RS Aggarwal 2019,2020 solution class 7 chapter 20 Mensuration Exercise 20E

Exercise 20E

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Question 1:

Find the circumference of a circle whose radius is

(i) 28 cm.
(ii) 1.4 m.

Answer 1:

(i) Here, r = 28 cm
     ∴ Circumference = 2π r
                                  = $\left(2\times \frac{22}{7}\times 28\right)$cm
                                  =  176 cm
    Hence, the circumference of the given circle is 176 cm.

(ii) Here, r = 1.4 m
      ∴ Circumference = 2π r
                                   = $\left(2\times \frac{22}{7}\times 1.4\right)$ m
                                   =  $\left(2\times 22\times 0.2\right)$ m = 8.8 m
    Hence, the circumference of the given circle is 8.8 m.

Question 2:

Find the circumference of a circle whose diameter is

(i) 35 cm.
(ii) 4.9 m.

Answer 2:

(i) Here, d = 35 cm
     Circumference = 2π r
                              = $\left(\mathrm{\pi }d\right)$    [since 2r = d]
                              =  $\left(\frac{22}{7}\times 35\right)$ cm = (22 $\times$ 5) = 110 cm
     Hence, the circumference of the given circle is 110 cm.

(ii) Here, d = 4.9 m
      Circumference =2π r
                              = $\left(\mathrm{\pi }d\right)$    [since 2r = d]
                              =  $\left(\frac{22}{7}\times 4.9\right)$ m = (22 $\times$ 0.7) = 15.4 m
      Hence, the circumference of the given circle is 15.4 m.

Question 3:

Find the circumference of a circle of radius 15 cm. (Take π = 3.14.)

Answer 3:

Here, r = 15 cm
∴ Circumference = $2\mathrm{\pi r}$
                             = ( 2 $\times$ 3.14 $\times$ 15) cm
                             = 94.2 cm
Hence, the circumference of the given circle is 94.2 cm

Question 4:

Find the radius of a circle whose circumference is 57.2 cm.

Answer 4:

Circumference of the given circle = 57.2 cm
∴ C = 57.2 cm
Let the radius of the given circle be r cm.
C = $2\mathrm{\pi r}$
⇒ r = $\frac{\mathrm{C}}{2\mathrm{\pi }}$ cm
⇒ r = $\left(\frac{57.2}{2}\times \frac{7}{22}\right)$ cm = 9.1 cm
Thus, radius of the given circle is 9.1 cm.

Question 5:

Find the diameter of a circle whose circumference is 63.8 m.

Answer 5:

Circumference of the given circle = 63.8 m
∴ C = 63.8 m
Let the radius of the given circle be r cm.
C = $2\mathrm{\pi r}$
⇒ r = $\frac{\mathrm{C}}{2\mathrm{\pi }}$
⇒ r = $\left(\frac{63.8}{2}\times \frac{7}{22}\right)$m =10.15 m
∴ Diameter of the given circle = 2r = (2 $\times$ 10.15) m = 20.3 m

Question 6:

The circumference of a circle exceeds its diameter by 30 cm. Find the radius of the circle.

Answer 6:

Let the radius of the given circle be r cm.
Then, its circumference = $2\mathrm{\pi r}$

Given:
(Circumference) - (Diameter) = 30 cm     
∴ ($2\mathrm{\pi r}$ - 2r ) =  30
⇒ $2r\left(\mathrm{\pi }-1\right)=30$
⇒ $2r\left(\frac{22}{7}-1\right)=30$
⇒ $2r\times \frac{15}{7}=30$
⇒ $r=\left(30\times \frac{7}{30}\right)=7$
∴ Radius of the given circle = 7 cm

Question 7:

The ratio of the radii of two circle is 5 : 3. Find the ratio of their circumferences.

Answer 7:

Let the radii of the given circles be 5x and 3x, respectively.
Let their circumferences be C₁ and C_2, respectively.

C₁ = $2\times \pi \times 5x=10\mathrm{\pi }x$

C₂ = $2\times \pi \times 3x=6\mathrm{\pi }x$
∴ $\frac{C_1}{C_2}=\frac{10\pi x}{6\pi x}=\frac{5}{3}$
⇒ C₁:C₂ = 5:3
Hence, the ratio of the circumference of the given circle is 5:3.

Question 8:

How long will a man take to make a round of a circular field of radius 21 m, cycling at the speed of 8 km/h?

Answer 8:

Radius of the circular field, r = 21 m.
Distance covered by the cyclist = Circumference of the circular field
                                             = $2\mathrm{\pi r}$
                                             = $\left(2\times \frac{22}{7}\times 21\right)$ m = 132 m
Speed of the cyclist = 8 km per hour = $\frac{8000 \mathrm{m}}{(60\times 60) \mathrm{s}}=\left(\frac{8000}{3600}\right)\mathrm{m}/\mathrm{s}$ = $\left(\frac{20}{9}\right)\mathrm{m}/\mathrm{s}$
                                                                                                
Time taken by the cyclist to cover the field = $\frac{\mathrm{Distance} \mathrm{covered} \mathrm{by} \mathrm{the} \mathrm{cyclist}}{\mathrm{Speed} \mathrm{of} \mathrm{the} \mathrm{cyclist}}$
                                                                     = $\left[\frac{132}{\left({\displaystyle \frac{20}{9}}\right)}\right]\mathrm{s}$
                                                                     = $\left(\frac{132\times 9}{20}\right)\mathrm{s}$
                                                                     = 59.4 s

Question 9:

A racetrack is in the form of a ring whose inner circumference is 528 m and the outer circumference is 616 m. Find the width of the track.

Answer 9:

Let the inner and outer radii of the track be r metres and R metres, respectively.

Then, $2\mathrm{\pi r}$ = 528
$2\mathrm{\pi R}$ = 616
⇒ $2\times \frac{22}{7}\times r=528$ 
$2\times \frac{22}{7}\times R=616$
⇒ r = $\left(528\times \frac{7}{44}\right)=84$ 
R = $\left(616\times \frac{7}{44}\right)=98$
⇒ (R - r) = (98 - 84) m = 14 m
Hence, the width of the track is 14 m.

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Question 10:

The inner circumference of a circular track is 330 m. The track is 10.5 m wide everywhere. Calculate the cost of putting up a fence along the outer circle at the rate of Rs 20 per metre.

Answer 10:

Let the inner and outer radii of the track be r metres and (r + 10.5) metres, respectively.

Inner circumference = 330 m
∴ $2\mathrm{\pi r}=330$ ⇒ $2\times \frac{22}{7}\times r=330$
                      ⇒ r = $\left(330\times \frac{7}{44}\right)=52.5 \mathrm{m}$
Inner radius of the track = 52.5 m
∴ Outer radii of the track = (52.5 + 10.5) m = 63 m


∴ Circumference of the outer circle = $\left(2\times \frac{22}{7}\times 63\right) \mathrm{m}=396 \mathrm{m}$
Rate of fencing = Rs. 20 per metre
∴ Total cost of fencing the outer circle = Rs. (396 $\times$ 20) = Rs. 7920

Question 11:

One circle has radius of 98 cm and a second concentric circle has a radius of 1 m 26 cm. How much longer is the circumference of the second circle than that of the first?

Answer 11:

We know that the concentric circles are circles that form within each other, around a common centre point.
Radius of the inner circle, r = 98 cm
∴  Circumference of the inner circle = $2\mathrm{\pi r}$
                                                         = $\left(2\times \frac{22}{7}\times 98\right)$ cm = 616 cm

Radius of the outer circle, R = 1 m 26 cm = 126 cm              [since 1 m = 100 cm]
∴  Circumference of the outer circle = $2\mathrm{\pi R}$
                                                           = $\left(2\times \frac{22}{7}\times 126\right)$ cm = 792 cm
∴ Difference in the lengths of the circumference of the circles = (792 - 616) cm = 176 cm
Hence, the circumference of the second circle is 176 cm larger than that of the first circle.

Question 12:

A piece of wire is bent in the shape of an equilateral triangle each of whose sides measures 8.8 cm. This wire is rebent to form a circular ring. What is the diameter of the ring?

Answer 12:

Length of the wire = Perimeter of the equilateral triangle
                              = 3 $\times$ Side of the equilateral triangle = (3 $\times$ 8.8) cm = 26.4 cm
Let the wire be bent into the form of a circle of radius r cm.
Circumference of the circle = 26.4 cm
⇒ $2\mathrm{\pi r}=26.4$
⇒ $2\times \frac{22}{7}\times r=26.4$
⇒ r = $\left(\frac{26.4\times 7}{2\times 22}\right)$ cm = 4.2 cm

∴ Diameter = 2r = (2 × 4.2) cm = 8.4 cm
Hence, the diameter of the ring is 8.4 cm.

Question 13:

A rhombus has the same perimeter as the circumference of a circle. If each side of the rhombus measures 33 cm, find the radius of the circle.

Answer 13:

Circumference of the circle = Perimeter of the rhombus
                                           = 4  × Side of the rhombus = (4  ×  33) cm = 132 cm

∴ Circumference of the circle = 132 cm
⇒ $2\mathrm{\pi r}=132$
⇒ $2\times \frac{22}{7}\times r=132$
⇒ r = $\left(\frac{132\times 7}{2\times 22}\right)$cm = 21 cm
Hence, the radius of the circle is 21 cm.

Question 14:

A wire in the form of a rectangle 18.7 cm long and 14.3 cm wide is reshaped and bent into the form of a circle. Find the radius of the circle so formed.

Answer 14:

Length of the wire = Perimeter of the rectangle
                               = 2(l + b) = 2  × (18.7 + 14.3) cm = 66 cm

Let the wire be bent into the form of a circle of radius r cm.

Circumference of the circle = 66 cm

⇒ $2\mathrm{\pi r}=66$
⇒ $\left(2\times \frac{22}{7}\times r\right)=66$
⇒ r = $\left(\frac{66\times 7}{2\times 22}\right)$ cm = 10.5 cm

Hence, the radius of the circle formed is 10.5 cm.

Question 15:

A wire is looped in the form of a circle of radius 35 cm. If it is rebent in the form of a square, what will be the length of each side of the square?

Answer 15:

It is given that the radius of the circle is 35 cm.
Length of the wire = Circumference of the circle
⇒ Circumference of the circle  = $2\mathrm{\pi r}$ = $\left(2\times \frac{22}{7}\times 35\right)$ cm = 220 cm
Let the wire be bent into the form of a square of side a cm.
Perimeter of the square = 220 cm
⇒ 4a = 220
⇒ a = $\left(\frac{220}{4}\right)$cm = 55 cm
Hence, each side of the square will be 55 cm.

Question 16:

A well of diameter 140 cm has a stone parapet around it. If the lenght of the outer edge of the parapet is 616 cm, find the width of the parapet.

Answer 16:

Given:
Diameter of the well (d) = 140 cm.
Radius of the well (r) = $\left(\frac{140}{2}\right)$cm = 70 cm

Let the radius of the outer circle (including the stone parapet) be R cm.
Length of the outer edge of the parapet = 616 cm
⇒ $2\mathrm{\pi R}=616$
⇒ $\left(2\times \frac{22}{7}\times R\right)=616$
⇒ R = $\left(\frac{616\times 7}{2\times 22}\right)$ cm = 98 cm

Now, width of the parapet = {Radius of the outer circle (including the stone parapet) - Radius of the well}
                                           = {98 -70} cm = 28 cm
Hence, the width of the parapet is 28 cm.

Question 17:

Find the distance covered by the wheel of a bus in 2000 rotations if the diameter of the wheel is 98 cm.

Answer 17:

It may be noted that in one rotation, the bus covers a distance equal to the circumference of the wheel.
Now, diameter of the wheel = 98 cm
∴ Circumference of the wheel = $\mathrm{\pi d}$ = $\left(\frac{22}{7}\times 98\right)$ cm = 308 cm
Thus, the bus travels 308 cm in one rotation.

∴ Distance covered by the bus in 2000 rotations = (308  × 2000) cm
                                                                          = 616000 cm 
                                                                          = 6160 m          [since 1 m = 100 cm]

Question 18:

The diameter of the wheel of a cycle is 70 cm. How far will it go in 250 revolutions?

Answer 18:

It may be noted that in one revolution, the cycle covers a distance equal to the circumference of the wheel.
Diameter of the wheel = 70 cm
∴ Circumference of the wheel = $\mathrm{\pi d}$ = $\left(\frac{22}{7}\times 70\right)$ cm = 220 cm
Thus, the cycle covers 220 cm in one revolution.

∴ Distance covered by the cycle in 250 revolutions = (220  × 250) cm
                                                                               = 55000 cm 
                                                                               = 550 m              [since 1 m = 100 cm]

Hence, the cycle will cover 550 m in 250 revolutions.

Question 19:

The diameter of the wheel of a car is 77 cm. How many revolutions will it make to travel 121 km?

Answer 19:

Diameter of the wheel = 77 cm
⇒ Radius of the wheel = $\left(\frac{77}{2}\right)$ cm
Circumference of the wheel = $2\mathrm{\pi r}$
                                              = $\left(2\times \frac{22}{7}\times \frac{77}{2}\right)$cm = (22  × 11) cm = 242 cm
                                                                                                        = $\left(\frac{242}{100}\right)$m = $\left(\frac{121}{50}\right)$m
Distance covered by the wheel in 1 revolution = $\left(\frac{121}{50}\right)$ m
Now, $\left(\frac{121}{50}\right)$ m is covered by the car in 1 revolution.
(121  × 1000) m will be covered by the car in $\left(1\times \frac{50}{121}\times 121\times 1000\right)$ revolutions, i.e. 50000 revolutions.
∴ Required number of revolutions = 50000

Question 20:

A bicycle wheel makes 5000 revolutions in moving 11 km. Find the circumference and the diameter of the wheel.

Answer 20:

It may be noted that in one revolution, the bicycle covers a distance equal to the circumference of the wheel.
Total distance covered by the bicycle in 5000 revolutions = 11 km

⇒ 5000 × Circumference of the wheel = 11000 m                [since 1 km = 1000 m]

Circumference of the wheel = $\left(\frac{11000}{5000}\right)$ m =2.2 m = 220 cm               [since 1 m = 100 cm]

Circumference of the wheel = $\mathrm{\pi }\times \mathrm{Diameter} \mathrm{of} \mathrm{the} \mathrm{wheel}$
⇒ 220 cm = $\frac{22}{7}\times \mathrm{Diameter} \mathrm{of} \mathrm{the} \mathrm{wheel}$
⇒ Diameter of the wheel = $\left(\frac{220\times 7}{22}\right)$ cm = 70 cm
Hence, the circumference of the wheel is 220 cm and its diameter is 70 cm.

Question 21:

The hour and minute hands of a clock are 4.2 cm and 7 cm long respectively. Find the sum of the distances covered by their tips in 1 day.

Answer 21:

Length of the hour hand (r)= 4.2 cm.
Distance covered by the hour hand in 12 hours = $2\mathrm{\pi r}$ = $\left(2\times \frac{22}{7}\times 4.2\right)$ cm = 26.4 cm

∴ Distance covered by the hour hand in 24 hours = (2  × 26.4) = 52.8 cm
Length of the minute hand (R)= 7 cm
Distance covered by the minute hand in 1 hour = $2\mathrm{\pi R}$ = $\left(2\times \frac{22}{7}\times 7\right)$ cm = 44 cm

∴ Distance covered by the minute hand in 24 hours = (44  × 24) cm = 1056 cm

∴ Sum of the distances covered by the tips of both the hands in 1 day = (52.8 + 1056) cm
                                                                                                                = 1108.8 cm