These NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.3

Question 1.

Find the circumference of the circles with the following radius: (Take π = \(\frac{22}{7}\))

(a) 14 cm

(b) 28 mm

(c) 21 cm

Answer:

(a) Radius of the circle (r) = 14 cm

Circumference of the circle = 2πr units

= 2 × \(\frac{22}{7}\) x 14 cm

= 2 × 22 × 2 cm

= 88 cm

(b) Radius of the circle = 28 mm.

.’. Circumference of the circle = 2πr units

= 2 × \(\frac{22}{7}\) × 28 mm

= 2 × 22 × 4 mm

= 176 mm

(c) Radius of the circle = 21 cm

Circumference of the circle = 2πr 22

= 2 × \(\frac{22}{7}\) x 21

= 2 × 22 × 3 cm

= 132 cm

Question 2.

Find the area of the following circles, given that:

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

(b) Diameter = 49 m

(c) Radius = 5 cm

Answer:

(a) Radius of the circle = 14 mm

Area of the circle = πr^{2} sq.m

= \(\frac{22}{7}\) × 14^{2}

= \(\frac{22}{7}\) × 196

= 22 × 28

= 616 mm^{2}

(b) Diameter of the circle = 49 m.

Radius of the circle = \(\frac{49}{2}\) m

= 24.5 m

Area of the circle = πr^{2}

= \(\frac{22}{7}\) × (24.5)^{2}

= \(\frac{22}{7}\) × 600.25

= 22 × 85.75

= 1886.5 m^{2}

Area of the circle = πr^{2} sq units

(c) Radius of the circle = 5 cm

Area of the circle = πr^{2} sq.unit

= \(\frac{22}{7}\) × 5 × 5 cm^{2}

= \(\frac{22 \times 25}{7}\) cm^{2}

= \(\frac{550}{7}\) cm^{2} or 78.57 cm^{2}.

Question 3.

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

Answer:

Circumference of a circle = 154 m

2πr = 154

2 × \(\frac{22}{7}\) × r = 154

Area of the circle = πr^{2} sq.m

= \(\frac{22}{7}\) × (24.5)^{2}

= \(\frac{22}{7}\) × 600.25

= 22 × 85.75 m^{2}

= 1886.5 m^{2}

Question 4.

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

Answer:

Diameter of the circular garden = 21 m

radius of the circular garden = \(\frac{21}{2}\)

Circumference of the garden = 2πr units

= 2 × \(\frac{22}{7}\) × \(\frac{21}{2}\) m = 22 x 3m = 66m

Length of the rope required for one round fence = 66 m

Length of the rope required for two round fence = 66 m × 2 m = 132 m

Cost of rope per meter = ₹ 4

Total cost of the rope = ₹132 × 4 = ₹ 528

Question 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)

Answer:

Radius of outer circle (R) = 4 cm

Radius of inner circle (r) = 3 cm

= Area of the outer circle – Area of the inner circle

Area of the remaining sheet = n (R^{2} – r^{2})

= 3.14 (4^{2} – 3^{2})

= 3.14 × (16-9)

= 3.14 x 7 = 21.98 cm^{2}

So, the area of the remaining sheet is 21.98 cm^{2}.

Question 6.

Saima wants to put a 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)

Answer:

Diameter of the table cover = 1.5 m

Radius of the table cover = \(\frac { 1.5 }{ 2 }\) m

Circumference of the table cover = 2πr units

= 2 x 3.14 x \(\frac { 1.5 }{ 2 }\) = 3.14 x 1.5 m = 4.71 m

Cost of lace for one metre = ₹15

Cost of lace for table cover = ₹ 4.71 x 15

= ₹ 70.71

Question 7.

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

Answer:

Diameter of the semicircle = 10 cm

Radius of the semicircle = \(\frac { 10 }{ 2 }\) = 5 cm

Circumference of the semicircle = \(\frac { 1 }{ 2 }\) x 2πr unit

= πr = \(\frac { 22 }{ 7 }\) x 5

= 15.71 cm

∴ Perimeter of the semicircle

= 15.71 + 10 cm

= 25.71 cm

Question 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)

Answer:

Diameter of the table-top = 1.6 m

Radius of the table-top = \(\frac { 1.6 }{ 2 }\) = 0.8m

Area of the table-top = πr^{2} sq. m

= 3.14 × 0.8 × 0.8 m^{2} = 2.0096 m^{2}

Rate of Polishing = ₹ 15 per m^{2}

Cost of polishing the table-top = ₹ 2.0096 × 15 =₹ 30.14 (approx.)

Question 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 = \(\frac { 22 }{ 7 }\))

Answer:

Length of the wire = 44 cm

Let the radius of the circle be r.

Circumference of the circle = 44 cm

2πr = 44

2 × \(\frac { 22 }{ 7 }\) × r = 44

r = \(\frac{44 \times 7}{2 \times 22}\) cm =7 cm

Area of the circle = πr^{2} sq.m

= \(\frac { 22 }{ 7 }\) × 7 × 7 cm^{2}

= 154 cm

Since, the wire is rebent to form a square. Perimeter of the square = Length of the wire

4a = 44

[Perimeter of a square = 4a] 44

a = \(\frac { 44 }{ 6 }\) = 11 cm 6

Side of a square =11 cm

Area of the square = side × side

= 11 × 11 cm^{2} = 121 cm^{2}

154 cm^{2} > 121 cm^{2}

Area of the circle > Area of the square

∴ The circle encloses greater area.

Question 10.

From a circular card sheet of radius 14cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1 cm are removed, (as shown in the given figure). Find the area of the remaining 22

sheet. (Take π = \(\frac { 22 }{ 7 }\))

Answer:

Radius of the circular card sheet =14 cm

Area of the sheet = πr^{2} sq m

= \(\frac { 22 }{ 7 }\) × 14 × 14 cm^{2}

= 22 × 2 × 14 cm2 = 616 cm^{2}

Radius of a small circle = 3.5 cm

Area of 2 small circles = 2 × πr^{2} sq m

= 2 × \(\frac { 22 }{ 7 }\) × 3.5 × 3.5 cm^{2}

= 2 × 22 × 0.5 × 3.5 cm = 77 cm^{2}

Length of a small rectangle = 3 cm

Breadth of a small rectangle = 1 cm.

Area of small rectangle = l × b sq.m

= 3 × 1 cm^{2} = 3 cm^{2}

Area of the remaining sheet = Area of the circular sheet – (Area of two small circles + Area of rectangle)

= 616 cm^{2}– (77 + 3)cm^{2}

= 616 cm^{2} – 80 cm^{2} = 536 cm^{2}

∴ Required area of the remaining sheet = 536 cm^{2}

Question 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 left over aluminium sheet? (Take π = 3.14)

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 left over aluminium sheet? (Take π = 3.14)

Answer:

Side of the square = 6 cm

Area of the square = side × side

= 6 cm × 6 cm = 36 cm^{2}

Radius of the circle cut out from the sheet = 2 cm

Area of the circle = πr^{2} sq.m

= 3.14 × 2 × 2 cm^{2} = 12.56 cm^{2}

Area of the remaining sheet

= 36 cm^{2} – 12.56 cm^{2} = 23.44 cm^{2}

Question 12.

The circumference of a circle is 31.4 cm. Find the radius and the area of the circle? (Take n = 3.14)

Answer:

Let the radius of the circle be ‘r’

Circumference of a circle = 31.4 cm

2πr^{2} = 31.4

2 × 3.14 × r = 31.4

r = \(\frac{31.4}{2 \times 3.14}\)

= \(\frac{314 \times 10}{2 \times 314}\)

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

= 5 cm

Area of the circle = 2πr^{2} sq. units = 3.14 × 5 × 5 cm^{2} = 78.5 cm^{2}

Question 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)

Answer:

Diameter of the flower bed = 66 m

Radius of the flower bed = \(\frac { 66 }{ 2 }\) m

(r) = 33 m

width of the surrounding path = 4 m

Radius of the outer circle (R)

= 33 m + 4 m

(R) = 37 m

Area of the pathway = Area of the outer circle – Area of the inner circle

= πR^{2} – πr^{2}

= π(R^{2} – r^{2}) sq. units

= 3.14 (37^{2} – 33^{2} ) m^{2}

= 3.14 (37 + 33) (37 – 33) m^{2}

= 3.14 × 70 × 4 m^{2}

= 3.14 × 280 m^{2}

= 879.2 m^{2}

Thus, the area of the path = 879.2 m^{2}

Question 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)

Answer:

Let the radius of the garden be ‘r’

Area of the circular garden = 314 m^{2}

πr^{2} =314

3.14 × r^{2} = 314

r^{2} = \(\frac{314}{3.14}=\frac{314 \times 100}{314}\)

r^{2} = 100

r^{2}= 10^{2}

r = 10 m

Radius of the area covered by the sprinkler = 12 m

Since 12 m > 10 m

The sprinkler covers an area beyond the garden.

Yes, the entire garden is covered by the sprinkler.

Question 15.

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

Answer:

Radius of the outer circle (R) = 19 m.

Circumference of the outer circle

= 2πr unit

= 2 × 3.14 × 19 m

= 119.32 m

Radius of the inner circle (r)

= 19- 10 = 9m

Circumference of the inner circle

= 2π = 2 × 3.14 × 9 m = 56.52 m

Question 16.

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

Answer:

Radius of a wheel = 28 cm

∴ Circumference of the wheel = 2πr units

= 2 × \(\frac { 22 }{ 7 }\) × 28

= 2 × 22 × 4 cm = 176 cm

Total distance covered

= 352 m

= 352 × 100 cm

= 35200 cm

Number of rotations

= \(\begin{aligned}

&\frac{\text { Total distance }}{\text { umference of the wheel }}\\&\text { Circum }

\end{aligned}\)

= \(\frac{35200}{176}\) = 200

Thus, the distance of 352 m will be covered in 200 rotations.

Question 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)

Answer:

Length of the minute hand =15 cm

radius of the circle (r) = 15 cm

(made by the tip of the minute hand)

Perimeter of the circle = 2πr units

= 2 × 3.14 × 15 cm = 94.2 cm

Distance covered by the minute hand = 94.2 cm