Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 11 Perimeter and Area Ex 11.1 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.1

1. Find the perimeter and the area of a rectangle having :

(i) Length = 28 cm, Breadth = 15 cm

(ii) Length = 9.4 cm Breadth = 2.5 cm

Solution:

(i) Given length of rectangle = 28 cm

Breadth of rectangle = 15 cm

Perimeter of rectangle = 2 [length + Breadth]

= 2 [28 + 15]

= 2 × 43

= 86 cm

Area of rectangle = length × Breadth

= 28 × 15

= 420 cm^{2}

(ii) Perimeter of rectangle = 2 [9.4 + 2.5]

= 2 × 11.9

= 23.8 cm

Area of rectangle = 9.4 × 2.5

= 23.5 cm^{2}

2. Find the perimeter and the area of a square whose side measures

(i) 29 cm

(ii) 8.3 cm

Solution:

(i) Given side of square = 29 cm

Perimeter of square = 4 × side

= 4 × 29

= 116 cm

Area of square = (side)^{2}

= (29)^{2}

= 841 cm^{2}

(ii) Perimeter of square = 4 × 8.3

= 33.2 cm

Area of square = 8.3 × 8.3

= 68.89 cm^{2}

3. The perimeter of a square park is 148 m. Find its area.

Solution:

Given the perimeter of square park = 148 m

Side of the square park = \(\frac{perimeter}{4}\)

= \(\frac {148}{4}\)

Area of the square park = (side)^{2}

= (37)^{2}

= 1369 m^{2}

4. The area of a rectangle is 580 cm^{2}. Its length is 29 cm. Find its breadth and also, the perimeter.

Solution:

Given area of rectangle = 580 cm^{2}

Length of the rectangle = 29 cm

Let breadth of the rectangle = b cm

Area of the rectangle = length × breadth

580 = 29 × b

\(\frac {580}{29}\) = b

b = 20 cm

Perimeter of rectangle = 2 [length + breadth]

= 2 [29 + 20]

= 2 × 49

= 98 cm

5. A wire is in the shape of a rectangle. Its length is 48 cm and breadth is 32 cm. If the same wire is rebent into the shape of a square, what will be the measure of each side. Also, find which shape encloses more area and by how much ?

Solution:

Given length of the rectangle = 48 cm

Breadth of the rectangle = 32 cm

Perimeter of the rectangle = 2 [length + breadth]

= 2 [48 + 32]

= 2 × 80

= 160 cm

Let side of square = a cm

Perimeter of the square = 4 × a

Since wire is rebent into the shape of a square

Perimeter of square = Perimeter of rectangle

4 a = 160

Therefore, a = \(\frac {160}{4}\)

= 40 cm

Area of square = (side)^{2}

= 40 × 40

= 1600 cm^{2}

Area of rectangle = length × breadth

= 48 × 32

= 1536 cm^{2}

∴ Square encloses more area by 64 cm^{2}

6. The area of a square park is the same as that of a rectangular park. If the side of the square park is 75 m and the length of the rectangular park is 125 m, find the breadth of the rectangular park. Also, find the perimeter of rectangular park.

Solution:

Given side of square park = 75 m

Area of square park = (75)^{2}

= 75 × 75

= 5625 m^{2}

Length of rectangular park = 125 m

Let breadth of rectangular park = b m

Area of rectangular park = length × breadth

= 125 × b m^{2}

Given that

Area of rectangular park = Area of square park

125 × b = 5625

= 45 m

Perimeter of rectangular park = 2 [length + breadth]

= 2 [125 + 45]

= 2 × 170

= 340 m

7. A door of length 2.5 m and breadth 1.5 m is fitted in a wall. The length of wall, is 9 m and breadth is 6 m. Find the cost of painting the wall, if the rate of painting the wall is ₹ 30 per m^{2}.

Solution:

Length of door = 2.5 m

Breadth of door = 1.5 m

Area of door = length × breadth

= 2.5 × 1.5

= 3.75 m^{2}

Area of wall = 9 × 6

= 54 m^{2}

Area of wall painting = Area of wall including door – Area of door

= 54 – 3.75

= 50.25 m^{2}

Cost of painting 1 m^{2} of wall = ₹ 30

Cost of painting 50.25 m^{2} of wall = ₹ 50.25 × 30

= ₹ 1507.50

8. A door of dimensions 3 m × 2 m and a window of dimensions 2.5 m × 1.5 m is fitted in a wall. The length of the wall is 7.8 m and breadth is 3.9 m. Find the cost of painting the wall, if the rate of painting the wall is ₹ 25 per m^{2}.

Solution:

Area of door = 3 × 2 = 6 m^{2}

Area of window = 2.5 m × 1.5 m

= 3.75 m^{2}

Area of wall = 7.8 m × 3.9 m

= 30.42 m^{2}

Area of painting the wall = Area of wall – Area of door – Area of window

= 30.42 – 6 – 3.75

= 20.67 m^{2}

Cost of painting the wall = ₹ 25 × 20.67

= ₹ 516.75

9. Find the area and the perimeter of the following figures.

Solution:

(i) Perimeter of the given figure

= AB + BC + CD + DE + EF + FG + GH + HA

= 2 + 3.5 + 3 + 2 + 5 + 3.5 + 10 + 9

= 38 cm^{2}

Area of the figure = Area of rectangle ABCJ + Area of rectangle JDEI + Area of rectangle IFGH

= 2 × 3.5 + 5 × 2 + 10 × 3.5

= 7 + 10 + 35

= 52 cm^{2}

(ii) Perimeter of the given figure

= 8cm + 5 cm + 1.5 cm + 2.5 cm + 2.5 cm + 1.5 cm + 1.5 cm + 1.5 cm + 2.5 cm + 1.5 cm

= 29 cm

Area of the given figure = Area of rectangle I + Area of rectangle II + Area of rectangle III

= 8 cm × 1.5 cm + 3.5 cm × 1.5 cm + 1.5 cm × 1.5 cm

= 12 cm^{2} + 5.25 cm^{2} + 2.25 cm

= 19.5 cm^{2}

10. Multiple Choice Questions :

Question (i).

What is the area of a rectangle of dimensions 12 cm × 10 cm ?

(a) 44 cm^{2}

(b) 120 cm^{2}

(c) 1200 cm^{2}

(d) 1440 cm^{2}

Answer:

(b) 120 cm^{2}

Question (ii).

Find the breadth of a rectangle whose length is 12 cm and perimeter is 36 cm.

(a) 6 cm

(b) 3 cm

(c) 9 cm

(d) 12 cm

Answer:

(a) 6 cm

Question (iii).

If each side of a square is 1 m then its area is ?

(a) 10 cm^{2}

(b) 100 cm^{2}

(c) 1000 cm^{2}

(d) 10000 cm^{2}

Answer:

(d) 10000 cm^{2}

Question (iv).

Find the area of a square whose perimeter is 96 cm.

(a) 576 cm^{2}

(b) 626 cm^{2}

(c) 726 cm^{2}

(d) 748 cm^{2}.

Answer:

(a) 576 cm^{2}

Question (v).

The area of a rectangular sheet is 500 cm^{2}. If the length of the sheet is 25 cm, what is its breadth ?

(a) 30 cm

(b) 40 cm

(c) 20 cm

(d) 25 cm.

Answer:

(c) 20 cm

Question (vi).

What happens to the area of a square, if its side is doubled ?

(a) The area becomes 4 times, the area of original square.

(b) The area becomes \(\frac {1}{4}\) times, the area of original square.

(c) The area becomes 16 times, the area of original square.

(d) The area becomes \(\frac {1}{6}\) times, 6 the area of original square.

Answer:

(a) The area becomes 4 times, the area of original square.