## RD Sharma Solutions Class 7 Chapter 21 Exercise 21.2

**Q1) Find the area of a circle whose radius is **

**(i) 7 cmÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â **

**Solution: **The area of the circle = Ï€ r^{2}

r = 7 cm

Area of the circle = (22/7)Â Ã— 7^{2} =Â 154 cm^{2}

**(ii) 2.1 m Â Â Â Â Â Â **

**Solution:** The area of the circle = Ï€ r^{2}

r = 2.1 m

Area of the circle = (22/7)Â Ã— 2.1^{2}Â =Â 13.86 m^{2}

**(iii) 7 km**

**Solution:** The area of the circle = Ï€ r^{2}

r = 7 km

Area of the circle = (22/7)Â Ã— 7^{2}Â =Â 154 km^{2}

**Q2) Find the area of a circle whose diameter is **

**(i) 8.4 cmÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â **

**Solution:Â **The area of the circle = Ï€ r^{2}

diameter (d) = 8.4 cm

radius (r) = diameter/2

r = 8.4/2 = 4.2 cm

Area of the circle = (22/7)Â Ã— 4.2^{2}Â =Â 55.44 cm^{2}

**(ii) 5.6 mÂ Â Â **

**Solution:Â **The area of the circle = Ï€ r^{2}

diameter (d) = 5.6 m

radius (r) = diameter/2

r = 5.6/2 = 2.8 m

Area of the circle = (22/7)Â Ã— 2.8^{2}Â = 24.64 m^{2}

**(iii) 7 km**

**Solution:Â **The area of the circle = Ï€ r^{2}

diameter (d) = 7 km

radius (r) = diameter/2

r = 7/2 = 3.5 km

Area of the circle = (22/7)Â Ã— 3.5^{2}Â = 38.5Â km^{2}

**Q3) The area of a circle is 154 cm ^{2}. Find the radius of the circle. **

**Solution:Â **The area of the circle = Ï€ r^{2}

Hence, the radius of the circle is 7 cm.

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

**(i) 4 Ï€ cm ^{2}Â Â Â Â Â Â Â Â Â Â Â Â Â **

**Solution:Â **The area of the circle = Ï€ r^{2}

â‡’Â r = 2 cm

**(ii) 55.44 m ^{2}Â Â **

**Solution:Â **The area of the circle = Ï€ r^{2}

â‡’Â r = 4.2 m

**(iii) 1.54 km ^{2}**

**Solution:Â **The area of the circle = Ï€ r^{2}

â‡’ r = 0.7 km = 700 m

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

**Solution:Â **Circumference of the circle = 2 Ï€ rÂ Â Â [Taking Ï€ = 3.14]

â‡’ r = 0.5 m

Area of the circle = Ï€ r^{2} = (22/7)Â Ã— 0.5^{2} = 0.785 m^{2}

**Q6) If the area of a circle is 50.24 m ^{2}, find its circumference. **

**Solution:Â **Area of the circle = Ï€ r^{2}

â‡’ 50.24Â = (22/7)Â Ã—Â r^{2}

â‡’ r = 3.998 m

Circumference of circle (C) =Â 2 Ï€ r

â‡’Â C = 2Â Ã—Â (22/7)Â Ã—Â 3.998

â‡’Â C = 25.12 m

**Â ****Q7) A horse is tied to a pole with 28 m long string. Find the area where the horse can graze. (Take Ï€ = 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

AreaÂ where horse can graze = (22/7)Â Ã— 28^{2} =Â 2464 m^{2}

**Q8) A steel wire when bent in the form of a square encloses an area of 121 cm ^{2}. If the same wire is bent in the form of a circle, find the area of the circle. **

**Solution:Â **Area of the square = (side)^{2}

â‡’ 121Â = (side)^{2}

â‡’ side = 11 cm

So, the perimeter of the square = 4 Ã— 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

â‡’ 2 Ï€ r =Â 44

â‡’ 2Â Ã— (22/7)Â Ã— r = 44

â‡’ r = 7 cm

So,Â Area of the circle = Ï€ r^{2} = (22/7)Â Ã— 7^{2} = 154Â cm^{2}

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

**Solution:Â **Circumference of circular park =Â 2 Ï€ r

â‡’ 352 = 2Â Ã—Â (22/7)Â Ã— r

â‡’ r = 56 m

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

Area of the road =Â Ï€Â Ã— 63^{2} –Â Â Ï€Â Ã— 56^{2}

= Ï€ [63^{2} – 56^{2}]

=Â Ï€ [3969 â€“ 3136]

= (22/7)Â Ã— 833

= 2618 m^{2}

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

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

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

Area of the path = Ï€ (r + h)^{2} –Â Ï€ r^{2}

=Â Ï€ (r^{2} + h^{2} + 2rh) –Â Ï€ r^{2}

=Â Ï€ r^{2} + Ï€ h^{2} + 2 Ï€ r h –Â Ï€ r^{2}

=Â Ï€ h^{2} + 2 Ï€ r h

=Â Ï€ h (h + 2r)

=Â Ï€ h (2r + h)

Hence proved.

**Q11) The perimeter of a circle is 4Ï€rÂ cm. What is the area of the circle? **

**Solution:** It is given that,Â Perimeter of the circle = 4Ï€r

But according to the formula, perimeter or circumference of a circle is 2Ï€r.

â‡’ Perimeter of the circle = 2Ï€(2r)

Radius of the circle is 2r cm.

Area of the circle =Â Ï€ Ã— (2r)^{2} = 4Ï€r^{2} cm^{2}

**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,Â Perimeter of the square = 5024 m

â‡’ 4Â Ã— side = 5024

â‡’ side = 5024/4

â‡’ side = 1256 m

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

Circumference of the circle = Perimeter of the square

â‡’ 2Ï€r = 5024

â‡’ 2Â Ã— Ï€ Ã— r = 5024

â‡’ r = 2512/Ï€

Area of the square : Area of the circle = (side)^{2} :Â Ï€r^{2}

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:Â **Let the area of the circle whose radius is 14 cm be A_{1}.

A_{1} =Â Ï€ (14)^{2}

Let A_{2} and r_{2} be the area and radius of the another circle respectively whose area is double the area of circle A_{1}.

A_{2} = 2 A_{1}

â‡’Â Ï€ (r_{2})^{2}Â = 2 Ã— Ï€ (14)^{2}

â‡’Â (r_{2})^{2} =Â 2 Ã— (14)^{2}

â‡’ r_{2}Â = 14âˆš2 cm

Hence the radius of the circle A_{2} isÂ 14âˆš2 cm.

**Q14) The radius of one circularÂ 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:Â **Let A_{1Â }= TheÂ area of the circle whose radius is 20 m

A_{2Â }= TheÂ area of the circle whose radius is 48 m

A_{3}_{Â }=Â A_{1} + A_{2}

â‡’ Ï€ r^{2} =Â Ï€ (20)^{2} +Â Ï€ (48)^{2}

â‡’ Ï€ r^{2} =Â Ï€ [(20)^{2} + (48)^{2}]

â‡’ r^{2} = 400Â + 2304

â‡’ 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:Â **Let A_{1Â }= TheÂ area of the circular field whose radius is 5 m

A_{2Â }= TheÂ area of the circular field whose radius is 13Â m

A_{3}_{Â }=Â A_{2} – A_{1}

â‡’ Ï€ r^{2} =Â Ï€ (13)^{2}Â – Ï€ (5)^{2}

â‡’ Ï€ r^{2} =Â Ï€ (13)^{2}Â – Ï€ (5)^{2}

â‡’ Ï€ r^{2} =Â Ï€ [(13)^{2}Â – (5)^{2}]

â‡’ r^{2} = 169Â – 25

â‡’ r^{2} = 144

â‡’ 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 =Â Ï€ r^{2} =Â Ï€ (8)^{2} = 81Ï€ cm^{2}

Let d_{1}Â = (2/3)Â Ã— 18 = 12 cm

r_{1} = 6 cm

Area of the circle, A_{1} =Â Ï€ r^{2} =Â Ï€ (6)^{2} = 36Ï€ cm^{2}

d_{2}Â = (1/3)Â Ã— 18 = 6 cm

r_{2} = 3 cm

Area of the circle, A_{2} =Â Ï€ r^{2} =Â Ï€ (3)^{2} = 9Ï€Â cm^{2}

Area of the shaded portion = A – (A_{1}Â + A_{2} )

Area of the shaded portion = 81Ï€ – (36Ï€ – 9Ï€) = 36Ï€Â cm^{2}

**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 = Ï€ r^{2} = Ï€ (2)^{2} = 4Ï€

Radius of the quarter circular plot = 2 m

Area of the quarter circular plot = (Ï€ r^{2})/4

Area of 4 quarter circular plots = 4 Ã— (Ï€ r^{2})/4 =Â Ï€ r^{2}^{Â }= Ï€ (2)^{2} = 4Ï€

Area of the rectangular region = Length x Breadth

Area of the rectangular region = 8 x 6 = 48 m^{2}

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 – (4Ï€ + 4Ï€) = 48 – 25.14 = 22.86 m^{2}

**Q18) Four equal circles, each of radius 5 cm, touch each other as shown in Figure. Find the area included between them. (Take Ï€ = 3.14).**

**Solution: **As we can see from the figure;

Side of the square = 10 cm

So, Area of the square = side x side =Â 10 x 10 = 100 cm^{2}

Radius of theÂ quarter circle = 5 cm

Area of the quarter circle = (Ï€ r^{2})/4

Area of 4 quarter circle = 4 Ã— (Ï€ r^{2})/4 =Â Ï€ r^{2}^{Â }= 3.14Â Ã— (5)^{2} = 78.5Â cm^{2}

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

Area included in them = (100 – 78.5) cm^{2Â }= 21.5 cm^{2}

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

**Solution:Â **Let the area of the circles be A_{1} and A_{2} and their circumference be c_{1} and c_{2Â }respectively.

A_{1} = 100 A_{2}

â‡’ Ï€(r_{1})^{2} = 100Â Ã—Â Ï€(r_{2})^{2}

â‡’Â r_{1} = 10 r_{2}

â‡’ r_{1}/r_{2} = 10/1 ……………………..(i)

Finding the ratios of the circumference;

c_{1} : c_{2} =Â 2Ï€r_{1} :Â Â 2Ï€r_{2}

c_{1}/c_{2} = (2Ï€r_{1})/(2Ï€r_{2})

c_{1}/c_{2} = r_{1}/r_{2}

Putting the value of r_{1}/r_{2Â }from equation (i)

c_{1}/c_{2} = 10/1

c_{1} : c_{2} = 10 : 1

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