RD Sharma Solutions for Class 10 Chapter 15 Areas Related To Circles Exercise 15.3

Another special part of a circular region is known as the segment of a circle. In this exercise, students will be tested with problems about the segment of a circle and its area. The RD Sharma Solutions Class 10 can be used by students to learn the correct method of solving problems explained in a simple language. In addition, students can also download the RD Sharma Solutions for Class 10 Maths Chapter 15 Areas Related To Circles Exercise 15.3 PDF provided below.

RD Sharma Solutions for Class 10 Chapter 15 Areas Related To Circles Exercise 15.3 Download PDF

Access RD Sharma Solutions for Class 10 Chapter 15 Areas Related To Circles Exercise 15.3

1. AB is a chord of a circle with centre O and radius 4 cm. AB is of length 4 cm and divides the circle into two segments. Find the area of the minor segment. 

Solution:

Given,

Radius of the circle with centre ‘O’, r = 4 cm = OA = OB

Length of the chord AB = 4 cm

So, OAB is an equilateral triangle and angle AOB = 60°

Thus, the angle subtended at centre θ = 60°

Area of the minor segment = (Area of sector) – (Area of triangle AOB)

= (8π/3 – 4√3) = 8.37 – 6.92 = 1.45 cm²

Therefore, the required area of the segment is (8π/3 – 4√3) cm² 

2. A chord PQ of length 12 cm subtends an angle 120^o at the centre of a circle. Find the area of the minor segment cut off by the chord PQ. 

Solution:

Given, ∠POQ = 120^o and PQ = 12 cm 

Draw OV ⊥ PQ,

PV = PQ × (0.5) = 12 × 0.5 = 6 cm

Since, ∠POV = 120^o 

∠POV = ∠QOV = 60^o

In triangle OPQ, we have 

sin θ = PV/ OA

sin 60^o = 6/ OA

√3/2 = 6/ OA

OA = 12/ √3 = 4√3 = r

Now, using the above we shall find the area of the minor segment

We know that,

Area of the segment = area of sector OPUQO – area of △OPQ

= θ/360 x πr² – ½ x PQ x OV

= 120/360 x π(4√3)² – ½ x 12 x 2√3

= 16π – 12√3 = 4(4π – 3√3)

Therefore, the area of the minor segment = 4(4π – 3√3) cm²

3. A chord of a circle of radius 14 cm makes a right angle at the centre. Find the areas of the minor and major segments of the circle. 

Solution:

Given,

Radius (r) = 14 cm

Angle subtended by the chord with the centre of the circle, θ = 90°

Area of minor segment = θ/360 x πr² – ½ x r² sin θ

= 90/360 x π(14)² – ½ x 14² sin (90)

= ¼ x (22/7) (14)² – 7 x 14

= 56 cm²

Area of circle = πr²

= 22/7 x (14)² = 616 cm²

Thus,

The area of the major segment = Area of circle – Area of the minor segment

= 616 – 56

= 560 cm²

4. A chord 10 cm long is drawn in a circle whose radius is 5√2 cm. Find the area of both the segments.

Solution:

Given,

Radius of the circle, r = 5√2 cm = OA = OB

Length of the chord AB = 10 cm

In triangle OAB,

We see that the Pythagoras theorem is satisfied.

So, OAB is a right angle triangle.

Angle subtended by the chord with the centre of the circle, θ = 90°

Area of minor segment = area of sector – area of triangle

= θ/360 × πr² – ½ x r² sin θ

= 90/360 × (3.14) 5√2² – ½ x (5√2)² sin 90

= [¼ x 3.14 x 25 x 2] – [½ x 25 x 2 x 1]

= 25(1.57 – 1)

= 14.25 cm²

Area of circle = πr² = 3.14 x (5√2)² = 3.14 x 50 = 14.25 cm²

Thus, Area of major segment = Area of circle – Area of minor segment

= 157 – 14.25

= 142.75 cm²

5. A chord AB of circle of radius 14 cm makes an angle of 60° at the centre of a circle. Find the area of the minor segment of the circle.

Solution:

Given,

Radius of the circle (r) = 14 cm = OA = OB

Angle subtended by the chord with the centre of the circle, θ = 60°

In triangle AOB, angle A = angle B [angle opposite to equal sides OA and OB = x]

By angle sum property, 

∠A + ∠B + ∠O = 180 

x + x + 60° = 180°

2x = 120°, x = 60°

All angles are 60° so the triangle OAB is an equilateral with OA = OB = AB

Area of the minor segment = area of sector – area of triangle OAB

= θ/360 × πr² – √3/4 (side)²

= 60/360 x (22/7) 14² – √3/4 (14)²

= [1/6 x 22 x 2 x 14] – √3 x (7)²

= 102.7 – 84.9 = 17.8

Therefore, the area of the minor segment = 17.80 cm²