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

RD Sharma Class 7 Solutions Chapter 21 Ex 21.2 PDF Free Download

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

RD Sharma Solutions Class 7 Maths Chapter 21 Exercise 21.2 Ans 3

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

RD Sharma Solutions Class 7 Maths Chapter 21 Exercise 21.2 Qs 4 (i)

⇒ r = 2 cm

(ii) 55.44 m2   

Solution: The area of the circle = π r2

RD Sharma Solutions Class 7 Maths Chapter 21 Exercise 21.2 Qs 4 (ii)

⇒ r = 4.2 m

(iii) 1.54 km2

Solution: The area of the circle = π r2

RD Sharma Solutions Class 7 Maths Chapter 21 Exercise 21.2 Qs 4 (iii)

⇒ 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]

RD Sharma Solutions Class 7 Maths Chapter 21 Exercise 21.2 Qs 5

⇒ 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

RD sharma class 7 maths solution chapter 21 exercise 21.2 ans 12

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.

14

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.

15

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).

16

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.

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