2x = 110°     [vertically opposite angles]
⇒ x =(110°/2) = 55°

Q.6. If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio 2 : 3 then, what is the smaller angle?
Sol. ℓ || m and p is the transversal
∴ ‘a’ and ‘b’ are interior angles on the same side of the transversal p.

Let a = 2x and b = 3x
∴ a + b = 180°
⇒ 2x + 3x = 180°
⇒ 5x = 180°
⇒ x = (180/5)= 36°
∴ smaller angle = 2x = 2 x 36 = 72°

Q.7. In the figure, what is the measure of ∠ ABC?
Sol. ∠PAQ = ∠BAC     [vertically opp. angles]

Now, ext. ∠ACR= ∠ABC + ∠BAC = 105°
⇒ ∠ABC = 105° – 45° = 60°

Q.8. In the following figure AB || CD. Find the measure of ∠BOC.

Sol. Extending AB to intersect OC, we get the following figure.
ABF is a straight line
∴ ∠OBF = 180° – 165° = 15°

AB || CD ⇒ EF || CD
∴ ∠1 + 75° = 180°
⇒ ∠1 = 180° – 75° = 105°
⇒ ∠2 = 105°
Now, in Δ, ∠2 + 15° + ∠BOC = 180°
⇒ 105° + 15° + ∠BOC = 180°
⇒ ∠BOC = 180° – 105° – 15°
= 60°

Very Short Answer Type Questions: Lines & Angles

Q.1. What is the measure of an angle whose measure is 32° less than its supplement?
Sol. Let the required angle be x
∴ x = (180°- x) – 32°
⇒ x = 74°

Q.2. If the supplement of an angle is 4 times its complement, find the angle.
Sol. Let the required angle be x
∴  (180°- x) = 4 (90° – x)
⇒ x = 60°

Q.3. An exterior angle of a Δ is 110° and its two opposite interior angles are equal. What is the measure of each angle?
Sol. Let each of the interior opposite angle be x
∴ x + x  = 110°
⇒ x = 55°

Q.4. In a rt. ΔABC, ∠A = 90° and AB = AC. What are the values of ∠B and ∠C?
Sol. ∵ AB = AC
⇒ ∠B = ∠C
Also, ∠A = 90°
⇒ ∠B + ∠C = 90°
⇒ ∠B = ∠C = (90^o/2) = 45°

Q.5. In the figure, what is the value of x?
Sol. ∵ ℓ || m and p is a transversal

∴ ∠1 + 70° = 180°       [co-interior angles]
⇒ ∠1 = 180° – 70° = 110°
Now, 2x = 110°     [vertically opposite angles]
⇒ x =(110°/2) = 55°

Q.6. If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio 2 : 3 then, what is the smaller angle?
Sol. ℓ || m and p is the transversal
∴ ‘a’ and ‘b’ are interior angles on the same side of the transversal p.

Let a = 2x and b = 3x
∴ a + b = 180°
⇒ 2x + 3x = 180°
⇒ 5x = 180°
⇒ x = (180/5)= 36°
∴ smaller angle = 2x = 2 x 36 = 72°

Q.7. In the figure, what is the measure of ∠ ABC?
Sol. ∠PAQ = ∠BAC     [vertically opp. angles]

Now, ext. ∠ACR= ∠ABC + ∠BAC = 105°
⇒ ∠ABC = 105° – 45° = 60°

Q.8. In the following figure AB || CD. Find the measure of ∠BOC.

Sol. Extending AB to intersect OC, we get the following figure.
ABF is a straight line
∴ ∠OBF = 180° – 165° = 15°

AB || CD ⇒ EF || CD
∴ ∠1 + 75° = 180°
⇒ ∠1 = 180° – 75° = 105°
⇒ ∠2 = 105°
Now, in Δ, ∠2 + 15° + ∠BOC = 180°
⇒ 105° + 15° + ∠BOC = 180°
⇒ ∠BOC = 180° – 105° – 15°
= 60°

Question 1. What is the Euclid’s second axiom?
Solution: The 2nd axiom is: “If equals be added to equals, the wholes are equal.”

Question 2. What is a theorem?
Solution: A mathematical statement whose truth has been logically established is called a theorem.

Question 3. What is the Euclid’s fifth postulate?
Solution: It states that: “If a straight line falling on two straight lines make the interior angles on the same side of it taken together less than two right angles, then the two straight lines, if produced indefinitely, meet on the side on which the sum of the angles is less than two right angles.”

Euclid’s fifth postulate

Question 4. Is the following statement correct? “The axioms are Universal truths in all branches of Mathematics.”
Solution: Yes.

Question 5. Complete the following statement: The things which are double of the same thing are ………………….. .

Solution: The things which are double of the same thing are equal.

Question 6. In how many chapters Euclid divided his famous treatise, “The Elements”?  
Solution: 13 Chapters.

Question 7. J, M and R are friends, J is of the same age as M. R is also of the same age as M.
Which Euclid’s axiom illustrates the relative age of J and R?
Solution: First Axiom.

Question 8. If a straight line falling on two straight lines makes the interior angles on the same side of it, whose sum is 120°, then on which side the two straight lines meet, if produced indefinitely?
Solution: They meet on the side on which the sum of angles is less than 120°.

Question 9. How many dimensions does a solid has?
Solution: A solid has 3 dimensions.

Question 10. How many dimensions does a surface has?
Solution: A surface has 2 dimensions.
 

Question 11. How many dimensions does a point has?
 Solution: A point has zero (0) dimensions.
 

Question 12. Which of the following are the boundaries of a surface: lines; curves; surfaces?
Solution: The boundaries of a surface are curves.

Question 13. Johan is of the same age as Mohan. Raghu is also of the same age as Mohan. State Euclid’s axiom  that illustrate the relative ages of Johan and Raghu.
Solution: The first axiom explains the above statement.

Question 14. Which of the following is the 5th postulate of Euclid?
(i) The whole is greater than the part.
(ii) If a straight line falling on two straight lines make the interior angles on the same side of it taken together less than two right angles then the two straight lines if produced indefinitely meet on that side on which the sum of angles is less than two right angles.
(iii)“all right angles are equal to one another.”
Solution: The option (ii) is the Euclid’s fifth postulate.

Question 15. Which of the following are assumed as axioms?
(i) Universal Truths in all branches of Mathematics.
(ii) Universal Truths specific to geometry.
(iii) Theorems
(iv) Definitions.
Solution: (i) The universal truths in all branches of Mathematics.

Question 16. Complete the following statement: Pythagoras was a student of …………………………….. .
Solution: Pythagoras was the student of ‘Thales’.

Question 17. What do we call a closed figure formed by three line segments
Solution: A triangle.

Question 18. What do we call a figure formed by two straight lines having a common point?
Solution: An angle.

Question 19. What is the minumum number of lines required to make a closed figure?
Solution: Three lines.

Question 20. How many straight lines can be drawn through two given points
Solution: Only one line.