Std 5 Mathematics · Chapter 12

12. Perimeter and Area

Solutions to Problem Sets 48 to 50, with diagrams — perimeter, area, and word problems on fencing and flooring. Maharashtra State Board (Balbharati), Grade Five Mathematics, Part Two.

Key Ideas

Perimeter = the sum of the lengths of all the sides of a closed figure.

Rectangle: perimeter = 2 × length + 2 × breadth; area = length × breadth.

Square: perimeter = 4 × side; area = side × side.

Units: perimeter is a length (cm, m); area is measured in square units (sq cm, sq m, sq km).

Perimeter: Revision

RFill in the empty boxes.

Rectangle ABCD (5 cm, 4 cm): 5 + 4 + 5 + 4 = 18 cm
Rectangle with sides 10 cm and 7 cm: 10 + 7 + 10 + 7 = 34 cm
Square PQRS with side 5 cm: 5 + 5 + 5 + 5 = 20 cm
Triangle ABC (4 cm, 8 cm, 6 cm): 4 + 8 + 6 = 18 cm

Problem Set 48

1Write the perimeter of each figure in the box given below it.

Quadrilateral: 7 + 16 + 20 + 14 = 57 cm
Hexagon: 10 + 6 + 6 + 10 + 8 + 8 = 48 cm
House shape: 8 + 8 + 18 + 12 + 18 = 64 m

In the house shape, the side at the bottom is 12 m. The sides where the roof meets the walls are not part of the outline, so they are not added.

2If a square of side 1 cm is cut out of the corner of a larger square with side 3 cm, what will be the perimeter of the remaining shape?

Side lengths of the remaining shape (in cm)

Perimeter = 3 + 3 + 2 + 1 + 1 + 2 = 12 cm

This is the same as the perimeter of the original square (4 × 3 = 12 cm). The two sides removed with the corner (1 cm + 1 cm) are replaced by the two new inner sides (1 cm + 1 cm).

Problem Set 49

1How much wire will be needed to make a rectangle 7 cm long and 4 cm wide?

Wire needed = perimeter = 2 × 7 + 2 × 4 = 14 + 8 = 22

22 cm of wire will be needed.

2If the length of a rectangle is 20 m and its width is 12 m, what is its perimeter?

Perimeter = 2 × 20 + 2 × 12 = 40 + 24 = 64

The perimeter is 64 m.

3Each side of a square is 9 m long. Find its perimeter.

Perimeter = 4 × 9 = 36

The perimeter is 36 m.

4If we take 4 rounds around a field that is 160 m long and 90 m wide, what is the distance we walk in kilometres?

One round = perimeter = 2 × 160 + 2 × 90 = 320 + 180 = 500 m

4 rounds = 4 × 500 = 2000 m

2000 m = 2 km

We walk 2 km.

5Sanju completes 12 rounds around a square park every day. If one side of the park is 120 m, find out in kilometres and metres the distance that Sanju covers daily.

One round = 4 × 120 = 480 m

12 rounds = 12 × 480 = 5760 m

5760 m = 5000 m + 760 m = 5 km 760 m

Sanju covers 5 km 760 m daily.

6The length of a rectangular plot of land is 50 m and its width is 30 m. A triple fence has to be put along its edges. If the wire costs 60 rupees per metre, what will be the total cost of the wire needed for the fence?

Perimeter = 2 × 50 + 2 × 30 = 100 + 60 = 160 m

Wire for a triple fence = 3 × 160 = 480 m

Cost = 480 × 60 = 28,800 rupees

The total cost of the wire is ₹ 28,800.

7A game requires its players to run around a square playground. Each side of the playground is 20 m long. One player took 5 rounds around the playground. How many metres did he run altogether?

One round = 4 × 20 = 80 m

5 rounds = 5 × 80 = 400 m

He ran 400 m altogether.

8Four rounds of wire fence have to be put around a field. If the field is 60 m long and 40 m wide, how much wire will be needed?

Perimeter = 2 × 60 + 2 × 40 = 120 + 80 = 200 m

Four rounds = 4 × 200 = 800 m

800 m of wire will be needed.

9The sides of a triangle are 24.7 cm, 20.4 cm and 10.5 cm respectively. What is the perimeter of the triangle?

Perimeter = 24.7 + 20.4 + 10.5 = 55.6

The perimeter of the triangle is 55.6 cm.

10Look at the figures on the sheet of graph paper. Measure their sides with the help of the lines on the graph paper. Write the perimeter of each in the right box.

Rectangle ABCD: length 3.5 cm, breadth 2.5 cm → 2 × 3.5 + 2 × 2.5 = 7 + 5 = 12 cm
Rectangle EFGH: length 3.8 cm, breadth 1.3 cm → 2 × 3.8 + 2 × 1.3 = 7.6 + 2.6 = 10.2 cm
Square PQRS: side 2.4 cm → 4 × 2.4 = 9.6 cm
Rectangle STUV: length 3 cm, breadth 2 cm → 2 × 3 + 2 × 2 = 6 + 4 = 10 cm

On this graph paper each big square has a side of 1 cm and is divided into 10 small divisions of 1 mm (0.1 cm). Count the big squares, then the small divisions, to read each side.

Area: Revision

ACount the squares in each figure and write its area.

Area of MNRS: 3 rows of 3 squares = 9 sq cm
Area of EFGH: 2 rows of 2 squares = 4 sq cm
Area of PQRS: 1 row of 5 squares = 5 sq cm
Area of IJKL: 2 rows of 4 squares = 8 sq cm

Figure ABCD, given in the textbook as the example, has 3 rows of 2 squares = 6 sq cm.

BIn the rectangle with sides 4 cm and 2 cm, make squares of 1 sq cm each. Count the number of squares.

There are 2 rows of 4 squares, that is, 2 × 4 = 8 squares. So the area is 8 sq cm, which equals length × breadth = 4 × 2.

Problem Set 50

1The length of the side of each square is given below. Find its area.

12 metres: 12 × 12 = 144 sq m
6 cm: 6 × 6 = 36 sq cm
25 metres: 25 × 25 = 625 sq m
18 cm: 18 × 18 = 324 sq cm

2If the cost of 1 sq m of a plot of land is 900 rupees, find the total cost of a plot of land that is 25 m long and 20 m broad.

Area = 25 × 20 = 500 sq m

Cost = 500 × 900 = 4,50,000 rupees

The total cost of the plot is ₹ 4,50,000.

3The side of a square is 4 cm. The length of a rectangle is 8 cm and its width is 2 cm. Find the perimeter and area of both figures.

Square: perimeter = 4 × 4 = 16 cm; area = 4 × 4 = 16 sq cm
Rectangle: perimeter = 2 × 8 + 2 × 2 = 16 + 4 = 20 cm; area = 8 × 2 = 16 sq cm

Both figures have the same area (16 sq cm), but their perimeters are different. Figures with equal areas need not have equal perimeters.

4What will be the labour cost of laying the floor of an assembly hall that is 16 m long and 12 m wide if the cost of laying 1 sq m is 80 rupees?

Area of the floor = 16 × 12 = 192 sq m

Labour cost = 192 × 80 = 15,360 rupees

The labour cost will be ₹ 15,360.

5The picture shows some squares. Find out how many squares with the same measures will fit in the empty space in the figure.

The green dashed squares fill the empty space.

The whole figure is 4 squares long and 3 squares high, so it can hold 4 × 3 = 12 squares. 6 squares are already drawn, so 12 − 6 = 6 squares will fit in the empty space.

6Divide the figure into four parts in such a way that the area and shape of each part is the same. Colour the parts with different colours.

Four equal L-shaped parts of 3 squares each

The figure has 12 small squares, so each part must have 12 ÷ 4 = 3 squares. Each part is an L-shape made of 3 squares; the four parts are shown in four colours.

★Fair and Square: divide the square plot, with four houses and a well at the centre, among four persons.

Four parts of the same L-shape; the well is at the centre.

The plot has 16 small squares, so each person gets 16 ÷ 4 = 4 squares.
Each part is the same L-shape of 4 squares, turned in a different direction.
Each part contains exactly one house.
Each part has one square touching the well at the centre, so every person can reach the well from their own land.

AActivity: Using a graph paper, find out the area of different rectangles and squares.

Draw rectangles and squares along the 1 cm lines of the graph paper and count the 1 cm squares inside. For example, a rectangle 6 cm by 4 cm contains 24 squares (area 24 sq cm), and a square of side 5 cm contains 25 squares (area 25 sq cm). Check each count with the formula length × breadth.

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