NCERT Solutions For Class 10 Maths Chapter 12 Areas Related to Circles

📝 NCERT Solutions For Class 10 Maths Chapter 12 Areas Related to Circles

EXERCISE 12.1

1. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.

Solution

`r_1 = 19` cm

`r_2 = 9`cm

ATQ

`C = C_1 + C_2`

`2piR = 2pir_1 + 2pir_2`

`⇒ 2piR = 2pi (r_1 + r_2)`

`⇒ R = r_1 + r_2`

`⇒ R = 19 + 9`

`⇒ R = 28 cm`

2. The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.

Solution

`r_1 = 8`cm

`r_2 = 6`cm

ATQ

Area of the circle `=` Area of  the `1^{st}` circle `+` Area of the second circle

`piR^2 = pir_1^2 + pir_2^2`

`⇒ piR^2 = pi (r_1^2 + r_2^2)`

`⇒ R^2 = r_1^2 + r_2^2`

`⇒ R^2 = 8^2 + 6^2` 

 `⇒  R^2 = 64+ 34`

`⇒  R^2  = sqrt 100`

`⇒  R  = 10` cm 

Work in progress 

EXERCISE 12.2

1. Find the area of a sector of a circle with a radius of 6 cm if the angle of the sector is 60°.

Solution: 

`r = 6 cm` 

`\theta` = 60°

Area of the sector  = `\theta/360 \pir^2`

`= 60/360 × 22/7 × 6×6`

=  `132/7 cm^2`

2. Find the area of a quadrant of a circle whose circumference is 22 cm.

Solution:

Circumference = 22 cm 

`\theta` = 90°

Circumference = 22 cm 

`Rightarrow  2\pi r = 22`

`Rightarrow  2 × 22/7 × r = 22`

`Rightarrow  44/7  r = 22`

`Rightarrow    r = 22 × 7/44`

`\therefore r = 7/2 cm`

Area of the sector  = `\theta/360 \pir^2`

`= 90/360 × 22/7 × 7/2 × 7/2 `

`=  77/8 cm^2`

3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Solution:

`r = 14 cm`

Time = 5 minutes 

1 minute = 6°

5 minutes = 30°

`\therefore theta` =  30°

Area swept by the minute hand = `\theta/360 pir^2`

= `30/360 × 22/7 × 14 × 14`

= `154/3 m^2`

4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (i) minor segment (ii) major sector. (Use `pi` = 3.14).

Solution:

`r = 10` cm

`theta` = 90°

(i) minor segment

Area of the minor sector OACB = `\theta/360 pir^2`

= `90/360`  ×  3.14 × 10 × 10

= `78.5 cm^2`

Area of the `\triangle`OAB = `1/2× b ×h` 

= `1/2` × 10 × 10 

= `50 cm^2`

Area of the minor segment ACB = Area of the minor sector OACB `-` Area of the `triangle`OAB

=  `(78.5  - 50) cm^2`

= `28.5 cm^2`

(ii) major sector

Area of the major sector OAPB = `pir^2 -` Area of minor sector OACB 

= (3.14 × `10^2`) `-` 78.5

=  314 `-` 78.5

= 235.5 `cm^2`

5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: (i) the length of the arc (ii) area of the sector formed by the arc (iii) area of the segment formed by the corresponding chord. 

Solution:

`r = 21 cm`

`theta` = 60°

(i) the length of the arc APB

The length of the arc = `theta/360 × 2pir`

= `60/360 × 2 × 22/7 × 21`

= 22 cm 

 (ii) area of the sector formed by the arc

Area of the sector OAPB =  `\theta/360 pir^2`

= `60/360 × 22/7 × 21 × 21`

= `231 cm^2`

(iii) area of the segment formed by the corresponding chord. 

In `triangle ODA`

`cos` 30° = `frac{OD}{OA}`

`Rightarrow  sqrt3/2 = frac{OD}{21}`

`therefore` OD = `frac{21sqrt3}{2}` cm

`sin` 30° = `frac{AD}{OA}`

`Rightarrow 1/2 = frac{AD}{21}`

`therefore` AD = `21/2`

AB = AD + BD

`therefore` AB = `21/2 + 21/2`

= 21 cm 

Area of the `triangle`OAB = `1/2 × b × h`

= `1/2 × 21× frac{21sqrt3}{2}`

=  `frac{441sqrt3}{4} cm^2`

Area of the segment APB = Area of the sector OAPB  `-` Area of the `triangle`OAB

= `(231 - frac{441sqrt3}{4}) cm^2`

Work in progress