Std 7 Mathematics · Chapter 12

12. Perimeter and Area

Step-by-step solutions to Practice Sets 44 to 47: perimeter, area of rectangles, squares and triangles, and surface area of cubes and cuboids. Maharashtra State Board (Balbharati), Grade Seven Mathematics.

Practice Set 44

1If the length and breadth of a rectangle are doubled, how many times the perimeter of the old rectangle will that of the new rectangle be?

Old perimeter = 2(l + b)
New perimeter = 2(2l + 2b) = 2 × 2(l + b)

The new perimeter is 2 times the old perimeter.

2If the side of a square is tripled, how many times the perimeter of the first square will that of the new square be?

First perimeter = 4a
New perimeter = 4 × 3a = 3 × 4a

The new perimeter is 3 times the first perimeter.

3Given alongside is the diagram of a playground. It shows the length of its sides. Find the perimeter of the playground.

Lengths marked in the textbook: 10 m, 15 m, 5 m, 15 m. The other two are worked out below.

Left side = 15 + 15 = 30 m (it runs the full height)
Bottom side = 10 + 5 = 15 m (the top plus the step)
Perimeter = 10 + 15 + 5 + 15 + 15 + 30 = 90 m

The perimeter of the playground is 90 m.

4As shown in the figure, four napkins all of the same size were made from a square piece of cloth of length 1 m. What length of lace will be required to trim all four sides of all the napkins?

The 1 m square is cut into 2 × 2 napkins, so each napkin is a square of side 12 m = 0.5 m.
Perimeter of one napkin = 4 × 0.5 = 2 m
Lace for 4 napkins = 4 × 2 = 8 m

8 m of lace will be required.

Practice Set 45

1If the side of a square is 12 cm, find its area.

Area = side² = 12 × 12 = 144

Area = 144 cm²

2If the length of a rectangle is 15 cm and breadth is 5 cm, find its area.

Area = l × b = 15 × 5 = 75

Area = 75 cm²

3The area of a rectangle is 102 sqcm. If its length is 17 cm, what is its perimeter?

Area = l × b
102 = 17 × b ∴ b = 6 cm
Perimeter = 2(l + b) = 2(17 + 6) = 46

Perimeter = 46 cm

4*If the side of a square is tripled, how many times will its area be as compared to the area of the original square?

Original area = a²
New area = (3a)² = 9a²

The new area is 9 times the original area.

Practice Set 46

1A page of a calendar is 45 cm long and 26 cm wide. What is its area?

Area = 45 × 26 = 1170

Area = 1170 cm²

2What is the area of a triangle with base 4.8 cm and height 3.6 cm?

Area = 12 × base × height = 12 × 4.8 × 3.6 = 2.4 × 3.6 = 8.64

Area = 8.64 cm²

3What is the value of a rectangular plot of land 75.5 m long and 30.5 m broad at the rate of 1000 rupees per square metre?

Area = 75.5 × 30.5 = 2302.75 m²
Value = 2302.75 × 1000 = 2302750 rupees

The value of the plot is ₹ 23,02,750.

4A rectangular hall is 12 m long and 6 m broad. Its flooring is to be made of square tiles of side 30 cm. How many tiles will fit in the entire hall? How many would be required if tiles of side 15 cm were used?

Area of hall = 12 × 6 = 72 m² = 72 × 10000 = 720000 cm²
Tile of side 30 cm: area = 900 cm²; number = 720000900 = 800
Tile of side 15 cm: area = 225 cm²; number = 720000225 = 3200

800 tiles of side 30 cm, or 3200 tiles of side 15 cm.

(Check another way: 1200 cm ÷ 30 = 40 and 600 cm ÷ 30 = 20, so 40 × 20 = 800 tiles.)

5Find the perimeter and area of a garden with measures as shown in the figure alongside.

The garden is made of 5 squares of side 13 m (one in the centre and one on each side). It has 12 sides of 13 m each.

Perimeter = 12 × 13 = 156 m
Area = 5 × (13 × 13) = 5 × 169 = 845 m²

Perimeter = 156 m and area = 845 m².

Practice Set 47

1Find the total surface area of cubes having the following sides. (i) 3 cm (ii) 5 cm (iii) 7.2 m (iv) 6.8 m (v) 5.5 m

Total surface area of a cube = 6 × (side)².

(i) 6 × (3)² = 6 × 9 = 54 cm²

(ii) 6 × (5)² = 6 × 25 = 150 cm²

(iii) 6 × (7.2)² = 6 × 51.84 = 311.04 m²

(iv) 6 × (6.8)² = 6 × 46.24 = 277.44 m²

(v) 6 × (5.5)² = 6 × 30.25 = 181.5 m²

2Find the total surface area of the cuboids of length, breadth and height as given below: (i) 12 cm, 10 cm, 5 cm (ii) 5 cm, 3.5 cm, 1.4 cm (iii) 2.5 cm, 2 m, 2.4 m (iv) 8 m, 5 m, 3.5 m

Total surface area of a cuboid = 2(lb + bh + lh).

(i) 2(12 × 10 + 10 × 5 + 12 × 5) = 2(120 + 50 + 60) = 2 × 230 = 460 cm²

(ii) 2(5 × 3.5 + 3.5 × 1.4 + 5 × 1.4) = 2(17.5 + 4.9 + 7) = 2 × 29.4 = 58.8 cm²

(iii) Units must be the same: 2.5 cm = 0.025 m.
2(0.025 × 2 + 2 × 2.4 + 0.025 × 2.4) = 2(0.05 + 4.8 + 0.06) = 2 × 4.91 = 9.82 m²

In (iii) the textbook gives the length as 2.5 cm while the other two measures are in metres. If the length was meant to be 2.5 m, the answer would be 2(5 + 4.8 + 6) = 2 × 15.8 = 31.6 m².

(iv) 2(8 × 5 + 5 × 3.5 + 8 × 3.5) = 2(40 + 17.5 + 28) = 2 × 85.5 = 171 m²

3A matchbox is 4 cm long, 2.5 cm broad and 1.5 cm in height. Its outer sides are to be covered exactly with craft paper. How much paper will be required to do so?

Surface area = 2(4 × 2.5 + 2.5 × 1.5 + 4 × 1.5)
= 2(10 + 3.75 + 6) = 2 × 19.75 = 39.5

39.5 cm² of craft paper will be required.

4An open box of length 1.5 m, breadth 1 m, and height 1 m is to be made for use on a trolley for carrying garden waste. How much sheet metal will be required to make this box? The inside and outside surface of the box is to be painted with rust proof paint. At a rate of 150 rupees per sqm, how much will it cost to paint the box?

The box is open, so it has no top: it has the base and four side faces only.

Sheet metal = lb + 2(bh + lh)
= 1.5 × 1 + 2(1 × 1 + 1.5 × 1) = 1.5 + 2 × 2.5 = 1.5 + 5 = 6.5 m²
Area to paint (inside and outside) = 2 × 6.5 = 13 m²
Cost = 13 × 150 = 1950 rupees

Sheet metal required = 6.5 m²; cost of painting = ₹ 1,950.

← 11. Circle 13. Pythagoras’ Theorem →