## MP Board Class 7th Maths Solutions Chapter 11 Perimeter and Area Ex 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

Solution:

(a) r = 14 cm

∴ Circumference = 2πr = 2 × \(\frac{22}{7}\) × 14 =88 cm

(b) r = 28 mm

∴ Circumference = 2πr = 2 × \(\frac{22}{7}\) × 28 = 176 mm

(c) r = 21 cm

∴ Circumference = 2πr = 2 × \(\frac{22}{7}\) × 21 = 132 cm

Question 2.

Find the area of the following circles, given that: (Take π = \(\frac{22}{7}\))

(a) radius = 14 mm

(b) diameter = 49 m

(c) radius = 5 cm

Solution:

(a) r = 14 mm

Question 3.

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

Solution:

Circumference = 2πr = 154 m

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 costs ₹ 4 per meter. (Take π = \(\frac{22}{7}\))

Solution:

Diameter (d) = 21 m 21

∴ Radius (r) = \(\frac{21}{2}\)m

Circumference = 2πr = 2 × \(\frac{22}{7} \times \frac{21}{2}\) = 66 m

Length of rope required for fencing = 2 × 66 m = 132 m

Cost of 1 m rope = ₹ 4

Cost of 132 m rope = 4 × 132 = ₹ 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)

Solution:

Outer radius of circular sheet (R) = 4 cm

Inner radius of circular sheet (r) = 3 cm

Remaining area = πR^{2} – πr^{2}

= 3.14 × 4 × 4 – 3.14 × 3 × 3

= 50.24 – 28.26 = 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)

Solution:

The length of the lace required = circumference of circular table

Circumference = 2πr = 2 × 3.14 × \(\frac{d}{2}\)

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

Cost of 1 m lace = ₹ 15

Cost of 4.71 m lace = 4.71 × 15 = ₹ 70.65

Question 7.

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

Solution:

Diameter = 10 cm

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

Circumference of semicircle = \(\frac{2 \pi r}{2}\)

= 2 × \(\frac{1}{2} \times \frac{22}{7}\) × 5 = 15.71 cm

Total perimeter = Circumference of semicircle + Length of diameter

= 15.71 + 10 = 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)

Solution:

Diameter = 1.6 m

∴ Radius = \(\frac{1.6}{2}\) = 0.8 m

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

Cost for polishing 1 m^{2} area = ₹ 15

Cost for polishing 2.0096 m^{2} area

= 15 × 2.0096 = ₹ 30.14

Therefore, it will cost ₹ 30.14 for polishing circular table.

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

Solution:

If the wire is bent into a circle, then the length of wire = circumference of the circle

⇒ 2πr = 44 cm

⇒ 2 × \(\frac{22}{7}\) × r = 44

⇒ r = 7 cm

Area = πr^{2}= \(\frac{22}{7}\) × 7 × 7 = 154 cm^{2}

If the wire is bent into a square, then the length of the wire = perimeter of the square

⇒ 4 × side = 44cm ⇒ side = \(\frac{44}{4}\) = 11 cm

Area of square = (11)^{2} = 121 cm^{2}

As 154 > 121,

Therefore, circle encloses more area.

Question 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 1 cm are removed (as shown in the following figure). Find the area of the remaining sheet. (Take π = \(\frac{22}{7}\))

Solution:

Area of bigger circle = \(\frac{22}{7}\) × 14 × 14 = 616 cm^{2}

Area of 2 small circles = 2 × πr^{2}

= 2 × \(\frac{22}{7}\) × 3.5 × 3.5 = 77 cm^{2}

Area of rectangle = Length × Breadth = 3 × 1

= 3 cm^{2}

Area of remaining sheet = Area of bigger circle – (Area of 2 small circles + Area of rectangle)

= 616 – (77 + 3) = 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)

Solution:

Area of square-shaped sheet = (Side)^{2}

= (6)^{2} = 36 cm^{2}

Area of circle = 3.14 × 2 × 2= 12.56 cm^{2}

Area of remaining sheet = Area of square sheet – area of circle

= 36 – 12.56 = 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 π = 3.14)

Solution:

Let r be the radius of circle. Circumference = 2πr = 31.4 cm

⇒ 2 × 3.14 × r = 31.4 cm

⇒ r = 5 cm

Area = 3.14 × 5 × 5 = 78.50 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)

Solution:

Radius of flower bed = \(\frac{66}{2}\) = 33 m

Width of the path = 4 m

Radius of flower bed and path together = 33 + 4 = 37 m

Area of flower bed and path together

= 3.14 × 37 × 37 = 4298.66 m^{2}

Area of flower bed = 3.14 × 33 × 33 = 3419.46 m^{2}

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

= 4298.66 – 3419.46 = 879.20 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)

Solution:

Area = πr^{2} = 314 m^{2}

3.14 × r^{2} = 314 ⇒ r^{2} = 100 ⇒ r = 10 m

Yes, the sprinkler will water the whole garden.

Question 15.

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

Solution:

Radius of outer circle = 19 m

Circumference = 2πr =2 × 3.14 × 19 = 119.32 m

Radius of inner circle = 19 – 10 = 9 m

Circumference = 2πr = 2 × 3.14 × 9 = 56.52 m

Question 16.

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

Solution:

r = 78 cm

Therefore, it will rotate 200 times.

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)

Solution:

Distance travelled by the tip of minute hand = Circumference of the clock

= 2πr = 2 × 3.14 × 15 = 94.2 cm