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.