# RD Sharma Solutions Class 10 Areas Related To Circles Exercise 15.2

### RD Sharma Class 10 Solutions Chapter 15 Ex 15.2 PDF Free Download

#### Exercise 15.2

Q1.  Find in terms of π, the length of the arc that subtends an angle of 30 degrees, at the center of ‘O’ of the circle with a radius of 4 cm.

Soln:

Given Data :

Angle subtended at the centre ‘O’ =  30°

Formula to be used :

$Length of arc=\frac{\theta}{360}\ast 2\pi r cm$

$Length of arc=\frac{30}{360}\ast 2\pi\ast 4 cm$

$\frac{2\pi }{3} cm$

Therefore, the Length of arc the length of the arc that subtends an angle of 60 degrees is $\frac{2\pi }{3} cm$

Q2. Find the angle subtended at the centre of circle of radius 5 cm by an arc of length $\frac{ 5\pi }{ 3 } cm$.

Soln:

Given data:

Length of arc = $\frac{5\pi}{3} cm$

Formula to be used:

Length of arc =$\frac{\theta }{360}\ast 2\pi r cm$

$\frac {5\pi}{3} cm$ =$\frac{\theta }{360}\ast 2\pi r cm$

Solving the above equation, we have:

θ = 60°

Therefore, angle subtended at the centre of circle is60°

Q3. An arc of length cm subtends an angle of 144° at the center of the circle.

Soln:

Given Data : length of arc = cm

θ = angle subtended at the centre of circle = 144°

Formula to be used :

Length of arc = $\frac{\theta }{360}\ast 2\pi r cm$

$\frac{\theta }{360}\ast 2\pi r cm$ = $\frac{144 }{360}\ ast 2\pi r cm$

= $\frac {4\pi}{5}\ast r cm$

As given in the question, length of arc = cm ,

Therefore,   cm = $\frac{4\pi}{5}\ast r cm$

Solving the above equation, we have

r = 25 cm.

Therefore the radius of the circle is found to be 25 cm.

Q4.  An arc of length 25 cm subtends an angle of 55° at the center of a circle. Find in terms of radius of the circle.

Soln:

Given Data :

length of arc =25 cm

θ = angle subtended at the centre of circle = 55°

Formula to be used :

Length of arc = $\frac{\theta}{360}\ast2\pi r cm$

=$\frac{55}{36}\ast2\pi r cm$

As given in the question length of arc =25 cm ,hence,

25 cm =$\frac{55}{360}\ast2\pi\ast r cm$

25 = $\frac{11\pi r}{36}$

$radius = \frac{25\ast36}{11\ast\pi}$

$\frac{900}{11\pi }$

Therefore, the radius of the circle is $\frac{900}{11\pi }$

Q5. Find the angle subtended at the center of the circle of radius ‘a’ cm by an arc of $\ frac {a \ pi }{ 4 }$ length cm .

Soln:

Given data :

Length of arc = $\frac{a\pi }{4}$  cm

θ = angle subtended at the centre of circle

Formula to be used:

Length of arc = $\frac{\theta}{360}\ast2\pi r cm$

Length of arc = $\frac{\theta}{360}\ast2\pi a cm$

$\frac{\theta}{360}\ast 2\pi a cm$  = $\frac{a\pi }{4}$  cm

Solving the above equation, we have

θ = 45°

Therefore, the angle subtended at the centre of circle is 45°

Q6. A sector of the circle of radius 4 cm subtends an angle of 30°. Find the area of the sector.

Soln:

Given Data:

Angle subtended at the centre ‘O’ = 30°

Formula to be used :

$Area of the sector = \frac{\Theta }{360}\ast \pi r^{2}$

$Area of the sector = \frac{30}{ 360}\ast\pi 4^{2}$

Solving the above equation, we have:

Area of the sector = 4.9 cm2

Therefore, Area of the sector is found to be 4.9 cm2

Q7. A sector of a circle of radius 8 cm subtends an angle of 135°. Find the area of sector.

Soln:

Given Data:

Angle subtended at the centre ‘O’ = 135°

Formula to be used:

$Area of the sector = \frac{\Theta }{360}\ast \pi r^{2}$

$Area of the sector = \frac{ 135 }{ 360 }\ast\pi 8^{2}$

=$\frac{528}{ 7}$ cm2

Therefore, Area of the sector calculated is $\frac{ 528 }{ 7 }$ cm2

Q8. The area of sector of circle of radius 2 cm is  cm2. Find the angle subtended by the sector.

Soln:

Given Data:

Angle subtended at the centre ‘O’ =?

Area of sector of circle = cm2

Formula to be used:

$Area of the sector = \frac{\Theta }{360}\ast\pi r^{2}$

$Area of the sector = \frac{\Theta}{360}\ast\pi 3^{2}$

= $\frac{\pi\Theta }{90}$

As given in the question area of sector of circle = cm2

cm2   = $\frac{\pi\Theta }{90}$

Solving the above equation, we have

θ = 90°

Therefore, the angle subtended at the centre of circle is 90°

Q10. PQ is a chord of circle with centre ‘O’ and radius 4 cm. PQ is of the length 4 cm. Find the area of sector of the circle formed by chord PQ.

Soln:

Given Data: PQ is chord of length 4 cm.

Also, PO = QO= 4 cm

OPQ is an equilateral triangle.

Angle POQ = 60°

Area of sector ( formed by the chord (shaded region ) ) = ( area of sector )

Formula to be used:

$Area of the sector = \frac{\Theta }{360}\ast\pi r^{2}$

$Area of the sector = \frac{60 }{360}\ast\pi 4^{2}$

= $\frac{32\pi }{3}$

Therefore, Area of the sector is $\frac{32\pi }{ 3 }$ cm2

Q11. In a circle of radius 35 cm, an arc subtends an angle of 72° at the centre. Find the length of arc and area of sector.

Soln:

Given Data:

Angle subtended at the centre ‘O’ = 72°

Area of sector of circle =?

Formula to be used:

$Length of arc = \frac{\theta }{360}\ast 2\pi r cm$

$Length of arc = \frac{ 108 }{360}\ast 2\pi x 42 cm$

Solving the above equation we have,

Length of arc = 44 cm

We know that,

$Area of the sector = \frac{\Theta }{360}\ast \pi r^{2}$

$Area of the sector = \frac{ 72 }{360}\ast \pi 35^{2}$

Solving the above equation, we have, Area of the sector = ( 35 x 22 ) cm2

Therefore, Area of the sector is 770 cm2

Q12. The perimeter of a sector of a circle of radius 5.7 m is 27.2m. find the area of the sector.

Soln:

Given Data:

Radius = 5.7 cm = OA = OB [from the figure shown above]

Perimeter = 27.2 m

Let the angle subtended at the centre be θ

Perimeter = $\frac{\theta }{360}\ast 2\pi r cm$ + OA + OB

= $\frac{\theta }{360}\ast 2\pi x 5.7 cm$ + 5.7 + 5.7

Solving the above equation we have,

θ = 158.8°

We know that,

$Area of the sector = \frac{\Theta }{360}\ast\pi r^{2}$

Area of the sector = $\frac{ 158.8 }{ 360 }\ast\pi 5.7^{2}$

Solving the above equation we have,

Area of the sector = 45.048 cm2

Therefore, Area of the sector is 45.048 cm2

Q13.  The perimeter of a certain sector of a circle of radius is 5.6 m and 27.2 m. find the area of a sector.

Soln:

Given data:

Radius of the circle = 5.6 m = OA = OB

(AB arc length) + OA + OB = 27.2

Let the angle subtended at the centre be θ

We know that,

Length of arc = $\frac{\theta }{360}\ast 2\pi r cm$

$\frac{\theta }{360}\ast 2\pi r cm$ + OA + OB = 27.2 m

$\frac{\theta }{360}\ast 2\pi r cm$ + 5.6 + 5.6 = 27.2 m

Solving the above equation, we have,

θ = 163.64°

We know that,

$Area of the sector = \frac{\Theta }{360}\ast\pi r^{2}$

$Area of the sector = \frac{163.64 }{360}\ast\pi 5.6^{2}$

On solving the above equation, we have,

Area of the sector = 44.8 cm2

Therefore, Area of the sector is  44.8 cm2

Q14.  A sector was cut from a circle of radius 21 cm. The angle of sector is 120°. Find the length of its arc and its area.

Soln:

Given data:

Radius of circle ( r ) = 21 cm

θ = angle subtended at the centre of circle = 120°

Formula to be used:

$Length of arc =\frac{\theta }{360}\ast 2\pi r cm$

$Length of arc = \frac{ 120 }{360}\ast 2\pi x 21cm$

On solving the above equation, we get,

Length of arc = 44 cm

We know that,

$Area of the sector = \frac{\Theta }{360}\ast \pi r^{2}$

$Area of the sector = \frac{ 120 }{360}\ast\pi 21^{2}$

Area of the sector = ( 22 x 21 )  cm2

Therefore, Area of the sector is 462  cm2

Q15. The minute hand of a circle is $\sqrt{ 21 }$ cm long. Find the area described by the minute hand on the face of clock between 7:00 a.m to 7:05 a.m.

Soln:

Given data:

Radius of the minute hand (r) =$\sqrt{21}$ cm

Time between 7:00 a.m  to 7:05a.m  =  5 min

We know that, 1 hr = 60 min, minute hand completes

One revolution  =  360°

60 min  = 360°

θ = angle subtended at the centre of circle = 5 x 6° = 30°

$Area of the sector = \frac{\Theta }{360}\ast \pi r^{2}$

$Area of the sector = \frac{ 30}{360}\ast \ pi 35^{2}$

Area of the sector = 5.5 cm2

Therefore, Area of the sector is 5.5 cm2

Q 16. The minute hand of clock is 10 cm long. Find the area of the face of the clock described by the minute hand between 8 a.m to 8:25 a.m.

Soln:

Given data:

Radius of the circle = radius of the clock = length of the minute hand = 10 cm

We know that, 1 hr = 60 min

60 min = 360°

1 min = 6°

Time between 8:00 a.m to 8:25 a.m  =  25 min

Therefore, the subtended = 6° x 25 = 150°

Formula to be used :

$Area of the sector = \frac{\Theta }{360}\ast \pi r^{2}$

$Area of the sector = \frac{ 150 }{360}\ast \pi 10^{2}$

Area of the sector = 916.6 cm2= 917 cm2

Therefore, Area of the sector is 917 cm2

Q17. A sector of 56° cut out from a circle subtends area of 4.4 cm2. Find the radius of the circle.

Soln:

Given data:

Angle subtended by the sector at the centreof the circle, θ = 56°

Let the radius of the circle be = ‘r’ cm

Formula to be used:

$Area of the sector = \frac{ 56}{360}\ast\pi r^{2}$

On solving the above equation, we get,

r2 = $\sqrt{\frac{ 9 }{ 1}}$ cm

r = 3 cm

Therefore, radius of the circle is r = 3 cm

Q18. In circle of radius 6 cm.  Chord of length 10 cm makes an angle of 110° at the centre of circle. Find:

(i) Circumference of the circle

(ii) Area of the circle

(iii) Length of arc

(iv) The area of sector

Soln:

Given data:

Radius of the circle = 6 cm

Chord of length = 10 cm

Angle subtended by chord with the centre of the circle = 110°

Formulae to be used:

Circumference of a circle = 2

Area of a Circle =

$Area of the sector = \frac{\Theta }{360}\ast \pi r^{2}$

$Length of arc = \frac{ 90 }{360}\ast 2\pi x 28cm$

Circumference of a circle = 2 = 2 x 3.14 x 8 = 37.7 cm

Area of a Circle = = 3.14 x 6 x 6 = 113.14 cm2

$Area of the sector = \frac{\Theta }{360}\ast \pi r^{2}$

$Area of the sector = \frac{ 110 }{360}\ast\pi 6^{2}$

On solving the above equation we get,

Area of the sector = 33.1 cm2

$Length of arc = \frac{\theta }{360}\ast 2\pi r cm$

$Length of arc = \frac{ 110 }{ 360 }\ast 2\pi 6cm$

On solving the above equation we get,

Length of arc = 22.34 cm.

Therefore,Circumference = 37.7 cm

Area of a Circle = 113.14 cm2

Area of the sector = 33.1 cm2

Q19.  The given figure shows a sector of a circle with centre ‘O’ subtending an angle θ°. Prove that:

1. Perimeter of shaded region is $r\left(\tan\theta+\sec\theta + \left(\frac{\pi\ theta}{ 180 }\right) – 1\right)$

2. Area of the shaded region is $\frac{r^{ 2 }}{2}(\tan\theta -\frac{\pi\theta } {180} )$

Soln:

Given Data: Angle subtended at the centre of the circle = θ°

Angle OAB = 90° [ at point of contact, tangent is perpendicular to radius ]

OAB is a right angle triangle

Cos θ = $\frac{adj side}{hypotenuse} = \frac{r}{OB} = OB = r\sec\theta$

sec θ = $\frac{ opp side }{ adj side } = \frac {AB}{r} = AB = r\tan\theta$

Perimeter of the shaded region = AB + BC + CA ( arc )

=  r tanθ + ( OB – OC ) +  $\frac{\theta }{360}\ast2\pi r cm$

= $r\left( \tan\theta + \sec\theta + \frac{\pi\theta }{180} – 1\right )$

Area of the shaded region = ( area of triangle AOB ) – ( area of sector )

$\left(\frac{ 1 }{ 2 }\ast OA\ast AB\right ) -\frac{\theta }{360}\ast \pi r^{2}$

On solving the above equation we get,

$\frac{r^{2}}{2}\left [\tan\theta -\frac{\pi\theta }{180}\right ]$

Q 20. The diagram shows a sector of circle of radius ‘r’ cm subtends an angle θ. The area of sector is A cm2 and perimeter of sector is 50 cm. Prove that $\theta = \frac{360}{\pi }\left (\frac{25}{r} -1\right )$ and A = 25 r – r2

Soln:

Given Data:

Radius of circle = ‘r’ cm

Angle subtended at centre of the circle = θ

Perimeter = OA + OB + (AB arc)

$r + r + \frac{\theta }{360}\ast 2\pi r = 2r + 2r\left[\frac{\pi \theta }{360}\right]$

As given in the question, perimeter = 50

$\theta = \frac{360}{\pi }\left [ \frac{25}{r} – 1\right ]$

Therefore, $\theta = \frac{360}{\pi }\left [ \frac{25}{r} – 1\right ]$

$Area of the sector = \frac{\Theta }{360}\ast\pi r^{2}$

On solving the above equation, we have

A = 25r – r2

Hence, proved.