Std 6 Mathematics · Chapter 15

15. Perimeter and Area

Solutions to Practice Set 15.1, Practice Set 15.2 and Exercise 15. Maharashtra State Board (Balbharati), Grade Six Mathematics.

About the chapter

The total length of the boundary of a closed figure is its perimeter. The space taken up by a closed figure is its area. In this chapter, we find the perimeter and area of rectangles, squares and triangles. We also solve problems on plots, house plans and running tracks.

Key points

Perimeter is a length. It is measured in cm, m and so on.
Area is measured in square units, like sq. cm and sq. m.
If a wire is bent into a new shape, its perimeter stays the same.
If the perimeter of a square increases, its area also increases.
If the perimeter of a rectangle increases, its area may not increase.

Formulas and terms

Perimeter of a rectangle = 2 × (length + breadth)
Perimeter of a square = 4 × side
Perimeter of a triangle = sum of its three sides
Area of a rectangle = length × breadth, Area of a square = side × side
Area of a triangle = base × height/2

Practice Set 15.1

1Find the perimeter of a rectangular plot with length 20 meters and breadth 10 meters.

Perimeter of a rectangle = 2 × (length + breadth)
= 2 × (20 + 10)
= 2 × 30 = 60 m

The perimeter of the plot is 60 metres.

2The perimeter of a rectangle is 64 cm. If its length is 17 cm, then what is its breadth?

2 × (length + breadth) = 64
∴ length + breadth = 64 ÷ 2 = 32
∴ 17 + breadth = 32
∴ breadth = 32 − 17 = 15 cm

The breadth of the rectangle is 15 cm.

3The perimeter of a triangle is 55 cm and two of its sides are 20 cm and 14 cm. Find the length of the third side.

Perimeter = sum of all three sides
55 = 20 + 14 + third side
55 = 34 + third side
∴ third side = 55 − 34 = 21 cm

The third side is 21 cm.

4A rectangle with sides of 5 cm and 3 cm is formed by bending a wire. If a square is formed using the same wire, what will be the length of the side of that square?

The same wire is used. So, the perimeter of both figures is the same.

Perimeter of the rectangle = 2 × (5 + 3) = 2 × 8 = 16 cm
∴ Perimeter of the square = 16 cm
4 × side = 16
∴ side = 16 ÷ 4 = 4 cm

The side of the square will be 4 cm.

5Take a square piece of cloth one meter long and make four handkerchiefs of the same size as shown in the figure. How much total length of lace will be required to lace all the handkerchiefs?

1 m = 100 cm. The cloth is cut into 4 equal squares. Each square has a side of 50 cm.

One square metre of cloth cut into four 50 cm squares.

Perimeter of one handkerchief = 4 × 50 = 200 cm
Lace for 4 handkerchiefs = 4 × 200 = 800 cm = 8 m

8 metres of lace will be required.

6If the following figures are made using a rope of length 36 cm, what will be the length of each side? a) Square b) Rectangle

a) Square

4 × side = 36
∴ side = 36 ÷ 4 = 9 cm

Each side of the square will be 9 cm.

b) Rectangle

2 × (length + breadth) = 36
∴ length + breadth = 18 cm

So, the length and breadth can be 10 cm and 8 cm, or 12 cm and 6 cm, or 11 cm and 7 cm.

7Take a rectangular piece of paper 6 cm long and 4 cm wide. Cut it into two equal parts. Arrange the two pieces without overlapping as shown in the figure and find the perimeter.

Each of the two pieces is 6 cm long and 2 cm wide.

(i) 12 cm × 2 cm strip

(ii) L-shape

(iii) T-shape

(i) Perimeter = 2 × (12 + 2) = 2 × 14 = 28 cm

(ii) Perimeter = 6 + 8 + 2 + 6 + 4 + 2 = 28 cm

(iii) Perimeter = 6 + 2 + 2 + 6 + 2 + 6 + 2 + 2 = 28 cm

The perimeter of each figure is 28 cm.

Practice Set 15.2

1Find the area of the triangle.

i) Base 8 cm, height 5 cm

Area of a triangle = base × height/2 = 8 × 5/2 = 40/2 = 20 sq. cm

ii) Base 7 m, height 4 m

Area = 7 × 4/2 = 28/2 = 14 sq. m

2Given the length of the sides of a rectangle below, find the perimeter and area of the rectangle.

i) 8 cm, 3 cm
Perimeter = 2 × (8 + 3) = 2 × 11 = 22 cm
Area = 8 × 3 = 24 sq. cm

ii) 3.5 m, 2 m
Perimeter = 2 × (3.5 + 2) = 2 × 5.5 = 11 m
Area = 3.5 × 2 = 7 sq. m

iii) 6 m, 3.5 m
Perimeter = 2 × (6 + 3.5) = 2 × 9.5 = 19 m
Area = 6 × 3.5 = 21 sq. m

iv) 9.5 cm, 5.3 cm
Perimeter = 2 × (9.5 + 5.3) = 2 × 14.8 = 29.6 cm
Area = 9.5 × 5.3 = 50.35 sq. cm

3Given the length of the side of a square below, find the perimeter and area of the square.

i) Side 10 cm
Perimeter = 4 × 10 = 40 cm
Area = 10 × 10 = 100 sq. cm

ii) Side 14 m
Perimeter = 4 × 14 = 56 m
Area = 14 × 14 = 196 sq. m

iii) Side 5.5 m
Perimeter = 4 × 5.5 = 22 m
Area = 5.5 × 5.5 = 30.25 sq. m

4A rectangular mango orchard is 100 meters long and 50 meters wide. If each mango tree takes 25 sq. meters of space, then how many mango trees can be planted in the orchard at most?

Area of the orchard = 100 × 50 = 5000 sq. m
Space for one tree = 25 sq. m
Number of trees = 5000 ÷ 25 = 200

At most 200 mango trees can be planted.

Exercise 15

1Choose the correct option.

i) What is the perimeter of a square with side length 7 cm? — d) 28 cm
Perimeter = 4 × 7 = 28 cm
ii) If the length of a rectangle is 5 cm and the breadth is 2 cm, what are its perimeter and area respectively? — b) 14 cm, 10 sq. cm
Perimeter = 2 × (5 + 2) = 14 cm. Area = 5 × 2 = 10 sq. cm
iii) A square piece of paper is folded in half and cut along the fold to form two rectangles. Which statement is always true? — c) The perimeter of the square is smaller than the sum of the perimeters of the two rectangles.

Let the side of the square be s.
Perimeter of the square = 4s
Each rectangle is s long and s/2 wide.
Perimeter of one rectangle = 2 × (s + s/2) = 3s
Sum of both perimeters = 6s. So, 4s < 6s.

2Given the length of the sides of a triangle, find the perimeter.

i) Perimeter = 12 + 7 + 5 = 24 m

ii) Perimeter = 6.5 + 7 + 5 = 18.5 cm

3Given the length and breadth of three rectangles below, which rectangle has the largest and smallest area?

i) Area = 9 × 6 = 54 sq. m
ii) Area = 17 × 3 = 51 sq. m
iii) Area = 4 × 14 = 56 sq. m

Rectangle (iii) has the largest area. Rectangle (ii) has the smallest area.

4In the figure below, the length of some sides and areas are given. Find the area of the shaded region.

The lengths and areas given in the figure.

Length of the bottom rectangle = 10 ÷ 2 = 5 cm
Width of the middle rectangle = 5 − 3 = 2 cm
Height of the middle rectangle = 10 ÷ 2 = 5 cm
Height of the shaded region = 5 − 2 = 3 cm
Area of the shaded region = 3 × 3 = 9 sq. cm

The area of the shaded region is 9 sq. cm.

Note: This answer is based on the textbook figure. Check where the 2 cm mark is in your book.

5Frame two examples for each of the following statements.

i) When the perimeter of a rectangle is increased, the area increases.

Example 1: 4 cm × 3 cm → perimeter 14 cm, area 12 sq. cm
6 cm × 5 cm → perimeter 22 cm, area 30 sq. cm

Example 2: 5 cm × 2 cm → perimeter 14 cm, area 10 sq. cm
8 cm × 4 cm → perimeter 24 cm, area 32 sq. cm

ii) When the perimeter of a square is increased, the area increases.

Example 1: Side 3 cm → perimeter 12 cm, area 9 sq. cm
Side 5 cm → perimeter 20 cm, area 25 sq. cm

Example 2: Side 4 cm → perimeter 16 cm, area 16 sq. cm
Side 6 cm → perimeter 24 cm, area 36 sq. cm

iii) Increasing the perimeter of a rectangle does not necessarily increase its area.

Example 1: 5 cm × 4 cm → perimeter 18 cm, area 20 sq. cm
9 cm × 1 cm → perimeter 20 cm, area 9 sq. cm

Example 2: 6 cm × 5 cm → perimeter 22 cm, area 30 sq. cm
11 cm × 1 cm → perimeter 24 cm, area 11 sq. cm

In (iii), the perimeter increased, but the area decreased.

6The plan of Amir's house is given below. This plan is in a square plot. Observe the plan. i) Find the missing areas in it. ii) Find the area of Amir's entire house.

Amir’s house plan, drawn in a 35 ft × 35 ft square plot.

i) The missing areas

Hall = 20 × 20 = 400 sq. ft (given)
Balcony = 15 × 5 = 75 sq. ft
Master Bedroom = 15 × 15 = 225 sq. ft
Parking = 15 × 12 = 180 sq. ft
Kitchen = 20 × 12 = 240 sq. ft
Garden = 35 × 3 = 105 sq. ft

ii) Area of the entire house

400 + 75 + 225 + 180 + 240 + 105 = 1225 sq. ft

The area of Amir's entire house is 1225 sq. ft.

7Two tracks are drawn in a rectangular field, one on the inside and one on the outside. Anju ran 7 laps on the inside track and Manju ran 4 laps on the outside track. i) How much distance did Anju cover? ii) How much distance did Manju cover? iii) Who among the two covered the greater distance?

The outside track is 60 m × 40 m; the inside track is 40 m × 20 m.

i) The inside track is 40 m long and 20 m wide.
One lap = 2 × (40 + 20) = 120 m
Distance covered by Anju = 7 × 120 = 840 m

ii) The outside track is 60 m long and 40 m wide.
One lap = 2 × (60 + 40) = 200 m
Distance covered by Manju = 4 × 200 = 800 m

iii) 840 m > 800 m. So, Anju covered the greater distance.

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