**Exercise 15.2**

**Q1.Â Find in terms of Ï€****, the length of the arc that subtends an angle of 30 degrees, at the center of â€˜Oâ€™ of the circle with a radius of 4 cm.**

**Soln: **

Given Data :

Radius = 4 cm

Angle subtended at the centre â€˜Oâ€™ =Â 30Â°

Formula to be used :

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

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

=Â \(\frac{2\pi }{3} cm\)

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:

Radius = 5 cm

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:

Î¸ = 60Â°

Therefore, angle subtended at the centre of circle is60Â°

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

**Soln:**

Given Data : length of arc =Â cm

Î¸ = angle subtended at the centre of circle = 144Â°

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

**Soln:**

Given Data :

length of arc =25 cm

Î¸ = angle subtended at the centre of circle = 55Â°

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 = \(\frac{11\pi r}{36}\)

\(radius = \frac{25\ast36}{11\ast\pi}\)

=Â \(\frac{900}{11\pi }\)

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 :

Radius = a cm

Length of arc = \(\frac{a\pi }{4}\)Â cm

Î¸ = angle subtended at the centre of circle

Formula to be used:

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

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

\(\frac{\theta}{360}\ast 2\pi a cm\)Â = \(\frac{a\pi }{4}\)Â cm

Solving the above equation, we have

Î¸ = 45Â°

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

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

**Soln:**

Given Data:

Radius = 4 cm

Angle subtended at the centre â€˜Oâ€™ = 30Â°

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 cm^{2 }

Therefore, Area of the sector is found to be 4.9 cm^{2}

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

**Soln:**

Given Data:

Radius = 8 cm

Angle subtended at the centre â€˜Oâ€™ = 135Â°

Formula to be used:

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

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

=\(\frac{528}{ 7}\) cm^{2}

Therefore, Area of the sector calculated is \(\frac{ 528 }{ 7 }\) cm^{2}

^{Â }**Â **

**Q8. The area of sector of circle of radius 2 cm is ****Â cm ^{2}. Find the angle subtended by the sector.**

**Soln:**

Given Data:

Radius = 2 cm

Angle subtended at the centre â€˜Oâ€™ =?

Area of sector of circle = cm^{2}

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 = cm^{2}

cm^{2Â Â }= \(\frac{\pi\Theta }{90}\)

Solving the above equation, we have

Î¸ = 90Â°

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

**Â **

**Q10. PQ is a chord of circle with centre â€˜Oâ€™ 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 = 60Â°

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 \(\frac{32\pi }{ 3 }\) cm^{2}

^{Â }**Â **

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

**Soln:**

Given Data:

Radius = 35 cm

Angle subtended at the centre â€˜Oâ€™ = 72Â°

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 ) cm^{2}

Therefore, Area of the sector is 770 cm^{2}

^{Â }

^{Â }**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 Î¸

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,

Î¸ = 158.8Â°

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 cm^{2}

Therefore, Area of the sector is 45.048 cm^{2}

**Q13.Â 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 Î¸

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,

Î¸ = 163.64Â°

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 cm^{2}

Therefore, Area of the sector isÂ 44.8 cm^{2}

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

**Soln:**

Given data:

Radius of circle ( r ) = 21 cm

Î¸ = angle subtended at the centre of circle = 120Â°

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 )Â cm^{2}

Therefore, Area of the sector is 462Â cm^{2}

**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.mÂ to 7:05a.mÂ =Â 5 min

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

One revolutionÂ =Â 360Â°

60 minÂ = 360Â°

Î¸ = angle subtended at the centre of circle = 5 x 6Â° = 30Â°

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

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

Area of the sector = 5.5 cm^{2}

Therefore, Area of the sector is 5.5 cm^{2}

^{Â }

**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 = 360Â°

1 min = 6Â°

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

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

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 cm^{2}= 917 cm^{2}

Therefore, Area of the sector is 917 cm^{2}

**Â **

**Q17. A sector of 56Â° cut out from a circle subtends area of 4.4 cm ^{2}. Find the radius of the circle.**

**Soln:**

**Given data:**

Angle subtended by the sector at the centreof the circle, Î¸ = 56Â°

Let the radius of the circle be = â€˜râ€™ cm

**Formula to be used:**

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

On solving the above equation, we get,

r^{2} = \(\sqrt{\frac{ 9 }{ 1}}\) cm

r = 3 cm

Therefore, radius of the circle is r = 3 cm

**Q18. In circle of radius 6 cm.Â Chord of length 10 cm makes an angle of 110Â° 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 = 110Â°

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 cm^{2}

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

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

On solving the above equation we get,

Area of the sector = 33.1 cm^{2 }

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

\(Length of arc = \frac{ 110 }{ 360 }\ast 2\pi 6cm\)On solving the above equation we get,

Length of arc = 22.34 cm.

Therefore,Circumference = 37.7 cm

Area of a Circle = 113.14 cm^{2}

Area of the sector = 33.1 cm^{2}

**Q19.Â The given figure shows a sector of a circle with centre â€˜Oâ€™ subtending an angle Î¸Â°. 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 \(\frac{r^{ 2 }}{2}(\tan\theta -\frac{\pi\theta } {180} )\)**

**Â **

**Â **

**Soln:**

Given Data: Angle subtended at the centre of the circle = Î¸Â°

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

OAB is a right angle triangle

Cos Î¸ = \(\frac{adj side}{hypotenuse} = \frac{r}{OB} = OB = r\sec\theta\)

sec Î¸ = \(\frac{ opp side }{ adj side } = \frac {AB}{r} = AB = r\tan\theta\)

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

=Â r tanÎ¸ + ( OB â€“ OC ) + Â \(\frac{\theta }{360}\ast2\pi r cm\)

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

Area of the shaded region = ( area of triangle AOB ) â€“ ( 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 â€˜râ€™ cm subtends an angle Î¸. The area of sector is A cm ^{2} and perimeter of sector is 50 cm. Prove that **

**\(\theta = \frac{360}{\pi }\left (\frac{25}{r} -1\right )\) andÂ**

**A = 25 r â€“ r**

^{2}

**Soln:Â **

Given Data:

Radius of circle = â€˜râ€™ cm

Angle subtended at centre of the circle = Î¸

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 â€“ r^{2}

Hence, proved.