Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 12 Perimeter and Area Ex 12.1 Textbook Exercise Questions and Answers.
PSEB Solutions for Class 6 Maths Chapter 12 Perimeter and Area Ex 12.1
1. Find the perimeter of the following shapes:
Question (i)
Solution:
Perimeter = AB + BC + CD + DA
= 8 cm + 7 cm + 12 cm + 9 cm
= 36 cm
Question (ii)
Solution:
Perimeter = XY + YZ + ZX
= 10m + 10m + 8m
= 28m
Question (iii)
Solution:
Perimeter = PQ + QR + RS + SP
= 15 cm + 12 cm + 15 cm + 12 cm
= 54 cm
Question (iv)
Solution:
Perimeter = MN + NO + OP + PL + LM
= 8 cm + 7 cm + 5 cm + 6 cm + 7 cm
= 33 cm
Question (v)
Solution:
Perimeter = AB + BC + CD + DE + EF + FG + GM + MA
=8m + 2m + 6m + 4m + 6m + 2m + 8m + 8m
= 44 m
Question (vi)
Solution:
Perimeter = LM + MN + NO + OP + PQ + QR + RS + SL
= 3 cm + 3 cm + 3 cm + 3 cm + 3 cm + 3 cm + 3 cm + 3 cm
= 24 cm
2. Find the perimeter of the triangle with sides:
Question (i)
5 cm, 6 cm and 7 cm
Solution:
Perimeter of a triangle
= Sum of lengths of its sides
= 5 cm + 6 cm + 7 cm = 18 cm
Question (ii)
10 m, 12 m, 18 m
Solution:
Sides of triangle
= 10 m, 12 m, 18 m
∴ Perimeter of a triangle
= Sum of lengths of its sides
= 10 m + 12 m + 18 m = 40 m
Question (iii)
4.6 cm, 3.2 cm and 5.8 cm.
Solution:
Sides of triangle
= 4.6 cm, 3.2 cm and 5.8 cm
∴ Perimeter of triangle
= Sum of lengths of its sides
= 4.6 cm + 3.2 cm + 5.8 cm
= 13.6 cm
3. Find the perimeter of an isosceles triangle with 15 cm as length of equal side and 18 cm as base.
Solution:
Sides of isosceles triangle = 15 cm, 15 cm, 18 cm
Area of isosceles triangle = Sum of lengths of its sides
= (15 + 15 + 18) cm = 48 cm
4. Find the perimeter of a square with side:
Question (i)
16 cm
Solution:
Side of square = 16 cm
∴ Perimeter of square = 4 × side
= 4 × 16 cm
= 64 cm
Question (ii)
4.8 mm
Solution:
Side of square = 4.8 mm
∴ Perimeter of square = 4 × side
= 4 × 4.8 mm
= 19.2 mm
Question (iii)
125 cm
Solution:
Side of square = 125 cm
∴ Perimeter of square = 4 × side
= 4 × 125 cm
= 500 cm
Question (iv)
45 m
Solution:
Side of square = 45 m
∴ Perimeter of square = 4 × side
= 4 × 45 m
= 180 m
Question (v)
39 cm.
Solution:
Side of square = 39 cm
∴ Perimeter of square = 4 × side
= 4 × 39 cm
= 156 cm
5. Find the perimeter of a rectangle with:
Question (i)
Length 20 m and breadth 15 m
Solution:
Length of rectangle = 20 m
Breadth of rectangle = 15 m
∴ Perimeter of rectangle = 2 × (length + breadth)
= 2 × (20 + 15) m
= 2 × 35 m
= 70 m
Question (ii)
Length 25 m and breadth 35 m
Solution:
Length of rectangle = 25 m
Breadth of rectangle = 35 m
∴ Perimeter of rectangle = 2 × (length + breadth)
= 2 × (25 + 35) m
= 2 × 60 m
= 120 m
Question (iii)
Length 40 cm and breadth 28 cm
Solution:
Length of rectangle = 40 cm Breadth of rectangle = 28 cm
∴ Perimeter of rectangle = 2 × (length + breadth)
= 2 × (40 + 28) cm
= 2 × 68 cm
= 136 cm
Question (iv)
Length 18.3 cm and breadth 6.8 cm
Solution:
Length of rectangle = 18.3 cm
Breadth of rectangle = 6.8 cm
∴ Perimeter of rectangle
= 2 × (length + breadth)
= 2 × (18.3 + 6.8) cm
= 2 × 25.1 cm = 50.2 cm
Question (v)
Length 0.125 m and breadth 15 cm.
Solution:
Length of rectangle
= 0.125 m = 12.5 cm
Breadth of rectangle = 15 cm
∴ Perimeter of rectangle = 2 × (length + breadth)
= 2 × (12.5 + 15) cm
= 2 × 27.5 cm
= 55 cm
6. Find the perimeter of a regular hexagon with side:
Question (i)
5 cm
Solution:
Side of a regular hexagon = 5 cm
Perimeter of a regular hexagon = 6 × side
= 6 × 5 cm
= 30 cm
Question (ii)
12 cm
Solution:
Side of a regular hexagon = 12 cm
Perimeter of a regular hexagon = 6 × side
= 6 × 12 cm
= 72 cm
Question (iii)
7.2 cm.
Solution:
Side of a regular hexagon = 7.2 cm
Perimeter of a regular hexagon = 6 × side
= 6 × 7.2 cm
= 43.2 cm
7. Find the perimeter of an equilateral triangle with side:
Question (i)
10 cm
Solution:
Side of an equilateral triangle
= 10 cm
∴ Perimeter of an equilateral triangle = 3 × side
= 3 × 10 cm
= 30 cm
Question (ii)
8 m
Solution:
Side of an equilateral triangle = 8 m
∴ Perimeter of an equilateral triangle = 3 × side
= 3 × 8 m
= 24 m
Question (iii)
24 m
Solution:
Side of an equilateral triangle = 24 m
∴ Perimeter of an equilateral triangle = 3 × side
= 3 × 24 m
= 72 m
Question (iv)
5.6 m
Solution:
Side of an equilateral triangle = 5.6 m
∴ Perimeter of an equilateral triangle = 3 × side
= 3 × 5.6 m
= 16.8 m
Question (v)
12.1 cm.
Solution:
Side of an equilateral triangle = 12.1 cm
∴ Perimeter of an equilateral triangle = 3 × side
= 3 × 12.1 cm
= 36.3 cm
8. If the perimeter of a triangle is 48 cm and two sides are 12 cm and 17 cm. Find the third side.
Solution:
Perimeter of a triangle = 48 cm Sum of length of two sides = (12 + 17) cm = 29 cm
∴ Third side = 48 cm – 29 cm
= 19 cm
9. Find the side of an equilateral triangle, if the perimeter is:
Question (i)
45 cm
Solution:
Given perimeter of an equilateral triangle = 45 cm
Perimeter of an equilateral triangle = 3 × (side of the triangle)
⇒ 45 cm = 3 × side
⇒ Side of the triangle
= 15cm
Question (ii)
69 mm
Solution:
Given perimeter of an equilateral triangle = 69 mm
Perimeter of an equilateral triangle = 3 × (side of the triangle)
⇒ 69 = 3 × (side of the triangle)
⇒ Side of the triangle = \(\frac{69 \mathrm{~mm}}{3}\)
= 23 mm
Question (iii)
117 cm.
Solution:
Given perimeter of an equilateral triangle = 117 cm
Perimeter of an equilateral triangle = 3 × (side of the triangle)
⇒ 117 = 3 × (side of the triangle)
⇒ Side of the triangle = \(\frac{117 \mathrm{~cm}}{3}\)
= 39 cm
10. Find the side of a square if the perimeter is:
Question (i)
52 cm
Solution:
Given Perimeter of a square = 52 cm
Perimeter of a square = 4 × (side of square)
⇒ Side of square
= \(\frac{\text { perimeter of a square }}{\text { 4 }}\)
= \(\frac{52 \mathrm{~cm}}{4}\)
= 13 cm
Question (ii)
60 cm
Solution:
Given perimeter of a square = 60 cm
Side of a square
= \(\frac{\text { perimeter of a square }}{\text { 4 }}\)
= \(\frac{60 \mathrm{~cm}}{4}\)
= 15 cm
Question (iii)
112 cm.
Solution:
Given perimeter of a square = 112 cm
Side of a square
= \(\frac{\text { perimeter of a square }}{\text { 4 }}\)
= \(\frac{112 \mathrm{~cm}}{4}\)
= 28 cm
11.
Question (i)
The perimeter of rectangular field is 260 m. If its length is 80 m then find its breadth.
Solution:
Given perimeter of rectangular field = 260 m
and Length of the rectangular field = 80 m
∴ Perimeter of rectangular field = 2 × (length + breadth)
⇒ 260 = 2 × (80 + breadth)
⇒ \(\frac {260}{2}\) = 80 + breadth
⇒ 80 + breadth = 130
⇒ breadth = 130 – 80 = 50 m
Hence breadth of rectangular field = 50 m
Question (ii)
The perimeter of a rectangular garden is 140 m. If its breadth is 45 m then find its length.
Solution:
Given perimeter of rectangular garden = 140 m
and breadth of rectangular garden = 45 m
∴ Perimeter of rectangular garden = 2 × (length + breadth)
⇒ 140 = 2 × (length + 45)
⇒ \(\frac {260}{2}\) = length + 45
⇒ length = 70 – 45 = 25 m
Hence length of rectangular garden = 25 m
Question (iii)
The perimeter of a rectangle is 114 cm. If its length is 32 cm then find its breadth.
Solution:
Given perimeter of rectangle = 114 cm
and length of rectangle = 32 cm
∴ Perimeter of rectangle
= 2 × (length + breadth)
⇒ 114 = 2 × (32 + breadth)
⇒ \(\frac {114}{2}\) = 32 + breadth
⇒ breadth = 57 – 32 = 25 cm
Hence breadth of rectangle = 25 cm
12. The side of a triangular field are 15 m, 20 m and 18 m. Find the total distance travelled by a boy in taking 2 complete rounds of this field.
Solution:
Sides of a triangular fields = 15 m, 20 m and 18 m
Distance covered in one round of a triangular field = Perimeter of rectangular field = Sum of the length of the sides of a rectangular field
= 15 m + 20 m + 18 m
= 53 m
∴ Distance covered in taking 2 complete rounds of this field
= 2 × 53 m
= 106 m
13. Find the cost of fencing a square field of side 26 m at the rate of ₹ 3 per metre.
Solution:
Given side of the square field = 26 m
∴ Perimeter of the square field = 4 × side
= 4 × 26 m = 104 m
Perimeter of fencing = 104 m
Cost of 1 m of fencing = ₹ 3
Cost of 104 m of fencing
= 104 × ₹ 3
= ₹ 312
14. Mani runs around a square park of side 75 m. Kush runs around a rectangular park of length 60 m and breadth 45 m. Who covers less distance?
Solution:
Side of a square park = 75 m
Perimeter of square park = 4 × (side)
= 4 × (75 m) = 300 m
∴ Distance covered by Mani = 300 m
Length of rectangular park = 60 m
Breadth of rectangular park = 45 m
Perimeter of rectangular park = 2 (length + breadth)
= 2 × (60 + 45) m
= 2 × (105) m
= 210 m
∴ Distance covered by Kush = 210 m
Kush covers less distance.
15. Find the cost of framing a rectangular whiteboard with length 240 cm and breadth 150 cm at the rate of ₹ 6 per cm.
Solution:
Length of rectangular white board = 240 cm
Breadth of rectangular white board = 150 cm
Perimeter of rectangular white board = 2 × (length + breadth)
= 2 × (240 + 150) cm
= 2 × (390) cm
= 780 cm
Cost of fencing 1 cm = ₹ 6 × 780
= ₹ 4680
16. If length of a rectangle is ‘a’ units and breadth is 5 units. Find the perimeter of the rectangle.
Solution:
Given length of rectangle = ‘a’ units,
and Breadth of rectangle = 5 units
Perimeter of rectangle = 2 × (length + breadth)
= 2 × (a + 5) units
= 2 (a + 5) units
17. Fill in the blanks:
Question (i)
The sum of lengths of all sides of a polygon is called ……………. .
Solution:
perimeter
Question (ii)
Perimeter of Square = ……………. × side.
Solution:
4
Question (iii)
Perimeter of Rectangle = 2 × (………. +………) .
Solution:
length, breadth
Question (iv)
Side of a square = (……………) ÷ 4.
Solution:
perimeter
Question (v)
Perimeter of an equilateral triangle = …………….. × side.
Solution:
3.