 # RD Sharma Solutions Class 7 Mensuration Ii Area Of Circle Exercise 21.2

## RD Sharma Solutions Class 7 Chapter 21 Exercise 21.2

Q1) Find the area of a circle whose radius is

(i) 7 cm

Solution: The area of the circle = π r2

r = 7 cm

Area of the circle = (22/7) × 72154 cm2

(ii) 2.1 m

Solution: The area of the circle = π r2

r = 2.1 m

Area of the circle = (22/7) × 2.12 = 13.86 m2

(iii) 7 km

Solution: The area of the circle = π r2

r = 7 km

Area of the circle = (22/7) × 72 = 154 km2

Q2) Find the area of a circle whose diameter is

(i) 8.4 cm

Solution: The area of the circle = π r2

diameter (d) = 8.4 cm

radius (r) = diameter/2

r = 8.4/2 = 4.2 cm

Area of the circle = (22/7) × 4.22 = 55.44 cm2

(ii) 5.6 m

Solution: The area of the circle = π r2

diameter (d) = 5.6 m

radius (r) = diameter/2

r = 5.6/2 = 2.8 m

Area of the circle = (22/7) × 2.82 = 24.64 m2

(iii) 7 km

Solution: The area of the circle = π r2

diameter (d) = 7 km

radius (r) = diameter/2

r = 7/2 = 3.5 km

Area of the circle = (22/7) × 3.52 = 38.5 km2

Q3) The area of a circle is 154 cm2. Find the radius of the circle.

Solution: The area of the circle = π r2 Hence, the radius of the circle is 7 cm.

Q4) Find the radius of a circle, if its area is

(i) 4 π cm2

Solution: The area of the circle = π r2 ⇒ r = 2 cm

(ii) 55.44 m2

Solution: The area of the circle = π r2 ⇒ r = 4.2 m

(iii) 1.54 km2

Solution: The area of the circle = π r2 ⇒ r = 0.7 km = 700 m

Q5) The circumference of a circle is 3.14 m, find its area.

Solution: Circumference of the circle = 2 π r    [Taking π = 3.14] ⇒ r = 0.5 m

Area of the circle = π r2 = (22/7) × 0.52 = 0.785 m2

Q6) If the area of a circle is 50.24 m2, find its circumference.

Solution: Area of the circle = π r2

⇒ 50.24 = (22/7) × r2

⇒ r = 3.998 m

Circumference of circle (C) = 2 π r

⇒ C = 2 × (22/7) × 3.998

⇒ C = 25.12 m

Q7) A horse is tied to a pole with 28 m long string. Find the area where the horse can graze. (Take π = 22/7).

Solution: It is given that the length of the string = 28 m

Hence, we can conclude that the area where horse can graze = Area of a circle with radius 28 m

Area where horse can graze = (22/7) × 282 = 2464 m2

Q8) A steel wire when bent in the form of a square encloses an area of 121 cm2. If the same wire is bent in the form of a circle, find the area of the circle.

Solution: Area of the square = (side)2

⇒ 121 = (side)2

⇒ side = 11 cm

So, the perimeter of the square = 4 × side = 4 x 11 = 44 cm

As the same wire is bent into the form of the circle,

Circumference of the circle = Perimeter of the square

⇒ 2 π r = 44

⇒ 2 × (22/7) × r = 44

⇒ r = 7 cm

So, Area of the circle = π r2 = (22/7) × 72 = 154 cm2

Q9) A road which is 7 m wide surrounds a circular park whose circumference is 352 m. Find the area of road.

Solution: Circumference of circular park = 2 π r

⇒ 352 = 2 × (22/7) × r

⇒ r = 56 m

Radius of the path including the 7 m wide road = r + 7 = 56 + 7 = 63 m

Area of the road = π × 632 –  π × 562

= π [632 – 562]

= π [3969 – 3136]

= (22/7) × 833

= 2618 m2

Q10) Prove that the area of a circular path of uniform width h surrounding a circular region of radius r is π h (2r + h).

Solution: Let us consider the radius of the circular region = r

Then the radius of the circular path = (r + h)

Area of the path = π (r + h)2 – π r2

= π (r2 + h2 + 2rh) – π r2

= π r2 + π h2 + 2 π r h – π r2

= π h2 + 2 π r h

= π h (h + 2r)

= π h (2r + h)

Hence proved.

Q11) The perimeter of a circle is 4πr cm. What is the area of the circle?

Solution: It is given that, Perimeter of the circle = 4πr

But according to the formula, perimeter or circumference of a circle is 2πr.

⇒ Perimeter of the circle = 2π(2r)

Radius of the circle is 2r cm.

Area of the circle = π × (2r)2 = 4πr2 cm2

Q12) A wire of 5024 m length is in the form of a square. It is cut and made a circle. Find the ratio of the area of the square to that of the circle.

Solution: It is given that, Perimeter of the square = 5024 m

⇒ 4 × side = 5024

⇒ side = 5024/4

⇒ side = 1256 m

The same wire is converted into the form of a circle. Therefore,

Circumference of the circle = Perimeter of the square

⇒ 2πr = 5024

⇒ 2 × π × r = 5024

⇒ r = 2512/π

Area of the square : Area of the circle = (side)2 : πr2 Area of the square : Area of the circle = 11 : 14

Q13) The radius of a circle is 14 cm. Find the radius of the circle whose area is double of the area of the circle.

Solution: Let the area of the circle whose radius is 14 cm be A1.

A1 = π (14)2

Let A2 and r2 be the area and radius of the another circle respectively whose area is double the area of circle A1.

A2 = 2 A1

⇒ π (r2)2 = 2 × π (14)2

⇒ (r2)2 = 2 × (14)2

⇒ r2 = 14√2 cm

Hence the radius of the circle A2 is 14√2 cm.

Q14) The radius of one circular field is 20 m and that of another is 48 m. Find the radius of the third circular field whose area is equal to the sum of the areas of two fields.

Solution: Let A= The area of the circle whose radius is 20 m

A= The area of the circle whose radius is 48 m

A3 = A1 + A2

⇒ π r2 = π (20)2 + π (48)2

⇒ π r2 = π [(20)2 + (48)2]

⇒ r2 = 400 + 2304

⇒ r = 52 m

Q15) The radius of one circular field is 5 m and that of the other is 13 m. Find the radius of the circular field whose area is the difference of the areas of first and second field.

Solution: Let A= The area of the circular field whose radius is 5 m

A= The area of the circular field whose radius is 13  m

A3 = A2 – A1

⇒ π r2 = π (13)2 – π (5)2

⇒ π r2 = π (13)2 – π (5)2

⇒ π r2 = π [(13)2 – (5)2]

⇒ r2 = 169 – 25

⇒ r2 = 144

⇒ r = 12 m

Hence, the radius of the circular field is 12 m.

Q16) Two circles are drawn inside a big circle with diameters $\frac{2}{3}$rd and $\frac{1}{3}$rd of the diameter of the big circle as shown in Figure. Find the area of the shaded portion, if the length of the diameter of the circle is 18 cm. Solution: It is given that, diameter of the big circle = 18 cm

Radius of the big circle = 9 cm

Area of the big circle, A = π r2 = π (8)2 = 81π cm2

Let d1 = (2/3) × 18 = 12 cm

r1 = 6 cm

Area of the circle, A1 = π r2 = π (6)2 = 36π cm2

d2 = (1/3) × 18 = 6 cm

r2 = 3 cm

Area of the circle, A2 = π r2 = π (3)2 = 9π cm2

Area of the shaded portion = A – (A1 + A2 )

Area of the shaded portion = 81π – (36π – 9π) = 36π cm2

Q17) In fig, the radius of quarter circular plot taken is 2 m and radius of the flower bed is 2 m. Find the area of the remaining field. Solution: It is given that;

Radius of flower bed = 2 m

Area of flower bed = π r2 = π (2)2 = 4π

Radius of the quarter circular plot = 2 m

Area of the quarter circular plot = (π r2)/4

Area of 4 quarter circular plots = 4 × (π r2)/4 = π r2 = π (2)2 = 4π

Area of the rectangular region = Length x Breadth

Area of the rectangular region = 8 x 6 = 48 m2

Area of the remaining field = Area of the rectangular region – (Area of 4 quarter circular plots + Area of the flower bed)

Area of the remaining field = 48 – (4π + 4π) = 48 – 25.14 = 22.86 m2

Q18) Four equal circles, each of radius 5 cm, touch each other as shown in Figure. Find the area included between them. (Take π = 3.14). Solution: As we can see from the figure;

Side of the square = 10 cm

So, Area of the square = side x side = 10 x 10 = 100 cm2

Radius of the quarter circle = 5 cm

Area of the quarter circle = (π r2)/4

Area of 4 quarter circle = 4 × (π r2)/4 = π r2 = 3.14 × (5)2 = 78.5 cm2

Area included in the quarter circle = Area of the square – Area of the four quarter circles

Area included in them = (100 – 78.5) cm= 21.5 cm2

Q19) The area of circle is 100 times the area of another circle. What is the ratio of their circumferences?

Solution: Let the area of the circles be A1 and A2 and their circumference be c1 and crespectively.

A1 = 100 A2

⇒ π(r1)2 = 100 × π(r2)2

⇒ r1 = 10 r2

⇒ r1/r2 = 10/1 ……………………..(i)

Finding the ratios of the circumference;

c1 : c2 = 2πr1 :  2πr2

c1/c2 = (2πr1)/(2πr2)

c1/c2 = r1/r2

Putting the value of r1/rfrom equation (i)

c1/c2 = 10/1

c1 : c2 = 10 : 1

Hence, the ratio of their circumferences is 10 : 1.