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, as they are explained in 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 Maths Chapter 15 Areas Related to Circles Exercise 15.3
Access RD Sharma Solutions for Class 10 Maths 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 the sector) – (Area of the 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 120o at the centre of a circle. Find the area of the minor segment cut off by the chord PQ.
Solution:
Given, ∠POQ = 120o and PQ = 12 cm
Draw OV ⊥ PQ,
PV = PQ × (0.5) = 12 × 0.5 = 6 cm
Since, ∠POV = 120o
∠POV = ∠QOV = 60o
In triangle OPQ, we have
sin θ = PV/ OA
sin 60o = 6/ OA
√3/2 = 6/ OA
OA = 12/ √3 = 4√3 = r
Now, using the above equations, 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
The angle subtended by the chord with the centre of the circle, θ = 90°
The 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 the 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 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 Pythagoras’ theorem is satisfied.
So, OAB is a right-angle triangle.
The angle subtended by the chord with the centre of the circle, θ = 90°
The area of minor segment = Area of the sector – Area of the 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 = 157 cm2
Thus, the Area of the major segment = Area of the circle – Area of the minor segment
= 157 – 14.25
= 142.75 cm2
5. A chord AB of a 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
The 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 the sector – Area of the triangle OAB
= θ/360 × πr2 – 1/2 r2 sin θ
= θ/360 × πr2 – 1/2 x (14)2 sin 60°
= 60/360 × (22/7) 142 – 1/2 x (14)2 x √3/2
= 60/360 x (22/7) 142 – √3/4 x (14)2
= 142 [(1/6) x (22/7)] – 0.4330
= 142 [(22/42)] – 0.4330
= 142 [(11-9.093)/21]
= 142 [0.09080]
= 17.79
Therefore, the area of the minor segment = 17.79 cm2
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