Class 7 Perimeter and Area Ex 9.2

1. Find the circumference of the circle with the following radius. (Take \(π = \dfrac{22}{7}\))

(a) 14 cm

Solution:-

The radius of the circle = 14 cm

Circumference of the circle = 2πr

\(= 2 × \dfrac{22}{7} × 14 \)

= 2 × 22 × 2

= 88 cm

(b) 28 mm

Solution:-

The radius of the circle = 28 mm

Circumference of the circle = 2πr

\(= 2 × \dfrac{22}{7}× 28 \)

= 2 × 22 × 4

= 176 mm

(c) 21 cm

Solution:-

The radius of the circle = 21 cm

Circumference of the circle = 2πr

\(= 2 ×\dfrac{22}{7} × 21\)

= 2 × 22 × 3

= 132 cm

2. Find the area of the following circles, given that

(a) Radius = 14 mm (Take \(π = \dfrac{22}{7}\))

Solution:

The radius of the circle = 14 mm

Area of the circle \(= πr^2\)

\(=\dfrac{22}{7} × 142\)

\(= \dfrac{22}{7} × 196\)

= 22 × 28

= 616 mm\(^2\)

(b) Diameter = 49 m

Solution:

The diameter of the circle (d) = 49 m

We know that radius (r) = \(\dfrac{d}{2}\)

= \(\dfrac{49}{2}\)

= 24.5 m

Area of the circle = πr\(^2\)

= \(\dfrac{22}{7} × (24.5)^2\)

=\( \dfrac{22}{7} × 600.25\)

= 22 × 85.75

= 1886.5 m\(^2\)

(c) Radius = 5 cm

Solution:

The radius of the circle = 5 cm

Area of the circle = πr\(^2\)

= \(\dfrac{22}{7} × 5^2\)

= \(\dfrac{22}{7} × 25\)

=\(\dfrac{550}{7} \)

= 78.57 cm\(^2\)

3. If the circumference of a circular sheet is 154 m, find its radius. Also, find the area of the sheet. (Take \(\pi = \dfrac{22}{7}\)).

Solution:-

Circumference of the circle = 154 m

The circumference of the circle = 2πr

\(154 = 2 × (\dfrac{22}{7}) × r\)

\(154 = \dfrac{44}{7} × r\)

\(r = \dfrac{(154 × 7)}{44}\)

\(r = \dfrac{(14 × 7)}{4}\)

\(r = \dfrac{49}{2}\)

r = 24.5 m

Area of the circle = πr\(^2\)

\(= \dfrac{22}{7} × (24.5)^2\)

\(= \dfrac{22}{7} × 600.25\)

\(= 22 × 85.75\)

\(= 1886.5\) m\(^2\)

So, the radius of the circle is 24.5, and the area of the circle is 1886.5.

4. A gardener wants to fence a circular garden of diameter 21m. Find the length of the rope he needs to purchase, if he makes 2 rounds of the fence. Also, find the cost of the rope, if it costs ₹ 4 per meter. (Take \(π = \dfrac{22}{7}\))



Solution:-

Diameter of the circular garden = 21 m

We know that radius (r) = d/2

=\( \dfrac{21}{2}\)

= 10.5 m

Circumference of the circle = 2πr

=\( 2 × \dfrac{22}{7} × 10.5\)

= \(\dfrac{462}{7}\)

= \(66\) m

So, the length of rope required = \(2 × 66 = 132\) m

Cost of \(1\) m rope = ₹ 4 [given]

Cost of \(132\) m rope = ₹ \(4 × 132\)

= ₹ \(528\)

5. From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)

Solution:-

Radius of circular sheet R = 4 cm

A circle of radius to be removed \(r = 3\) cm

The area of the remaining sheet = πR\(^2\) – πr\(^2\)

= \(π (R^2 – r^2\))

= \(3.14 (4^2 – 3^2)\)

= \(3.14 (16 – 9)\)

= \(3.14 × 7\)

= \(21.98\) cm\(^2\)

So, the area of the remaining sheet is \(21.98\) cm\(^2\).

6. Saima wants to put lace on the edge of a circular table cover of diameter \(1.5\) m. Find the length of the lace required, and also, find its cost if one meter of the lace costs ₹ \(15\). (Take \(π = 3.14\))

Solution:-

Diameter of the circular table = \(1.5\) m

We know that radius (r) = \(\dfrac{d}{2}\)

= \(\dfrac{1.5}{2}\)

= \(0.75\) m

Circumference of the circle = \(2πr\)

= \(2 × 3.14 × 0.75\)

=\( 4.71\) m

So, the length of the lace =\( 4.71\) m

Cost of \(1\) m lace = ₹ 15\) [given]

Cost of \(4.71\) m lace = ₹ 15 × 4.71\)

= ₹ \(70.65\)

7. Find the perimeter of the adjoining figure, which is a semicircle, including its diameter.



Solution:-

Diameter of semi-circle = 10 cm

We know that radius (r) = \(\dfrac{d}{2}\)

= \(\dfrac{10}{2}\)

= \(5\) cm

Circumference of the semi-circle =\( πr + 2r\)

= \(3.14(5) + 2(5)\)

= \(5 [3.14+ 2]\)

= \(5 [5.14]\)

Therefore, the perimeter of the semicircle = \(25.7\) cm

8. Find the cost of polishing a circular table top of diameter 1.6 m, if the rate of polishing is ₹\(15\)/m\(^2\). (Take π = 3.14)

Solution:-

Diameter of the circular table-top = 1.6 m

We know that radius (r) = \(\dfrac{d}{2}\)

= \(\dfrac{1.6}{2}\)

= 0.8 m

Area of the circular table-top = πr\(^2\)

= 3.14 × 0.82

= 3.14 × 0.8 ×0.8

= 2.0096 m\(^2\)

Cost for polishing 1 m\(^2\) area = ₹ 15 [given]

Cost for polishing 2.0096 m\(^2\) area = ₹ 15 × 2.0096

= ₹ 30.144

Hence, the cost of polishing 2.0096 m\(^2\) area is ₹ 30.144.

9. Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also, find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square? (Take π = 22/7)

Solution:-

Length of wire that Shazli took =44 cm

If the wire is bent into a circle,

The circumference of the circle = \(2πr\)

\(44 = 2 × (\dfrac{22}{7}) × r\)

\(44 = \dfrac{44}{2}\) × r\)

\(\dfrac{(44 × 7)}{44} = r\)

\(r = 7\) cm

Area of the circle = \(πr^2\)

\(= \dfrac{22}{7} × 7^2\)

\(= \dfrac{22}{7} × 7 ×7\)

\(= 22 × 7\)

= 154 cm\(^2\)

If the wire is bent into a square,

The length of each side of the square = \dfrac{44}{4}\)

= 11 cm

Area of the square = Length of the side of square\(^2\)

= \(11^2\)

= 121 cm\(^2\)

By comparing the two areas of the square and circle,

Clearly, the circle encloses more area.

10. From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1cm are removed. (As shown in the adjoining figure.) Find the area of the remaining sheet. (Take \(π = \dfrac{22}{7}\))



Solution:-

Radius of the circular card sheet = 14 cm

Radius of the two small circles = 3.5 cm

Length of the rectangle = 3 cm

Breadth of the rectangle = 1 cm

First, we have to find out the area of the circular card sheet, two circles and the rectangle to find out the remaining area.

Area of the circular card sheet = πr\(^2\)

=\( \dfrac{22}{7} × 14^2\)

= \(\dfrac{22}{7} × 14 × 14\)

= \(22 × 2 × 14\)

= 616 cm\(^2\)

Area of the 2 small circles \(= 2 × πr^2\)

\(= 2 × (\dfrac{22}{7} × 3.52)\)

\(= 2 × (\dfrac{22}{7} × 3.5 × 3.5)\)

\(= 2 × ((\dfrac{22}{7}) × 12.25)\)

\(= 2 × 38.5\)

= \(77\) cm\(^2\)

Area of the rectangle = Length × Breadth = 3 × 1 = 3 cm\(^2\)

The area of the remaining part = Card sheet area – (Area of two small circles + Rectangle area)

=\( 616 – (77 + 3)\)

=\( 616 – 80\)

= \(536\) cm\(^2\)

11. A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side

6 cm. What is the area of the leftover aluminium sheet? (Take π = 3.14)

Solution:-

Radius of circle = 2 cm

Square sheet side = 6 cm

First, we have to find out the area of the square aluminium sheet and circle to find out the remaining area.

Area of the square = side\(^2\) = 6\(^2\)= 36 cm\(^2\)

Area of the circle = πr\(^2\)

= 3.14 × 22

= 3.14 × 2 × 2

= 3.14 × 4

= 12.56 cm\(^2\)

The area of the remaining part = Area of the aluminium square sheet – The area of the circle

= 36 – 12.56

= 23.44 cm\(^2\)

12. The circumference of a circle is \(31.4\) cm. Find the radius and the area of the circle. (Take \(π = 3.14\))

Solution:-

Circumference of a circle = \(31.4\) cm

Circumference of a circle = \(2πr\)

\(31.4 = 2 × 3.14 × r\)

\(31.4 = 6.28 × r\)

\(\dfrac{31.4}{6.28} = r\)

\(r = 5\) cm

Area of the circle \(= πr^2\)

\(= 3.14 × 5^2\)

\(= 3. 14 × 25\)

\(= 78.5\) cm

13. A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? (π = 3.14)



Solution:-

Diameter of the flower bed = 66 m

Radius of the flower bed = \(\dfrac{d}{2}\)

= \(\dfrac{66}{2}\)

= 33 m

Area of flower bed = πr\(^2\)

= 3.14 × 332

= 3.14 × 1089

= 3419.46 m

Now, we have to find the area of the flower bed and path together.

So, the radius of the flower bed and path together = 33 + 4 = 37 m

Area of the flower bed and path together = πr\(^2\)

= \(3.14 × 372\)

= \(3.14 × 1369\)

= \(4298.66\) m

Area of the path = Area of the flower bed and path together – Area of the flower bed

= \(4298.66 – 3419.46\)

= \(879.20\) m\(^2\)

14. A circular flower garden has an area of 314 m\(^2\). A sprinkler at the centre of the garden can cover an area that has a radius of \(12\) m. Will the sprinkler water the entire garden? (Take \(π = 3.14\))

Solution:-

Area of the circular flower garden = 314 m\(^2\)

The sprinkler at the centre of the garden can cover an area that has a radius = 12 m

Area of the circular flower garden = πr\(^2\)

314 = 3.14 × r\(^2\)

\(\dfrac{314}{3.14} = r^2\)

\(r^2 = 100\)

\(r = \sqrt{100}\)

\(r = 10\) m

∴ Radius of the circular flower garden is \(10\) m.

The sprinkler can cover an area of a radius of \(12\) m.

Hence, the sprinkler will water the whole garden.

15. Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take π = 3.14)



Solution:-

Radius of inner circle = outer circle radius – 10

= \(19 – 10\)

=\( 9\) m

Circumference of the inner circle = \(2πr\)

= \(2 × 3.14 × 9\)

= \(56.52\) m

Radius of outer circle = \(19\) m

Circumference of the outer circle = \(2πr\)

= \(2 × 3.14 × 19\)

= \(119.32\) m

16. How many times a wheel of radius \(28\) cm must rotate to go 352 m? (Take \(π = \dfrac{22}{7}\))

Solution:-

Radius of the wheel = \(28\) cm

Circumference of the wheel = \(2πr\)

= \(2 × \dfrac{22}{7} × 28\)

= \(2 × 22 × 4\)

= \(176\) cm

Now, we have to find the number of rotations of the wheel.

= \(\dfrac{\text{Total distance to be covered}}{ \text{Circumference of the wheel}}\)

= \(\dfrac{352 \text{m}}{176 \text{cm}}\) 

= \(\dfrac{35200 \text{cm}}{176 \text{cm}}\) 

= \(200\)

17. The minute hand of a circular clock is \(15\) cm long. How far does the tip of the minute hand move in \(1\) hour? (Take \(π = 3.14\))

Solution:-

Length of the minute hand of the circular clock = \(15\) cm

Distance travelled by the tip of minute hand in \(1\) hour = Circumference of the clock

= \(2πr\)

= \(2 × 3.14 × 15\)

= \(94.2\) cm