RD Sharma solution class 7 chapter 21 Mensuration II Objective Type Questions

Objective Type Questions

Page-21.17

Question 1:

The ratio of the perimeter (circumference) and diameter of a circle is

(a) $\pi$                              (b) 2$\pi$                               (c) $\frac{\pi }{2}$                               (d) $\frac{\pi }{4}$

Answer 1:

Let r be the radius of the circle. Then
Perimeter of circle = 2$\pi$r
Diameter of circle = 2r
Now
$\frac{\mathrm{Perimeter} \mathrm{of} \mathrm{circle}}{\mathrm{Diameter} \mathrm{of} \mathrm{circle}}=\frac{2\pi r}{2r}=\pi$
Thus, the required ratio is $\pi$.
Hence, the correct option is (a)

Question 2:

The ratio of the area and circumference of a circle of diameter d is

(a) d                              (b) $\frac{d}{2}$                               (c) $\frac{d}{4}$                               (d) 2d
 

Answer 2:

Let r and d be respectively the radius and diameter of the circle. Then
d = 2r
Circumference of circle = 2$\pi$r = $2\pi \times \frac{d}{2}=\pi d$
Area of circle = $\pi$r² = $\pi {\left(\frac{\mathit{d}}{\mathit{2}}\right)}^{\mathit{2}}=\frac{\pi d^2}{4}$
Now
$\frac{\mathrm{Area} \mathrm{of} \mathrm{circle}}{\mathrm{Circumference} \mathrm{of} \mathrm{circle}}=\frac{{\displaystyle \frac{\pi d^2}{4}}}{\pi d}=\frac{d}{4}$
Hence, the correct option is (c)

Question 3:

The cost of fencing a circular garden of radius 21 m at ₹10 per metre is

(a) ₹1320                           (b) ₹132                               (c) ₹1200                               (d) ₹660

Answer 3:

Radius (r) = 21 m
Cost per metre = ₹10
Circumference of circle = $2\pi r=2\times \frac{22}{7}\times 21=132 \mathrm{m}$
Cost of fencing = Circumference $\text{×}$ Cost per metre
                          = 132 $\text{×}$ ₹10
                          = ₹1320
Hence, the correct option is (a).

Question 4:

If the diameter of a circle is equal to the diagonal of a square, then the ratio of their areas is

(a) 7 : 1                       (b) 1 : 1                           (c) 11 : 7                           (d) 22 : 7

Answer 4:

Let r and a be the diameter of the circle and side of the square respectively. Then
Diameter of circle = 2r
Diagonal of square = $a\sqrt{2}$
Now, as per the question
Diameter of circle = Diagonal of square
2r = $a\sqrt{2}$  $\Rightarrow a=\sqrt{2}r$
Therefore
$\frac{\mathrm{Area} \mathrm{of} \mathrm{circle}}{\mathrm{Area} \mathrm{of} \mathrm{square}}=\frac{\pi r^2}{a^2}=\frac{{\displaystyle \frac{22}{7}}\times r^2}{{\left(\sqrt{2}r\right)}^2}=\frac{{\displaystyle \frac{22}{7}\times r^2}}{2r^2}=\frac{{\displaystyle 11}}{7}$
Hence, the correct option is (c).

Question 5:

A circle is inscribed in a square of side 14 m.  The ratio of the area of the circle and that of
the square is

(a) $\mathit{\text{π}}$ : 3                       (b) $\mathit{\text{π}}$ : 4                           (c) $\mathit{\text{π}}$ : 2                           (d) $\mathit{\text{π}}$ : 1

Answer 5:

Let a and r be the side of the square and radius of the circle respectively.
Here, the diameter of the circle is equal to the side of the square. So
Diameter of circle = 2r = a
Therefore
$\frac{\mathrm{Area} \mathrm{of} \mathrm{circle}}{\mathrm{Area} \mathrm{of} \mathrm{square}}=\frac{\pi r^2}{a^2}=\frac{{\displaystyle \pi \times r^2}}{{\left(2r\right)}^2}=\frac{{\displaystyle \pi }}{4}$
Hence, the correct option is (b).

Question 6:

How many times should a wheel of radius 7 m rotate to go around the perimeter
of a rectangular field of length 60 m and breadth 50 m?

(a) 3                       (b) 4                           (c) 5                           (d) 6

Answer 6:

Here, Radius (r) = 7 m, Length (l) = 60 m and Breadth (b) = 50 m.
Perimeter of circle = $2\pi r=2\times \frac{22}{7}\times 7=44 \mathrm{m}$
Perimeter of rectangle = $2\left(l+b\right)=2\left(60+50\right)=220 \mathrm{m}$
Therefore
Number of turns = $\frac{\mathrm{Perimeter} \mathrm{of} \mathrm{rectangle}}{\mathrm{Perimeter} \mathrm{of} \mathrm{circle}}=\frac{220}{44}=5$
Hence, the correct option is (c).

Question 7:

The minute hand of a clock is 14 cm long. How far does the tip of the minute hand
move in 60 minutes?
(a) 22 cm                    (b) 44 cm                           (c) 33 cm                           (d) 88 cm

Answer 7:

Length of minute hand = 14 cm
Distance covered by minute hand in one round = $2\pi r=2\times \frac{22}{7}\times 14=88 \mathrm{cm}$
Thus, the minute hand move 88 cm in 60 minutes.
Hence, the correct option is (d).

Question 8:

The cost of fencing a semi-circular garden of radius 14 m at ₹10 per metre is 

(a) ₹1080                   (b) ₹1020                       (c) ₹700                        (d) ₹720

Answer 8:

Radius of circle (r) = 14 m
Perimeter of semi-circular garden
   $$\begin{gathered} =\pi r+2r \\ =\frac{22}{7}\times 14+2\times 14 \\ =44+28 \\ =72 \mathrm{m} \end{gathered}$$
Cost of fencing = 72 $\text{×}$ ₹10 = ₹720
Hence, the correct option is (d).

Question 9:

The area of a square is equal to the area of a circle. The ratio between the side of the square
and the radius of the circle is

(a) $\sqrt{\pi }$ : 1               (b) 1 : $\sqrt{\pi }$                       (c) 1 : $\pi$                  (d) $\pi$ : 1

Answer 9:

Let a and r be respectively the side of the square and radius of the circle.
Here, the area of square is equal to the area of the circle. So
                                                      $$\begin{gathered} a^2=\pi r^2 \\ \Rightarrow \frac{{\mathit{a}}^{\mathit{2}}}{{\mathit{r}}^{\mathit{2}}}\mathit{=}\pi \\ \Rightarrow \frac{a}{r}=\sqrt{\pi } \end{gathered}$$
Hence, the correct option is (a).

Question 10:

If A is the area and C be the circumference of a circle, then its radius is
(a) $\frac{A}{C}$                (b) $\frac{2A}{C}$                       (c) $\frac{3A}{C}$                  (d) $\frac{4A}{C}$

Answer 10:

Let r be the radius of the circle. Then
$$\begin{gathered} A=\pi r^2 and C=2\pi r \\ \Rightarrow \frac{A}{C}=\frac{\pi r^2}{\mathit{2}\pi r} \\ \Rightarrow \frac{A}{C}=\frac{r}{\mathit{2}} \\ \Rightarrow r=\frac{2A}{C} \end{gathered}$$
Hence, the correct option is (b).

Question 11:

The area of a circle of circumference C is

(a) $\frac{C^2}{4\pi }$                (b) $\frac{C^2}{2\pi }$                       (c) $\frac{C^2}{\pi }$                  (d) $\frac{4C^2}{\pi }$

Answer 11:

Let r be the radius of the circle. Then
$C=2\pi r \Rightarrow r=\frac{C}{\mathit{2}\pi }$
Therefore
Area of circle = $\pi r^2=\pi {\left(\frac{C}{2\pi }\right)}^{\mathit{2}}=\frac{C^2}{4\pi }$
Hence, the correct option is (a).

Question 12:

The circumference of a circle is 44 cm. Its area is

(a) 77 cm²                        (b) 154 cm²                       (c) 208 cm²                  (d) 144 cm²
 

Answer 12:

Let r be the radius of the circle. Then
$44=2\pi r \Rightarrow r=\frac{44}{2\times {\displaystyle \frac{22}{7}}}=7 \mathrm{cm}$
Therefore
Area of circle = $\pi r^2=\frac{22}{7}\times 7^2=154 {\mathrm{cm}}^2$
Hence, the correct option is (b).

Question 13:

Each side of an equilateral triangle is equal to the radius of a circle whose area is 154 cm².
The area of the triangle is

(a) $\frac{7\sqrt{3}}{4}$ cm²                        (b) $\frac{49\sqrt{3}}{2}$ cm²                       (c) $\frac{49\sqrt{3}}{4}$ cm²                  (d) $\frac{7\sqrt{3}}{2}$ cm²

Answer 13:

Let r be the radius of the circle and a be the side of the equilateral triangle. Then
Area of circle = 154 cm²
$\Rightarrow 154=\frac{22}{7}\times r^2 \Rightarrow r=\sqrt{\frac{154\times 7}{22}}=7 \mathrm{cm}$
Therefore
Area of equilateral triangle = $\frac{a^2\sqrt{3}}{4}=\frac{7^2\sqrt{3}}{4}=\frac{49\sqrt{3}}{4} \left(\because a=r=7 \mathrm{cm}\right)$
Hence, the correct option is (c).

Page-21.18

Question 14:

The area of a circle is 9$\mathit{\text{π}}$ cm². Its circumference is

(a) 6$\mathit{\text{π}}$ cm                        (b) 36$\mathit{\text{π}}$ cm                       (c) 9$\mathit{\text{π}}$ cm                  (d) 36${\mathit{\text{π}}}^{\mathit{2}}$ cm

Answer 14:

Let r be the radius of the circle. Then
Area of circle = 9$\mathit{\text{π}}$ cm²
$\Rightarrow \pi r^2=9\pi \Rightarrow r=3 \mathrm{cm}$
Therefore
Circumference of the circle = $2\pi r\mathit{=}\mathit{2}\pi \times 3=6\pi \mathit{ }\mathrm{cm}$
Hence, the correct option is (a).

Question 15:

The area of a circle is increased by 22 cm² when its radius is increased by 1 cm.
The original radius of the circle is

(a) 6 cm                        (b) 3 cm                       (c) 4 cm                  (d) 3.5 cm

Answer 15:

Let r be the radius of the circle. Then
Area of original circle = $\mathit{\text{π}}$r² cm²
Radius of circle after increment = (r + 1) cm
Thus,as per the question
$$\begin{gathered} \pi {\left(r+1\right)}^2\mathit{-}\pi r^2=22 \\ \Rightarrow {\left(r+1\right)}^2-r^2=\frac{22}{{\displaystyle \frac{22}{7}}}=7 \\ \Rightarrow r^2+2r+1-r^2=7 \\ \Rightarrow r=3 \mathrm{cm} \end{gathered}$$
Thus, the original radius of the circle is 3 cm.
Hence, the correct option is (b).

Question 16:

The radii of two circles are in the ratio 2 : 3. The ratio of their areas is

(a) 2 : 3                        (b) 4 : 9                       (c) 3 : 2                          (d) 9 : 4

Answer 16:

Let r₁ and r₂ be the radius of the two circles. So
$\frac{r_1}{r_2}=\frac{2}{3}$
Now
$\mathrm{Ratio} \mathrm{of} \mathrm{areas}=\frac{\pi {r_1}^2}{\pi {r_2}^2}={\left(\frac{r_1}{r_2}\right)}^2={\left(\frac{2}{3}\right)}^2=\frac{4}{9}$
Thus, the required ratio is 4 : 9.
Hence, the correct option is (b).

Question 17:

The areas of two circles are in the ratio 49 : 36. The ratio of their circumferences is

(a) 7 : 6                        (b) 6 : 7                       (c) 3 : 2                          (d) 2 : 3

Answer 17:

Let r₁ and r₂ be the radius of the two circles. Then
$$\begin{gathered} \frac{\pi {r_1}^2}{\pi {r_2}^2}=\frac{49}{36} \\ \Rightarrow {\left(\frac{r_1}{r_2}\right)}^2={\left(\frac{7}{6}\right)}^2 \\ \Rightarrow \frac{r_1}{r_2}=\frac{7}{6} \end{gathered}$$
Now
$\mathrm{Ratio} \mathrm{of} \mathrm{circumferences}=\frac{2\pi r_1}{2\pi r_2}=\frac{r_1}{r_2}=\frac{7}{6}$
Thus, the required ratio is 7 : 6.
Hence, the correct option is (a).

Question 18:

The circumferences of two circles are in the ratio 3 : 4. The ratio of their areas is

(a) 3 : 4                        (b) 4 : 3                       (c) 9 : 16                          (d) 16 : 9

Answer 18:

Let r₁ and r₂ be the radius of the two circles. Then
$\frac{2\pi r_1}{2\pi r_2}=\frac{3}{4}\Rightarrow \frac{r_1}{r_2}=\frac{3}{4}$
Now
$\mathrm{Ratio} \mathrm{of} \mathrm{areas}=\frac{\pi {r_1}^2}{\pi {r_2}^2}={\left(\frac{r_1}{r_2}\right)}^2={\left(\frac{3}{4}\right)}^2=\frac{9}{16}$
Thus, the required ratio is 9 : 16.
Hence, the correct option is (c).

Question 19:

The difference between the circumference and radius of a circle is 37 cm. The area of the circle is

(a) 111 cm²                        (b) 148 cm²                       (c) 154 cm²                          (d) 258 cm²

Answer 19:

Let r₁ and r₂ be the radius of the two circles. Then
$$\begin{gathered} 2\pi r-r=37 \\ \Rightarrow 2\times \frac{22}{7}\times r-r=37 \\ \Rightarrow \frac{44r-7r}{7}=37 \\ \Rightarrow r=\frac{37\times 7}{37}=7 \mathrm{cm} \end{gathered}$$
Now
$\mathrm{Area} \mathrm{of} \mathrm{circle}=\pi r^2=\frac{22}{7}\times 7\times 7=154 {\mathrm{cm}}^2$
Hence, the correct option is (c).