**Chapter 11 Perimeter and Area**

### Very Short Answer Type

Question 1.

The side of a square is 2.5 cm. Find its perimeter and area.

Solution:

Side of the square = 2.5 cm

Perimeter = 4 × Side = 4 × 2.5 = 10 cm

Area = (side)^{2} = (4)^{2} = 16 cm^{2}

Question 2.

If the perimeter of a square is 24 cm. Find its area.

Solution:

Perimeter of the square = 24 cm

Side of the square = cm = 6 cm

Area of the square = (Side)^{2} = (6)^{2} cm^{2} = 36 cm^{2}

Question 3.

If the length and breadth of a rectangle are 36 cm and 24 cm respectively. Find

(i) Perimeter

(ii) Area of the rectangle.

Solution:

Length = 36 cm, Breadth = 24 cm

(i) Perimeter = 2(l + b) = 2(36 + 24) = 2 × 60 = 120 cm

(ii) Area of the rectangle = l × b = 36 cm × 24 cm = 864 cm^{2}

Question 4.

The perimeter of a rectangular field is 240 m. If its length is 90 m, find:

(i) it’s breadth

(ii) it’s area.

Solution:

(i) Perimeter of the rectangular field = 240 m

2(l + b) = 240 m

l + b = 120 m

90 m + b = 120 m

b = 120 m – 90 m = 30 m

So, the breadth = 30 m.

(ii) Area of the rectangular field = l × b = 90 m × 30 m = 2700 m^{2}

So, the required area = 2700 m^{2}

Question 5.

The length and breadth of a rectangular field are equal to 600 m and 400 m respectively. Find the cost of the grass to be planted in it at the rate of ₹ 2.50 per m^{2}.

Solution:

Length = 600 m, Breadth = 400 m

Area of the field = l × b = 600 m × 400 m = 240000 m^{2}

Cost of planting the grass = ₹ 2.50 × 240000 = ₹ 6,00,000

Hence, the required cost = ₹ 6,00,000.

Question 6.

The perimeter of a circle is 176 cm, find its radius.

Solution:

The perimeter of the circle = 176 cm

Question 7.

The radius of a circle is 3.5 cm, find its circumference and area.

Solution:

Radius = 3.5 cm

Circumference = 2πr

Question 8.

Area of a circle is 154 cm^{2}, find its circumference.

Solution:

Area of the circle = 154 cm^{2}

Question 10.

The length of the diagonal of a square is 50 cm, find the perimeter of the square.

### Short Answer Type

Question 11.

A wire of length 176 cm is first bent into a square and then into a circle. Which one will have more area?

Solution:

Length of the wire = 176 cm

Side of the square = 176 ÷ 4 cm = 44 cm

Area of the square = (Side)^{2} = (44)^{2} cm^{2} = 1936 cm^{2}

Circumference of the circle = 176 cm

Since 2464 cm^{2} > 1936 cm^{2}

Hence, the circle will have more area.

Question 14.

A rectangle park is 45 m long and 30 m wide. A path 2.5 m wide is constructed outside the park. Find the area of the path.

Solution:

Length of the rectangular park = 45 m

Breadth of the park = 30 m

Area of the park = l × 6 = 45m × 30m = 1350 m^{2}

Length of the park including the path = 45 m + 2 × 2.5 m = 50 m

Breadth of the park including the path = 30 m + 2 × 2.5 m = 30m + 5m = 35m

Area of the park including the path = 50 m × 35 m = 1750 m^{2}

Area of the path = 1750 m^{2} – 1350 m^{2} = 400 m^{2}

Hence, the required area = 400 m^{2}.

### Long Answer Type

Question 17.

A nursery school playground is 160 m long and 80 m wide. In it 80 m × 80 m is kept for swings and in the remaining portion, there are 1.5 m wide path parallel to its width and parallel to its remaining length as shown in Figure. The remaining area is covered by grass. Find the area covered by grass. (NCERT Exemplar)

Solution:

Area of the playground = l × b = 160 m × 80 m = 12800 m^{2}

Area left for swings = l × b = 80m × 80m = 6400 m^{2}

Area of the remaining portion = 12800 m^{2} – 6400 m^{2} = 6400 m^{2}

Area of the vertical road = 80 m × 1.5 m = 120 m^{2}

Area of the horizontal road = 80 m × 1.5 m = 120 m^{2}

Area of the common portion = 1.5 × 1.5 = 2.25 m^{2}

Area of the two roads = 120 m^{2} + 120 m^{2} – 2.25 m^{2} = (240 – 2.25) m^{2} = 237.75 m^{2}

Area of the portion to be planted by grass = 6400 m^{2} – 237.75 m^{2} = 6162.25 m^{2}

Hence, the required area = 6162.25 m^{2}.

Question 18.

Rectangle ABCD is formed in a circle as shown in Figure. If AE = 8 cm and AD = 5 cm, find the perimeter of the rectangle. (NCERT Exemplar)

Solution:

DE (Radius) = AE + AD = 8 cm + 5 cm = 13 cm

DB = AC = 13 cm (Diagonal of a rectangle are equal)

In right ∆ADC,

AD^{2} + DC^{2} = AC^{2} (By Pythagoras Theorem)

⇒ (5)^{2} + DC^{2} = (13)^{2}

⇒ 25 + DC^{2} = 169

⇒ DC^{2} = 169 – 25 = 144

⇒ DC = √144 = 12 cm

Perimeter of rectangle ABCD = 2(AD + DC)

= 2(5 cm + 12 cm)

= 2 × 17 cm

= 34 cm

Question 19.

Find the area of a parallelogram-shaped shaded region. Also, find the area of each triangle. What is the ratio of the area of shaded portion to the remaining area of the rectangle?

Question 20.

A rectangular piece of dimension 3 cm × 2 cm was cut from a rectangular sheet of paper of dimensions 6 cm × 5 cm. Find the ratio of the areas of the two rectangles. (NCER T Exemplar)

Solution:

Length of the rectangular piece = 6 cm

Breadth = 5 cm

Area of the sheet = l × b = 6 cm × 5 cm = 30 cm^{2}

Area of the smaller rectangular piece = 3 cm × 2 cm = 6 cm^{2}

Ratio of areas of two rectangles = 30 cm^{2} : 6 cm^{2} = 5 : 1

### Higher Order Thinking Skills (HOTS) Type

Question 22.

Find the area of the following polygon if AB = 12 cm, AC = 2.4 cm, CE = 6 cm, AD = 4.8 cm, CF = GE = 3.6 cm, DH = 2.4 cm.

Solution:

BE = AB – AE

= 12 cm – (AC + CE)

= 12 cm – (2.4 cm + 6 cm)

= 12 cm – 8.4 cm

= 3.6 cm

Area of the polygon AFGBH = Area of ∆ACF + Area of rectangle FCEG + Area of ∆GEB + Area of ∆ABH

= 3.6 cm^{2} + 4.32 cm^{2} + 21.6 cm^{2} + 6.48 cm^{2} + 14.4 cm^{2}

= 50.40 cm^{2}

Hence, the required area = 50.40 cm^{2}.