RD Sharma solution class 7 chapter 21 Mensuration II Exercise 21.2

Exercise 21.2

Page-21.15

Question 1:

Find the area of a circle whose radius is
(i) 7 cm
(ii) 2.1 m
(iii) 7 km

Answer 1:

(i) We know that the area A of a circle of radius r is given by A = $\mathrm{\pi }{r}^2$.
     Here, r  = 7 cm
∴ $A = \frac{22}{7}\times {\left(7\right)}^2 {\mathrm{cm}}^2$.
$\Rightarrow A = \left(\frac{22}{7}\times 7\times 7\right) {\mathrm{cm}}^2 = \left(22\times 7\right) {\mathrm{cm}}^2 = 154 {\mathrm{cm}}^2$.
(ii) ​We know that the area A of a circle of radius r is given by A = $\mathrm{\pi }{r}^2$.
     Here, r  = 2.1 m
∴ $A = \frac{22}{7}\times {\left(2.1\right)}^2 {\mathrm{m}}^2$.
$\Rightarrow A = \left(\frac{22}{7}\times 2.1\times 2.1\right) {\mathrm{m}}^2 = \left(22\times 0.3\times 2.1\right) {\mathrm{m}}^2 = 13.86 {\mathrm{m}}^2$.
(iii) ​We know that the area A of a circle of radius r is given by A =  $\mathrm{\pi }{r}^2$.
     Here, r  = 7 km
∴ $A = \frac{22}{7}\times {\left(7\right)}^2 {\mathrm{km}}^2$.
$\Rightarrow A = \left(\frac{22}{7}\times 7\times 7\right) {\mathrm{km}}^2 = \left(22\times 7\right) {\mathrm{km}}^2 = 154 {\mathrm{km}}^2$.

Question 2:

Find the area of a circle whose diameter is
(i) 8.4 cm
(ii) 5.6 m
(iii) 7 km

Answer 2:

(i) Let r be the radius of the circle. Then, r = 8.4 $÷$ 2 = 4.2 cm.
∴ Area of the circle  = ${\mathrm{\pi r}}^2$
    $$\begin{gathered} \Rightarrow A = \frac{22}{7}\times {\left(4.2\right)}^2 {\mathrm{cm}}^2 \\ \Rightarrow A =\left(\frac{22}{7}\times 4.2 \times 4.2\right){\mathrm{cm}}^2 = \left(22\times 0.6\times 4.2\right) {\mathrm{cm}}^2 = 55.44 {\mathrm{cm}}^2. \end{gathered}$$.

(ii) Let r be the radius of the circle. Then, r = 5.6 $÷$ 2 = 2.8 m.
 Area of the circle  = ${\mathrm{\pi r}}^2$
    $$\begin{gathered} \Rightarrow A = \frac{22}{7}\times {\left(2.8\right)}^2 {\mathrm{m}}^2 \\ \Rightarrow A =\left(\frac{22}{7}\times 2.8 \times 2.8\right) {\mathrm{m}}^2 = \left(22\times 0.4\times 2.8\right) {\mathrm{m}}^2 = 24.64 {\mathrm{m}}^2. \end{gathered}$$​
(iii) ​​Let r be the radius of the circle. Then, r = 7 $÷$ 2 = 3.5 km.
     Area of the circle  = ${\mathrm{\pi r}}^2$
    $$\begin{gathered} \Rightarrow A = \frac{22}{7}\times {\left(3.5\right)}^2 {\mathrm{km}}^2 \\ \Rightarrow A =\left(\frac{22}{7}\times 3.5 \times 3.5\right) {\mathrm{km}}^2 = \left(22\times 0.5\times 3.5\right) {\mathrm{km}}^2 = 38.5 {\mathrm{km}}^2. \end{gathered}$$

Question 3:

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

Answer 3:

Let the radius of the circle be r cm. 
Area of the circle (A) = 154 cm²
    $$\begin{gathered} \Rightarrow 154 = \frac{22}{7}\times {\left(r\right)}^2 {\mathrm{cm}}^2 \\ \Rightarrow r^2 =\left(\frac{154\times 7}{22}\right) \\ \Rightarrow r^2 =\left(\frac{1078}{22}\right) \\ \Rightarrow r^2 =\left(49\right) \\ \Rightarrow r =7 \mathrm{cm}. \end{gathered}$$

Hence, the radius of the circle is 7 cm.

Question 4:

Find the radius of a circle, if its area is
(i) 4 π cm²
(ii) 55.44 m²
(iii) 1.54 km²

Answer 4:

(i) ​Let the radius of the circle be r cm. 

∴ Area of the circle (A) = 4$\mathrm{\pi }$ cm²
    $$\begin{gathered} \Rightarrow 4\mathrm{\pi } = \mathrm{\pi }\times {\left(r\right)}^2 {\mathrm{cm}}^2 \\ \Rightarrow r^2 =\left(\frac{4\mathrm{\pi }}{\mathrm{\pi }}\right) = 4 \\ \Rightarrow r =2 \mathrm{cm}. \end{gathered}$$

(ii) ​Let the radius of the circle be r cm. 

∴ Area of the circle (A) = 55.44 m²
    $$\begin{gathered} \Rightarrow 55.44 = \frac{22}{7}\times {\left(r\right)}^2 {\mathrm{m}}^2 \\ \Rightarrow r^2 =\left(\frac{55.44\times 7}{22}\right) \\ \Rightarrow r^2 =\left(\frac{5.04\times 7}{2}\right) \\ \Rightarrow r^2 =\left(17.64\right) \\ \Rightarrow r =4.2 \mathrm{m} \end{gathered}$$

(iii) ​Let the radius of the circle be r cm. 

∴ Area of the circle (A) = 1.54 km²
    $$\begin{gathered} \Rightarrow 1.54 = \frac{22}{7}\times {\left(r\right)}^2 {\mathrm{km}}^2 \\ \Rightarrow r^2 =\left(\frac{1.54\times 7}{22}\right) \\ \Rightarrow r^2 =\left(\frac{10.78}{22}\right) \\ \Rightarrow r^2 =\left(0.49\right) \\ \Rightarrow r =0.7 \mathrm{km} = 700 \mathrm{m} \end{gathered}$$

Question 5:

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

Answer 5:

We have :

Circumference of the circle = 3.14  ​m = $2\mathrm{\pi r}$
 $$\begin{gathered} \Rightarrow 3.14 \mathrm{m} =\left(2\times \frac{22}{7} \mathrm{x} \mathrm{r}\right) \mathrm{m} \\ \Rightarrow r=\left(\frac{3.14\times 7}{2\times 22}\right) \mathrm{m} = \left(\frac{1}{2}\right) \mathrm{m}. \end{gathered}$$

Area of the circle (A) = ${\mathrm{\pi r}}^2$ 
    $$\begin{gathered} \Rightarrow A = \frac{22}{7}\times {\left(\frac{1}{2}\right)}^2 {\mathrm{m}}^2 \\ \Rightarrow A =\left(\frac{22}{7}\times \frac{1}{2}\times \frac{1}{2}\right) {\mathrm{m}}^2 = \left(\frac{22}{28}\right) {\mathrm{m}}^2 = 0.785 {\mathrm{m}}^2. \end{gathered}$$

Question 6:

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

Answer 6:

We have :
Area of the circle (A) = ​${\mathrm{\pi r}}^2$ = 50.24 m²
$$\begin{gathered} \Rightarrow 50.24 {\mathrm{m}}^2= \frac{22}{7}\times {\left(r\right)}^2 \\ \Rightarrow r^2 =\left(\frac{50.24\times 7}{22}\right) {\mathrm{m}}^2 = \left(\frac{351.68}{22}\right) {\mathrm{m}}^2 = 15.985 {\mathrm{m}}^2 \\ \Rightarrow r = 3.998 \mathrm{m}. \end{gathered}$$

Circumference of circle (C) = $2\mathrm{\pi r}$
 $$\begin{gathered} \Rightarrow C =\left(2\times \frac{22}{7} \mathrm{x} 3.998\right) \mathrm{m} \\ \Rightarrow C=\left(44\times 0.571\right) \mathrm{m} = 25.12 \mathrm{m}. \end{gathered}$$ 

Question 7:

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

Answer 7:

We have :
Length of the string = ​28 m
The area over which the horse can graze is the same as the area of a circle of radius 28 m.
Hence, required area = π​r² = $\left(\frac{22}{7}\times 28\times 28\right){\mathrm{m}}^2 = \left(22\times 4\times 28\times \right){\mathrm{m}}^2 = 2464 {\mathrm{m}}^2$.

Question 8:

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.

Answer 8:

We have :
Area of the square = 121 cm² 
⇒ (side)² = (11)² cm² 
⇒ side =  11 cm.
So, the perimeter of the square = 4(side) = (4 x 11) cm = 44 cm.

Let r be the radius of the circle. Then,
Circumference of the circle = Perimeter of the square
  ⇒     2πr =  44
  ⇒     2 x $\frac{22}{7}$x r =  44
 ⇒ r = 7 cm.
∴ Area of the circle = π​r²  = $\left(\frac{22}{7}\times 7\times 7\right) = \left(22\times 7\right) = 154 {\mathrm{cm}}^2$ .

Question 9:

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

Answer 9:

We have :
Circumference of the circular park = 2πr = 352 m
  ​  ⇒     2 x $\frac{22}{7}$ x r =  352
   ⇒ r = 56 m.
Radius of the path including the 7m wide road = (r +7) =  56 +7 = 63 m.
∴ Area of the road :
 $$\begin{gathered} =\left[\mathrm{\pi }{\left(63\right)}^2 - \mathrm{\pi }{\left(56\right)}^2\right] m^2 \\ =\mathrm{\pi }\left[{\left(63\right)}^2 - {\left(56\right)}^2\right] m^2 \\ =\mathrm{\pi }\left[\left(63 +56\right)\left(63 - 56\right)\right] {\mathrm{m}}^2 \\ =\mathrm{\pi }\left[\left(119\right)\left(7\right)\right] \\ =\left(\frac{22}{7}\times 119\times 7\right) {\mathrm{m}}^2 = \left(22\times 119\right) {\mathrm{m}}^2 = 2618 {\mathrm{m}}^2. \end{gathered}$$

 

Question 10:

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

Answer 10:

Radius  of the circular region = r
​Radius of the circular path of uniform width h surrounding the circular region of radius r = (r + h).
​∴ Area of the path
 $$\begin{gathered} =\left[\mathrm{\pi }{\left(r+h\right)}^2-{\mathrm{\pi r}}^2\right] \\ =\mathrm{\pi }\left[{\left(\mathrm{r}+\mathrm{h}\right)}^2-{\mathrm{r}}^2\right] \\ =\mathrm{\pi }\left[{\mathrm{r}}^2 + 2\mathrm{rh} +{\mathrm{h}}^2 -{\mathrm{r}}^2\right] \\ =\mathrm{\pi }\left[2\mathrm{rh}+{\mathrm{h}}^2\right] \\ =\mathrm{\pi h}\left[2\mathrm{r}+\mathrm{h}\right] \end{gathered}$$

Page-21.16

Question 11:

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

Answer 11:

We have :
    Given perimeter of the circle = 4$\mathrm{\pi }$r cm = 2​$\mathrm{\pi }$ (2r) cm

    We know that, the perimeter of a circle = 2$\mathrm{\pi }$r

    ∴ Radius of the circle = 2r cm

    Area of the circle= $\mathrm{\pi }$r²   = ​$\mathrm{\pi }$ (2r)² = 4​$\mathrm{\pi }$r^{2 .}

Question 12:

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.

Answer 12:

We have:
    Perimeter of the square = 5024 m = Circumference of the circle
⇒​ 4 x Side of the square = 5024
∴ Side of the square = $\frac{5024}{4} = 1256 \mathrm{m}$.
Let the area of the square be A₁ and the area of the circle be A₂.
Area of the square (A₁)= side x side =$\left(\frac{5024}{4} \times \frac{5024}{4}\right){\mathrm{m}}^2$ .
Circumference of the circle = 5024 m
    ⇒ 2 $\mathrm{\pi }$r = 5024 m

  $$\begin{gathered} \Rightarrow \left(2\times \frac{22}{7}\times r\right)=5024 \mathrm{m} \\ \Rightarrow r = \frac{5024\times 7}{2\times 22}. \end{gathered}$$
 

Area of the circle (A₂)= $\mathrm{\pi }$r²   = ​$\left[\frac{22}{7}\times \frac{5024\times 7}{2\times 22}\times \frac{5024\times 7}{2\times 22}\right]=\left[\frac{5024\times 5024\times \times 7}{2\times 2\times 22}\right]{\mathrm{m}}^2$.

 $$\begin{gathered} \therefore A_1:A_2= \left[\frac{5024}{4}\times \frac{5024}{4}\right] : \left[\frac{5024\times 5024\times \times 7}{2\times 2\times 22}\right] \\ \Rightarrow \frac{A_1}{A_2}= \left[\frac{5024}{4}\times \frac{5024}{4}\right] ÷ \left[\frac{5024\times 5024\times \times 7}{2\times 2\times 22}\right] \\ \Rightarrow \frac{A_1}{A_2}=\left[\frac{5024}{4}\times \frac{5024}{4}\right]\times \left[\frac{2\times 2\times 22}{5024\times 5024\times \times 7}\right] \\ \Rightarrow \frac{A_1}{A_2}=\frac{11}{14}. \\ \Rightarrow A_1:A_{2 }= 11:14. \end{gathered}$$
Hence, the ratio of the area of the square to the area of the circle is 11:14.

Question 13:

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

Answer 13:

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 r₂ and A₂, respectively.

Thus,
A₁ = $\mathrm{\pi }{\left(\mathrm{r}\right)}^2 = \mathrm{\pi }{\left(14\right)}^2 = \left(\frac{22}{7}\times 14\times 14\right) {\mathrm{cm}}^2=\left(44\times 14\right) {\mathrm{cm}}^2= 616 {\mathrm{cm}}^2$.
A₂ = 2 $\times$ A₁ = 2 $\times$ 616 = 1232 cm²
A₂ = $\mathrm{\pi }{\left({\mathrm{r}}_2\right)}^2$ = 1232 cm^{2
 $$\begin{gathered} \Rightarrow \left(\frac{22}{7}\times {r_2}^2\right) = 1232 {\mathrm{cm}}^2 \\ \Rightarrow {r_2}^2 = \frac{1232\times 7}{22} = (56\times 7) {\mathrm{cm}}^2 \\ \Rightarrow r_2 = \sqrt{56\times 7} = \sqrt{7\times 8\times 7} =\sqrt{7\times 7\times 4\times 2} = 14\sqrt{2} \mathrm{cm} \end{gathered}$$}

Hence, the radius of the circle A₂ is 14$\sqrt{2}$ cm.

Question 14:

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.

Answer 14:

Let the area of the circle whose radius is 20 m be A₁ , and the area of the circle whose radius
is 48 m be A₂. Let A₃ 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.
∴ A₃ =  A₁ + A₂

 $$\begin{gathered} A_1 = \mathrm{\pi }{\left(20\right)}^2 = \left(\frac{22}{7}\times 20\times 20\right) {\mathrm{m}}^2=\left(400\mathrm{\pi }\right) {\mathrm{m}}^2 \\ {A}_2 = \mathrm{\pi } (48{)}^2 = \left(\frac{22}{7}\times 48\times 48\right) {\mathrm{m}}^2=\left(2304\mathrm{\pi }\right) {\mathrm{m}}^2 \\ {A}_3 = A_1+A_2 = \left(400\mathrm{\pi }\right) +\left(2304\mathrm{\pi }\right) = \mathrm{\pi }\left(400 +2304\right) {\mathrm{m}}^2 \\ \Rightarrow A_3 = \mathrm{\pi }{\left(r\right)}^2 = \mathrm{\pi }\left(400 +2304\right) {\mathrm{m}}^2 \\ \Rightarrow {\left(r\right)}^2 = \left(400 +2304\right) {\mathrm{m}}^2 \\ \Rightarrow r = \sqrt{2704} \mathrm{m} = 52 \mathrm{m} \end{gathered}$$

Question 15:

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.

Answer 15:

Let the area of the circular field whose radius is 5 m be A₁ , and the area of the circular field whose radius is 13 m be A₂. Let A₃ 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.
∴ A₃ = A₂ - A₁

 $$\begin{gathered} A_1 = \mathrm{\pi }{\left(5\right)}^2 = \left(25\mathrm{\pi }\right) {\mathrm{m}}^2 \\ {A}_2 = \mathrm{\pi } (13{)}^2 = \left(169 \mathrm{\pi }\right) {\mathrm{m}}^2 \\ {A}_3 = A_2-A_1 = \left(169\mathrm{\pi }\right) -\left(25 \mathrm{\pi }\right) = 144\mathrm{\pi } {\mathrm{m}}^2 \\ \Rightarrow A_{\mathit{3}} = \mathrm{\pi }{\left(r\right)}^2 = 144 \mathrm{\pi } {\mathrm{m}}^2 \\ \Rightarrow {\left(r\right)}^2 = 144 {\mathrm{m}}^2 \\ \Rightarrow r = \sqrt{144} \mathrm{m} = 12 \mathrm{m} \end{gathered}$$

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

Question 16:

Two circles are drawn inside a big circle with diameters $\frac{2}{3}\mathrm{rd}$ and $\frac{1}{3}\mathrm{rd}$ 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.

Answer 16:

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 = $\frac{2}{3}\times 18=12 \mathrm{cm}$
Radius of the 1st circle = 6 cm
Diameter of the 2nd circle = $\frac{1}{3}\times 18=6 \mathrm{cm}$
Radius of the 2nd circle = 3 cm
Area of the 1st circle = $\mathrm{\pi }(6{)}^2=36{\mathrm{\pi cm}}^2$
Area of the 2nd circle = $\mathrm{\pi }(3{)}^2=\mathrm{\pi }\times 3\times 3=9\mathrm{\pi } {\mathrm{cm}}^2$
Area of the big circle = $\mathrm{\pi }(9{)}^2=\mathrm{\pi }\times 9\times 9=81\mathrm{\pi } {\mathrm{cm}}^2$
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 = $81\mathrm{\pi }-(36\mathrm{\pi }+9\mathrm{\pi })=36\mathrm{\pi }$ cm².

Question 17:

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.

Answer 17:

Radius of the quarter circular plot = 2 m
Area of the quarter circular plot = $\mathrm{\pi }(2{)}^2=\frac{22}{7}\times 4==12.57 {\mathrm{m}}^2$
Radius of each flower bed = 2 m
Area of four flower beds = $4\times \frac{1}{4}\times \mathrm{\pi }(2{)}^2=12.57 {\mathrm{m}}^2$
Area of the rectangular region = Length $\times$ Breadth
Area of the rectangular region = 8 $\times$ 6 = 48 m²
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)] m² = 22.86 m^{2 .}

Question 18:

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

Answer 18:

Side of the square = 10 cm
Area of the square = Side $\times$ Side
Area of the square = ^{$10\times 10=100 {\mathrm{cm}}^2$}

Area of the four quarter circles = ^{$4\times \frac{1}{4}\times \frac{22}{7}\times 5^2=78.57 {\mathrm{cm}}^2$}
Area included in them = Area of the square - Area of the four quarter circles
Area included in them = ( 100 - 78.57 ) cm² = 21.43 cm² .

Question 19:

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

Answer 19:

Let the area of the first circle be A₁, the circumference be  C₁ and the radius be r₁.
Let the area of the second circle be A_2 , the circumference be C₂  and the radius be r₂.

Thus,

$$\begin{gathered} C_1: C_2=2\mathrm{\pi }{r}_{1 }: 2\mathrm{\pi }{r}_2 \\ \Rightarrow \frac{C_1}{C_2} = \frac{2\mathrm{\pi }{r}_1}{2{\mathrm{\pi r}}_2} = \frac{r_1}{r_2} \end{gathered}$$

We know that :
         
   A₁ =  100A_{2
∴ $$\begin{gathered} \Rightarrow \left({{\mathrm{\pi r}}_1}^2\right) = 100\times \left({{\mathrm{\pi r}}_2}^2\right) \\ \Rightarrow {{\mathrm{r}}_1}^2 = 100 \times {{\mathrm{r}}_2}^2 \\ \Rightarrow {\mathrm{r}}_1 = 10 \times {\mathrm{r}}_2 \\ \Rightarrow \frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}= 10 \end{gathered}$$}

 Substituting the values, we get:
∴​ $$\begin{gathered} \frac{C_1}{C_2}=\frac{r_1}{r_2} = \frac{10}{1} \\ {C}_1:C_2 =10:1 \end{gathered}$$ 

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