## CBSE Class 10 Maths Notes Chapter 12 Areas related to Circles

- Introduction:
- This chapter is about calculating the areas of different figures that are related to circles.
- The different topics covered in this chapter are arc of a circle, sector of a circle, segment of a circle, length of an arc, area of a sector, and area of a segment.
- These concepts are used in various fields like engineering, architecture, and more.

- Arc of a Circle:
- An arc of a circle is a portion of the circumference of the circle.
- The length of an arc of a circle can be found using the formula L = (θ/360) × 2πr, where L is the length of the arc, r is the radius of the circle, and θ is the angle subtended by the arc at the center of the circle.

- Sector of a Circle:
- A sector of a circle is a region bounded by two radii and an arc.
- The area of a sector of a circle can be found using the formula A = (θ/360) × πr², where A is the area of the sector, r is the radius of the circle, and θ is the angle subtended by the sector at the center of the circle.

- Segment of a Circle:
- A segment of a circle is a region bounded by a chord and the arc that it intersects.
- The area of a segment of a circle can be found using the formula A = (θ/360) × πr² – (1/2) × r² × sin(θ), where A is the area of the segment, r is the radius of the circle, and θ is the angle subtended by the segment at the center of the circle.

- Length of an Arc:
- The length of an arc can be found using the formula L = (θ/360) × 2πr, where L is the length of the arc, r is the radius of the circle, and θ is the angle subtended by the arc at the center of the circle.

- Area of a Sector:
- The area of a sector can be found using the formula A = (θ/360) × πr², where A is the area of the sector, r is the radius of the circle, and θ is the angle subtended by the sector at the center of the circle.

- Area of a Segment:
- The area of a segment can be found using the formula A = (θ/360) × πr² – (1/2) × r² × sin(θ), where A is the area of the segment, r is the radius of the circle, and θ is the angle subtended by the segment at the center of the circle.

- Some important theorems:
- The angle subtended by an arc at the center of a circle is double the angle subtended by it at any point on the remaining part of the circle.
- The perpendicular from the center of a circle to a chord bisects the chord.
- The line joining the midpoint of a chord to the center of a circle is perpendicular to the chord.

- Summary:
- This chapter deals with the areas of different figures related to circles, such as arc, sector, and segment.
- The formulas for finding the length of an arc, area of a sector, and area of a segment have been discussed.
- Some important theorems related to circles have also been introduced in this chapter.

**CBSE Class 10 Maths Important Questions Chapter 12 Areas related to Circles**

- Find the area of the segment of a circle with radius 21 cm and central angle 120°.
- Find the area of the major and minor segments of a circle of radius 10.5 cm and the length of the corresponding arc is 63 cm.
- A chord of length 10 cm is drawn at a distance of 6 cm from the centre of a circle. Find the length of one of the arcs cut by the chord from the corresponding major sector.
- Find the area of a sector of a circle with radius 28 cm and central angle 150°.
- In the given figure, O is the centre of the circle with PQ as diameter. If PS ⊥ QR and ∠PSR = 60°, find the area of the shaded region.

- In the given figure, a circle is inscribed in a quadrant of radius 14 cm. Find the area of the shaded region.

- Two circles with radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
- A square is inscribed in a circle of radius 21 cm. Find the area of the square.
- A chord of a circle of radius 18 cm subtends an angle of 90° at the centre. Find the area of the corresponding minor segment.
- Find the area of the largest possible rectangle that can be inscribed in a circle of radius 8 cm.

**CBSE Class 10 Maths Important Questions Answers Chapter 12 Areas related to Circles**

- Find the area of the shaded region in the given figure.

Answer: The area of the shaded region can be found by subtracting the area of the smaller circle from the area of the larger circle.

Radius of larger circle = 14 cm Radius of smaller circle = 7 cm

Area of larger circle = πr^2 = 22/7 x 14 x 14 = 616 sq cm Area of smaller circle = πr^2 = 22/7 x 7 x 7 = 154 sq cm

Area of shaded region = Area of larger circle – Area of smaller circle = 616 – 154 = 462 sq cm

Hence, the area of the shaded region is 462 sq cm.

- Find the area of the segment of a circle with radius 21 cm and central angle 120°.

Answer: The area of the segment of a circle can be found using the formula A = (θ/360)πr^2 – (1/2)rsin(θ), where θ is the central angle, r is the radius of the circle and s is the length of the chord.

Given, radius of the circle = 21 cm and central angle = 120°.

Area of the segment = (120/360) x π x 21^2 – (1/2) x 21 x sin(120°) = (1/3) x 22/7 x 21 x 21 – (1/2) x 21 x √3/2 = 462 – 220.5 = 241.5 sq cm

Hence, the area of the segment of the circle is 241.5 sq cm.

- In the given figure, O is the centre of the circle and ∠BAC = 60°. Find the area of the shaded region.

Answer: We can find the area of the shaded region by subtracting the area of the triangle from the area of the sector.

Given, radius of the circle = 14 cm and ∠BAC = 60°.

Area of the sector ABO = (60/360) x π x 14^2 = 308/3 sq cm

Area of the triangle ABO = (1/2) x AB x OB x sin∠BAC = (1/2) x 14 x 14 x sin60° = 49√3 sq cm

Area of the shaded region = Area of the sector ABO – Area of the triangle ABO = 308/3 – 49√3 = (924 – 147√3)/3 sq cm

Hence, the area of the shaded region is (924 – 147√3)/3 sq cm.

- The perimeter of a sector of a circle of radius 10.5 cm is 31.4 cm. Find the area of the sector.

Answer: The perimeter of the sector of the circle can be found using the formula P = r + l, where r is the radius of the sector and l is the arc length.

Given, radius of the circle = 10.5 cm and the perimeter of the sector = 31.4 cm.

Perimeter of the sector = r + l 31.4 = 10.5 + l l = 20.9 cm

**CBSE Class 10 Maths Important Questions Answers MCQs Chapter 12 Areas related to Circles**

In a circle with radius 7 cm, the area of the sector is 38.5 sq cm. What is the measure of the central angle of the sector?

a) 50°

b) 55°

c) 60°

d) 65°

Answer: c) 60°

Explanation:

Given, radius of the circle = 7 cm and area of the sector = 38.5 sq cm.

We know that the area of a sector of a circle is given by the formula A = (θ/360)πr^2, where θ is the central angle and r is the radius of the circle.

Plugging in the given values, we have:

38.5 = (θ/360) x π x 7^2

θ/360 = 38.5 / (22/7) x 7 x 7

θ = 60°

Hence, the measure of the central angle of the sector is 60°.

In a circle with radius 6 cm, the length of the arc is 9 cm. What is the measure of the central angle of the arc?

a) 30°

b) 45°

c) 60°

d) 90°

Answer: c) 60°

Explanation:

Given, radius of the circle = 6 cm and length of the arc = 9 cm.

We know that the length of an arc of a circle is given by the formula l = (θ/360) x 2πr, where θ is the central angle and r is the radius of the circle.

Plugging in the given values, we have:

9 = (θ/360) x 2 x π x 6

θ/360 = 9 / (2 x π x 6)

θ = 60°

Hence, the measure of the central angle of the arc is 60°.

In a circle with radius 8 cm, the length of the arc is 12.8 cm. What is the area of the sector formed by the arc?

a) 32 sq cm

b) 36 sq cm

c) 48 sq cm

d) 64 sq cm

Answer: c) 48 sq cm

Explanation:

Given, radius of the circle = 8 cm and length of the arc = 12.8 cm.

We know that the length of an arc of a circle is given by the formula l = (θ/360) x 2πr, where θ is the central angle and r is the radius of the circle.

Plugging in the given values, we have:

12.8 = (θ/360) x 2 x π x 8

θ/360 = 12.8 / (2 x π x 8)

θ = 120°

Now, we can use the formula for the area of a sector of a circle, which is A = (θ/360)πr^2, to find the area of the sector.

Plugging in the values, we have:

A = (120/360) x π x 8^2

A = 48 sq cm

Hence, the area of the sector formed by the arc is 48 sq cm.

The area of a sector of a circle with radius 10 cm is 50 sq cm. What is the measure of the central angle of the sector?

a) 18°

b) 30°

c) 36°

d) 45°

Answer: b) 30°

Explanation:

Given, radius of the circle = 10 cm

## CBSE Class 10 Maths Notes Chapter 12 Areas related to Circles

Circumference of a circle = 2πr

Area of a circle = πr^{2} …[where r is the radius of a circle]

Area of a semi-circle = πr22

Area of a circular path or ring:

Let ‘R’ and ‘r’ he radii of two circles

Then area of shaded part = πR^{2} – πr^{2} = π(R^{2} – r^{2}) = π(R + r)(R – r)

**Minor arc and Major Arc:** An arc length is called a major arc if the arc length enclosed by the two radii is greater than a semi-circle.

If the arc subtends angle ‘θ’ at the centre, then the

Length of minor arc = θ360×2πr=θ180×πr

Length of major arc = (360−θ360)×2πr

**Sector of a Circle and its Area**

A region of a circle is enclosed by any two radii and the arc intercepted between two radii is called the sector of a circle.

(i) A sector is called a minor sector if the minor arc of the circle is part of its boundary.

OAB^ is minor sector.

Area of minor sector = θ360(πr2)

Perimeter of minor sector = 2r+θ360(2πr)

(ii) A sector is called a major sector if the major arc of the circle is part of its boundary.

OACB^ is major sector

Area of major sector = (360−θ360)(πr2)

Perimeter of major sector = 2r+(360−θ360)(2πr)

**Minor Segment:** The region enclosed by an arc and a chord is called a segment of the circle. The region enclosed by the chord PQ & minor arc PRQ is called the minor segment.

Area of Minor segment = Area of the corresponding sector – Area of the corresponding triangle

**Major Segment:** The region enclosed by the chord PQ & major arc PSQ is called the major segment.

Area of major segment = Area of a circle – Area of the minor segment

Area of major sector + Area of triangle

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