# Class 10 RD Sharma Solutions – Chapter 15 Areas Related to Circles – Exercise 15.1 | Set 2

**Question 11. The radii of two circles are 19 cm and 9 cm respectively. Find the radius and area of the circle which has circumferences is equal to sum of the circumference of two circles. **

**Solution:**

Radius of circle 1 = r

_{1}= 19 cmRadius of circle 2 = r

_{2}= 9 cmSo, C

_{1}= 2πr_{1,}C_{2}= 2πr_{2}C = C

_{1}+ C_{2}2πr = 2πr

_{1}+ 2πr_{2}r = r

_{1}+ r_{2 }r = 19 + 9

r = 28 cm

Therefore, the radius of the circle = 28 cm

Therefore, area of required circle = πr

^{2}= (22/7) × 28 × 28

= 2464 cm

^{2}

**Question 12. The area of a circular playground is 22176 m**^{2}. Find the cost of fencing this ground at the rate of ₹**50 per meter.**

^{2}. Find the cost of fencing this ground at the rate of

**Solution:**

Area of the circular playground = 22176 m

^{2}Area = πr

^{2}πr

^{2}= 22176r

^{2}= 22176(7/22)= 7056

r = 84 m

Circumference of the ground = 2πr

= 2(22/7)84

= 528 m

Cost of fencing 528 m = ₹50 x 528

= ₹26400

Therefore, the cost of fencing the ground = ₹26400.

**Question 13. The side of a square is 10 cm. Find the area of the circumscribed and inscribed circles. **

**Solution:**

Diagonal of the square = AC = √2 x side

= 10√2 cm

Radius of circumscribed circle = Diagonal/2

R = 5√2cm

R = 7.07cm

Area= πR

^{2}= (22/7) × 7.07 × 7.07

= 157.09 cm

^{2 }Therefore, the Area of the Circumscribed circle = 157.09 cm

^{2 }For inscribed circle diameter of circle = side of square = AB

Radius = side of square/2

= 10/2

r = 5 m

Area = πr

^{2}= (22/7) × 5 × 5

= 78.5 cm

^{2 }Therefore, the area of the circumscribed circle = 157.09 cm

^{2 }and the area of the inscribed circle = 78.5 cm^{2}.

**Question 14. If a square is inscribed in a circle, find the ratio of areas of the circle and the square. **

**Solution:**

Let side of square AB be x cm which is inscribed in a circle.

Radius of circle (r) = 1/2 (diagonal of square)

= 1/2(a√2)

r = a/√2

Area of the square = a

^{2}Area of the circle = πr

^{2}= π(a

^{2}/2)Ratio of areas = Area of circle:Area of square

= π(a

^{2}/2) : a^{2}

^{ = }π : 2Therefore, the ratio of areas of the circle and the square = π : 2

**Question 15. The area of circle inscribed in an equilateral triangle is 154 cm**^{2}. Find the perimeter of the triangle.

^{2}. Find the perimeter of the triangle.

**Solution:**

Area of a Circle = πr

^{2}(22/7) × r

^{2}= 154r

^{2}= (154 x 7)/22= 7 × 7

= 49

r = 7 cm

OP is perpendicular bisector of BC (as BP is tangent and it is a equilateral triangle)

BP = ½ x BC

Consider the side of the equilateral triangle be a cm.

In right-angled triangle OPB

OB

^{2}= OP^{2}+ BP^{2}(By Pythagoras theorem)OB

^{2}= r^{2}+ (a/2)^{2 }(BP is half of a)OB

^{2 }= 49 + a^{2}/4OB = √(49 + a

^{2}/2) ….. (1)AP = (√3/2)a (height of an equilateral triangle)

OA = (√3/2)a – r

Similarly

OB = (√3/2)a – r …. (2)

From (1) and (2)

Squaring both sides

49 + a

^{2}/4 = (3/4)a^{2}+ r^{2}– √3arr = 7

49 + a

^{2}/4 = (3/4)a^{2}+ 49 – 7√3aa

^{2}/4- (3/4)a^{2}= -7√3aTaking 4 as LCM

(a

^{2}– 3a^{2}) / 4 = -7√3a-2a

^{2}/4 = -7√3aa = 14√3 cm

Perimeter of equilateral triangle = 3a

= 42√3

Therefore, the perimeter of the triangle = 42(1.73)

= 72.7 cm