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

Exercise 20F

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Question 1:

Find the area of a circle whose radius is

(i) 21 cm.
(ii) 3.5 m.

Answer 1:

(i) Given:
r = 21 cm
    
∴ Area of the circle = $\left({\mathrm{\pi r}}^2\right)$ sq. units
                                   = $\left(\frac{22}{7}\times 21\times 21\right)$ cm² = $\left(22\times 3\times 21\right)$ cm² = 1386 cm²

(ii) Given:
r = 3.5 m
    
Area of the circle = $\left({\mathrm{\pi r}}^2\right)$ sq. units
                                   = $\left(\frac{22}{7}\times 3.5\times 3.5\right)$ m² = $\left(22\times 0.5\times 3.5\right)$ m² = 38.5 m²

Question 2:

Find the area of a circle whose diameter is

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

Answer 2:

(i) Given:
d = 28 cm ⇒ r = $\left(\frac{d}{2}\right)$ = $\left(\frac{28}{2}\right)$ cm = 14 cm
Area of the circle = $\left({\mathrm{\pi r}}^2\right)$ sq. units
                                   = $\left(\frac{22}{7}\times 14\times 14\right)$ cm² = $\left(22\times 2\times 14\right)$ cm² = 616 cm²

(ii) Given:
r = 1.4 m ⇒ r = $\left(\frac{d}{2}\right)$ = $\left(\frac{1.4}{2}\right)$m = 0.7 m
      Area of the circle = $\left({\mathrm{\pi r}}^2\right)$ sq. units
                                   = $\left(\frac{22}{7}\times 0.7\times 0.7\right)$ m² = $\left(22\times 0.1\times 0.7\right)$ m² = 1.54 m²

Question 3:

The circumference of a circle is 264 cm. Find its area.

Answer 3:

Let the radius of the circle be r cm.
Circumference = $\left(2\mathrm{\pi r}\right)$cm
∴ $\left(2\mathrm{\pi r}\right)$ = 264
⇒ $\left(2\times \frac{22}{7}\times r\right)=264$
⇒ r = $\left(\frac{264\times 7}{2\times 22}\right)$ = 42
∴ Area of the circle = ${\mathrm{\pi r}}^2$
                                 = $\left(\frac{22}{7}\times 42\times 42\right)$ cm²
                                = 5544 cm²

Question 4:

The circumference of a circle is 35.2 m. Find its area.

Answer 4:

Let the radius of the circle be r m.
Then, its circumference will be $\left(2\mathrm{\pi r}\right)$m.
∴ $\left(2\mathrm{\pi r}\right)$ = 35.2
⇒ $\left(2\times \frac{22}{7}\times r\right)=35.2$
⇒ r = $\left(\frac{35.2\times 7}{2\times 22}\right)$ = 5.6
∴ Area of the circle = ${\mathrm{\pi r}}^2$
                                 = $\left(\frac{22}{7}\times 5.6\times 5.6\right)$ m² = 98.56 m²

Question 5:

The area of a circle is 616 cm². Find its circumference.

Answer 5:

Let the radius of the circle be r cm.
Then, its area will be ${\mathrm{\pi r}}^2$ cm².
∴ ${\mathrm{\pi r}}^2$ = 616
⇒ $\left(\frac{22}{7}\times r\times r\right)$ = 616
⇒ r² = $\left(\frac{616\times 7}{22}\right)$ = 196
⇒ r = $\sqrt{196}$ = 14
⇒  Circumference of the circle = $\left(2\mathrm{\pi r}\right)$ cm
                                                  = $\left(2\times \frac{22}{7}\times 14\right)$ cm = 88 cm

Question 6:

The area of a circle is 1381 m². Find its circumference.

Answer 6:

Let the radius of the circle be r m.
Then, area = ${\mathrm{\pi r}}^2$ m²
∴ ${\mathrm{\pi r}}^2$ = 1386
⇒ $\left(\frac{22}{7}\times r\times r\right)$ = 1386
⇒ r² = $\left(\frac{1386\times 7}{22}\right)$ = 441
⇒ r = $\sqrt{441}$ = 21
⇒  Circumference of the circle = $\left(2\mathrm{\pi r}\right)$ m
                                                  = $\left(2\times \frac{22}{7}\times 21\right)$ m = 132 m

Question 7:

The ratio of the radii of two circles is 4 : 5. Find the ratio of their areas.

Answer 7:

Let r₁ and r₂ be the radii of the two given circles and A₁ and A₂ be their respective areas.

$\frac{r_1}{r_2}=\frac{4}{5}$
∴ $\frac{A_1}{A_2}=\frac{\mathrm{\pi }{r_1}^2}{\mathrm{\pi }{r_2}^2}=\frac{{r_1}^2}{{r_2}^2}={\left(\frac{r_1}{r_2}\right)}^2={\left(\frac{4}{5}\right)}^2=\frac{16}{25}$
Hence, the ratio of the areas of the given circles is 16:25.

Question 8:

A horse is tied to a pole in a park with a string 21 m long. Find the area over which the horse can graze.

Answer 8:

If the horse is tied to a pole, then the pole will be the central point and the area over which the horse will graze will be a circle. The string by which the horse is tied will be the radius of the circle.
Thus,
Radius of the circle (r) = Length of the string = 21 m

Now, area of the circle = ${\mathrm{\pi r}}^2$ = $\left(\frac{22}{7}\times 21\times 21\right)$ m² = 1386 m²
∴ Required area = 1386 m²

Question 9:

A steel wire when bent in the form of a square encloses an area of 121 cm². The same wire is bent in the form of a circle. find the area of the circle.

Answer 9:

Let a be one side of the square.
Area of the square = 121 cm²                (given)
                 ⇒ a² = 121
                 ⇒ a = 11 cm   (since 11  × 11 = 121)
Perimeter of the square = 4  × side = 4a = (4  × 11) cm = 44 cm
Length of the wire = Perimeter of the square
                               = 44 cm
The wire is bent in the form of a circle.
Circumference of a circle = Length of the wire
∴ Circumference of a circle = 44 cm
⇒ $2\mathrm{\pi r}=44$
⇒ $\left(2\times \frac{22}{7}\times r\right)=44$
⇒ r = $\left(\frac{44\times 7}{2\times 22}\right)$= 7 cm
∴ Area of the circle = ${\mathrm{\pi r}}^2$
                                 = $\left(\frac{22}{7}\times 7\times 7\right)$ cm²
                                 = 154 cm²

Question 10:

A wire in a circular shape of radius 28 cm. If it is bent in the form of a square, what will be the area of the square formed?

Answer 10:

It is given that the radius of the circle is 28 cm.

Length of the wire = Circumference of the circle
⇒ Circumference of the circle  = $2\mathrm{\pi r}=\left(2\times \frac{22}{7}\times 28\right)$ cm = 176 cm
Let the wire be bent into the form of a square of side a cm.

Perimeter of the square = 176 cm

⇒ 4a = 176
⇒ a = $\left(\frac{176}{4}\right)$cm = 44 cm
Thus, each side of the square is 44 cm.

Area of the square = (Side)² = (a)² = (44 cm)²
                                                        = 1936 cm²
∴ Required area of the square formed = 1936 cm²

Question 11:

A rectangular sheet of acrylic is 34 cm by 24 cm. From it, 64 circular buttons, each of diameter 3.5 cm, have been cut out. Find the area of the remaining sheet.

Answer 11:

Area of the acrylic sheet = 34 cm  × 24 cm = 816 cm²
Given that the diameter of a circular button is 3.5 cm.
∴ Radius of the circular button (r)= $\left(\frac{3.5}{2}\right)$ cm = 1.75 cm
∴ Area of 1 circular button = ${\mathrm{\pi r}}^2$
                                                 = $\left(\frac{22}{7}\times 1.75\times 1.75\right)$ cm²
                                                 = 9.625 cm²
∴ Area of 64 such buttons = (64  × 9.625) cm² = 616 cm²
Area of the remaining acrylic sheet = (Area of the acrylic sheet - Area of 64 circular buttons)
                                                                    = (816 - 616) cm² = 200 cm²

Question 12:

A rectangular ground is 90 m long and 32 m broad. In the middle of the ground there is a circular tank of radius 14 metres. Find the cost of turfing the remaining portion at the rate of Rs 50 per square metre.

Answer 12:

Area of the rectangular ground = 90 m  × 32 m = (90  × 32) m² = 2880 m²
Given:
Radius of the circular tank (r) = 14 m
∴ Area covered by the circular tank = ${\mathrm{\pi r}}^2$ = $\left(\frac{22}{7}\times 14\times 14\right)$ m²
                                                               = 616 m²

∴ Remaining portion of the rectangular ground for turfing = (Area of the rectangular ground - Area covered by the circular tank)
                                                                                          = (2880 - 616) m² = 2264 m²
Rate of turfing = Rs 50 per sq. metre
∴ Total cost of turfing the remaining ground = Rs (50  × 2264) = Rs 1,13,200

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Question 13:

In the given figure, four equal circles are described about the four corners of a square so that each circle touches two of the circle as shown in the figure. find the area of the shaded region, each side of the square measuring 14 cm.

Answer 13:

Area of each of the four quadrants is equal to each other with radius 7 cm.

Area of the square ABCD = (Side)² = (14 cm)² = 196 cm²
Sum of the areas of the four quadrants = $\left(4\times \frac{1}{4}\times \frac{22}{7}\times 7\times 7\right)$ cm²
                                                        = 154 cm²
∴ Area of the shaded portion  = Area of square ABCD - Areas of the four quadrants
                                          = (196 - 154) cm²
                                          = 42 cm²

Question 14:

A horse is tethered to one corner of a rectangular field, 60 m by 40 m, by a rope 14 m long. On how much area can the horse graze?

Answer 14:

Let ABCD be the rectangular field.

Here, AB = 60 m
BC = 40 m

Let the horse be tethered to corner A by a 14 m long rope.

Then, it can graze through a quadrant of a circle of radius 14 m.
∴ Required area of the field = $\left(\frac{1}{4}\times \frac{22}{7}\times 14\times 14\right)$ m² = 154 m²
Hence, horse can graze 154 m² area of the rectangular field.

Question 15:

In the given figure, a circle of diameter 21 cm is given. Inside this circle, two circles with diameters $\frac{2}{3}$ and $\frac{1}{3}$ of the diameter of the big circle have been drawn, as shown in the given figure. Find the area of the shaded region.

Answer 15:

Diameter of the big circle = 21 cm
Radius = $\left(\frac{21}{2}\right)$ cm = 10.5 cm
∴ Area of the bigger circle = ${\mathrm{\pi r}}^2$ = $\left(\frac{22}{7}\times 10.5\times 10.5\right)$ cm²
                                          = 346.5 cm²


Diameter of circle I  = $\frac{2}{3}$ of the diameter of the bigger circle
                                 = $\frac{2}{3}$ of 21 cm = $\left(\frac{2}{3}\times 21\right)$ cm = 14 cm
Radius of circle I (r₁) = $\left(\frac{14}{2}\right)$ cm = 7 cm
∴ Area of circle I = ${{\mathrm{\pi r}}_1}^2$ = $\left(\frac{22}{7}\times 7\times 7\right)$ cm²
                                         = 154 cm²

Diameter of circle II  = $\frac{1}{3}$ of the diameter of the bigger circle
                                   = $\frac{1}{3}$ of 21 cm = $\left(\frac{1}{3}\times 21\right)$ cm = 7 cm
Radius of circle II (r₂) = $\left(\frac{7}{2}\right)$ cm = 3.5 cm
∴ Area of circle II = ${{\mathrm{\pi r}}_2}^2$ = $\left(\frac{22}{7}\times 3.5\times 3.5\right)$ cm²
                                          = 38.5 cm²

∴ Area of the shaded portion = {Area of the bigger circle - (Sum of the areas of circle I and II)}
                                      = {346.5 - (154 + 38.5)} cm²
                                      = {346.5 - 192.5} cm²
                                              = 154 cm²
Hence, the area of the shaded portion is 154 cm²

Question 16:

In the given figure a rectangular plot of land measures 8 m by 6 m. In each of the corners, there is a flower bed in the form of a quadrant of a circle of radius 2 m. Also, there is a flower bed in the area of the remaining plot.

Answer 16:

Let ABCD be the rectangular plot of land that measures 8 m by 6 m.
∴ Area of the plot = (8 m  × 6 m) = 48 m²
Area of the four flower beds = $\left(4\times \frac{1}{4}\times \frac{22}{7}\times 2\times 2\right)$ m² = $\left(\frac{88}{7}\right)$ m²
Area of the circular flower bed in the middle of the plot = ${\mathrm{\pi r}}^2$
                                                                                       = $\left(\frac{22}{7}\times 2\times 2\right)$ m² = $\left(\frac{88}{7}\right)$ m²

Area of the remaining part = $\left\{48-\left(\frac{88}{7}+\frac{88}{7}\right)\right\}$ m²
                                           = $\left\{48-\frac{176}{7}\right\}$ m²
                                           = $\left\{\frac{336-176}{7}\right\}$ m² = $\left(\frac{160}{7}\right)$ m² = 22.86 m²
∴ Required area of the remaining plot = 22.86 m²