The document RD Sharma Solutions (Part - 2) - Ex-21.2, Mensuration - II Area of Circle, Class 7, Math Class 7 Notes | EduRev is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.

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

We have :

Given perimeter of the circle = 4Ï€r cm = 2Ï€ (2r) cm

We know that, the perimeter of a circle = 2Ï€Ï€r

âˆ´ Radius of the circle = 2r cm

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

**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.**

We have:

Perimeter of the square = 5024 m = Circumference of the circle

â‡’ 4 x Side of the square = 5024

âˆ´ Side of the square = 5024/4 = 1256 m

Let the area of the square be *A*_{1} and the area of the circle be* **A*_{2}.

Area of the square (A_{1})= side x side =

Circumference of the circle = 5024 m

â‡’ 2 Ï€r = 5024 m

Area of the circle (A_{2})= Ï€r^{2} =

âˆ´A_{1 }: A_{2 }=

Hence, the ratio of the area of the square to the area of the circle is 11:14.

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

Let the area of the circle whose radius is 14 cm be *A*_{1}_{.}

Let the radius and area of the circle, whose area is twice the area of the circle A_{1}_{ , }be *r*_{2} and *A*_{2}, respectively.

Thus,*A*_{1} = Ï€(r)^{2} = Ï€(14)^{2} = = =(44Ã—14) cm^{2}= 616 cm^{2}

A_{1} = 2 Ã— *A*_{1} = 2 Ã— 616 = 1232 cm^{2}*A*_{2} = Ï€(r^{2})^{2} = 1232 cm^{2}

â‡’r^{2} =

Hence, the radius of the circle *A*_{2} is

**The radius of one circluar 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.**

Let the area of the circle whose radius is 20 m be *A*_{1} , and the area of the circle whose radius

is 48 m be *A*_{2}. Let *A*_{3} be the area of a circle that is equal to the sum of the areas of the two fields, with the radius of its field being *r* cm.

âˆ´ *A*_{3}_{ }= *A*_{1} + A_{2}

A_{1} = Ï€(20)^{2} =

A_{2} = Ï€ (48)^{2} =

A_{3} = A_{1}+A_{2} = (400Ï€) +(2304Ï€) = Ï€(400 +2304) m^{2}

â‡’A_{3} = Ï€(r)^{2} = Ï€(400 +2304) m^{2}

â‡’(r)^{2} = (400 +2304) m^{2}

â‡’ r =

**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.**

Let the area of the circular field whose radius is 5 m be *A*_{1} , and the area of the circular field whose radius is 13 m be *A*_{2}. Let *A*_{3}_{ }and *r* cm be the area and radius of the circular field, that is equal to the difference of the areas of the two fields.

âˆ´ *A*_{3}_{ }= *A*_{2}^{ }-A_{1}

A_{1} = Ï€(5)^{2} = (25Ï€) m_{2 }

A_{2} = Ï€ (13)^{2} = (169 Ï€) m^{2}

A_{3} = A_{2}âˆ’A_{1} = (169Ï€) âˆ’(25 Ï€) = 144Ï€ m^{2}

â‡’A_{3} = Ï€(r)^{2} = 144 Ï€ m^{2}

â‡’(r)^{2} = 144 m^{2}

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

**Two circles are drawn inside a big circle with diameters of the diameter of the big circle as shown in Fig. 18. Find the area of the shaded portion, if the length of the diameter of the circle is 18 cm.**

Let the left circle be denoted as the 1st circle and the right circle be denoted as the 2nd circle.

Diameter of the big circle = 18 cm

Radius of the big circle = 9 cm

Diameter of the 1st circle =

Radius of the 1st circle = 6 cm

Diameter of the 2nd circle =

Radius of the 2nd circle = 3 cm

Area of the 1st circle = Ï€(6)^{2}=36Ï€ cm^{2}

Area of the 2nd circle = Ï€(3)^{2}=Ï€Ã—3Ã—3=9Ï€ cm^{2}

Area of the big circle = Ï€(9)^{2}=Ï€Ã—9Ã—9=81Ï€ cm^{2}

Area of the shaded portion = Area of the big circle - (Area of the I^{st} circle + Area of the II^{nd} circle)

Area of the shaded portion = 81Ï€âˆ’(36Ï€+9Ï€)=36Ï€ cm^{2}.

**In Fig. 19, 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.**

Radius of the quarter circular plot = 2 m

Area of the quarter circular plot =

Radius of each flower bed = 2 m

Area of four flower beds =

Area of the rectangular region = Length Ã— Breadth

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

Area of the remaining field = Area of the rectangular region - (Area of the quarter circle + Area of the four flower beds)

Area of the remaining field = [48 - (12.57 + 12.57)] m^{2} = 22.86 m^{2}

**Four equal circles, each of radius 5 cm, touch each other as shown in Fig. 20. Find the area included between them. (Take Ï€ = 3.14).**

Side of the square = 10 cm

Area of the square = Side Ã— Side

Area of the square =^{ }10Ã—10=100 cm^{2}

Area of the four quarter circles =

Area included in them = Area of the square - Area of the four quarter circles

Area included in them = ( 100 - 78.57 ) cm^{2} = 21.43 cm^{2} .

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

Let the area of the first circle be *A*_{1}, the circumference be *C*_{1}_{ }and the radius be *r*_{1}.

Let the area of the second circle be *A*_{2}_{ , }the circumference be *C*_{2}_{ }and the radius be* **r*_{2}.

Thus,

We know that :

*A*_{1}_{ }= 100*A*_{2}

Substituting the values, we get:

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

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