Solutions:
(a) Perimeter = Sum of all the sides
= 1 + 2 + 4 + 5
= 12 cm
(b) Perimeter = Sum of all the sides
= 23 + 35 + 35 + 40
= 133 cm
(c) Perimeter = Sum of all the sides
= 15 + 15 + 15 + 15
= 60 cm
(d) Perimeter = Sum of all the sides
= 4 + 4 + 4 + 4 + 4
=20 cm
(e) Perimeter = Sum of all the sides
= 1 + 4 + 0.5 + 2.5 + 2.5 + 0.5 + 4
= 15 cm
(f) Perimeter = Sum of all the sides
= 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3
= 52 cm
2. The lid of a rectangular box, with sides 40 cm by 10 cm, is sealed all around with tape. What is the length of the tape required?
Solutions:
Length of required tape = Perimeter of rectangle
= 2 (Length + Breadth)
= 2 (40 + 10)
= 2 (50)
= 100 cm
∴ The required length of tape is 100 cm.
3. A table top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the tabletop?
Solutions:
Length of tabletop = 2 m 25 cm = 2.25 m
Breadth of tabletop = 1 m 50 cm = 1.50 m
Perimeter of tabletop = 2 (Length + Breadth)
= 2 (2.25 + 1.50)
= 2 (3.75)
= 2 × 3.75
= 7.5 m
∴ The perimeter of the table top is 7.5 m.
4. What is the length of the wooden strip required to frame a photograph of length and breadth, 32 cm and 21 cm, respectively?
Solutions:
The required length of the wooden strip = Perimeter of the photograph
= 2 (Length + Breadth)
= 2 (32 + 21)
= 2 (53)
= 2 × 53
= 106 cm
∴ The required length of the wooden strip is 106 cm.
5. A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?
Solutions:
Perimeter of the field = 2 (Length + Breadth)
= 2 (0.7 + 0.5)
= 2 (1.2)
= 2 × 1.2
= 2.4 km
Each side is to be fenced with 4 rows = 4 × 2.4
= 9.6 km
∴ The total length of the required wire is 9.6 km.
6. Find the perimeter of each of the following shapes:
(a) A triangle of sides 3 cm, 4 cm and 5 cm
(b) An equilateral triangle of side 9 cm
(c) An isosceles triangle with equal sides of 8 cm each and the third side of 6 cm.
Solutions:
(a) Perimeter of triangle = 3 + 4 + 5 = 12 cm
(b) Perimeter of an equilateral triangle = 3 × side = 3 × 9 = 27 cm
(c) Perimeter of isosceles triangle = 8 + 8 + 6 = 22 cm
7. Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
Solutions:
Perimeter of triangle = 10 + 14 + 15
= 39 cm
∴ The perimeter of the triangle is 39 cm.
8. Find the perimeter of a regular hexagon with each side measuring 8 m.
Solutions:
Perimeter of hexagon = 6 × 8
= 48 m
∴ The perimeter of the regular hexagon is 48 m.
9. Find the side of the square whose perimeter is 20 m.
Solutions:
Perimeter of square = 4 × side
20= 4 × side
Side \(= {20 \over 4}\)
Side = 5 m
∴ The side of the square is 5 m.
10. The perimeter of a regular pentagon is 100 cm. How long is its each side?
Solutions:
The perimeter of the regular pentagon = 100 cm
5 × side = 100 cm
Side \(= {100 \over 5}\)
Side = 20 cm
∴ The side of the pentagon is 20 cm.
11. A piece of string is 30 cm long. What will be the length of each side if the string is used to form:
(a) a square?
(b) an equilateral triangle?
(c) a regular hexagon?
Solutions:
(a) Perimeter of square = 30 cm
4 × side = 30
Side = \(= {30 \over 4}\)
Side = 7.5 cm
(b) Perimeter of equilateral triangle = 30 cm
3 × side = 30
Side \(= {30 \over 3}\)
Side = 10 cm
(c) Perimeter of regular hexagon = 30 cm
6 × side = 30
Side \(= {30 \over 6}\)
Side = 5 cm
12. Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?
Solutions:
Let x cm be the third side
Perimeter of triangle = 36 cm
12 + 14 + x = 36
26 + x = 36
x = 36 – 26
x = 10 cm
∴ The third side is 10 cm.
13. Find the cost of fencing a square park of side 250 m at the rate of ₹ 20 per metre.
Solutions:
Side of square = 250 m
Perimeter of square = 4 × side
= 4 × 250
= 1000 m
Cost of fencing = ₹ 20 per m
Cost of fencing for 1000 m = ₹ 20 × 1000
= ₹ 20,000
∴ The cost of fencing the square park is ₹ 20,000.
14. Find the cost of fencing a rectangular park of length 175 cm and breadth 125 m at the rate of ₹ 12 per metre.
Solutions:
Length = 175 cm
Breadth = 125 m
Perimeter of rectangular park = 2 (Length + Breadth)
= 2 (175 + 125)
= 2 (300)
= 2 × 300
= 600 m
Cost of fencing = 12 × 600
= 7200
∴ The cost of fencing is ₹ 7,200.
15. Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with a length of 60 m and a breadth of 45 m. Who covers less distance?
Solutions:
Perimeter of square = 4 × side
= 4 × 75
= 300 m
∴ The distance covered by Sweety is 300 m
Perimeter of the rectangular park = 2 (Length + Breadth)
= 2 (60 + 45)
= 2 (105)
= 2 × 105
= 210 m
∴ The distance covered by Bulbul is 210 m
Hence, Bulbul covers less distance than Sweety.
16. What is the perimeter of each of the following figures?
What do you infer from the answers?
Solutions:
(a) Perimeter of square = 4 × side
= 4 × 25
= 100 cm
(b) Perimeter of rectangle = 2 (40 + 10)
= 2 × 50
= 100 cm
(c) Perimeter of rectangle = 2 (Length + Breadth)
= 2 (30 + 20)
= 2 (50)
= 2 × 50
= 100 cm
(d) Perimeter of triangle = 30 + 30 + 40
= 100 cm
∴ All the figures have the same perimeter.
17. Avneet buys 9 square paving slabs, each with a side of 1 / 2 m. He lays them in the form of a square.
(a) What is the perimeter of his arrangement [fig 10.7(i)]?
(b) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement [(Fig 10.7 (ii)]?
(c) Which has a greater perimeter?
(d) Avneet wonders if there is a way of getting an even greater perimeter.
Can you find a way of doing this? (The paving slabs must meet along complete edges, i.e. they cannot be broken.)
Solutions:
(a) Side of square = 3 × side
\(= 3 \times {1 \over 2}\)
\(= {3 \over 2}\) m
Perimeter of Square = \(= 4 \times {3 \over 2}\)
= 2 × 3
= 6 m
(b) Perimeter = 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1
= 10 m
(c) The arrangement in the form of a cross has a greater perimeter.
(d) Perimeters greater than 10 m cannot be determined.
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