Students can download Chapter 10 Mensuration Ex 10.1 KSEEB Solutions for Class 6 Maths helps you to revise the complete syllabus.

## Karnataka State Syllabus Class 6 Maths Chapter 10 Mensuration Ex 10.1

Question 1.

Find the perimeter of each of the following figures:

Solution :

Perimeter of a polygon is equal to the sum of the lengths of all sides of that polygon.

(a) Perimeter = (4 + 2 +1 + 5) cm = 12 cm

(b) Perimeter = (23 + 35 + 40 + 35) cm = 133 cm

(c) Perimeter = (15 + 15 + 15 + 15) cm = 60 cm

(d) Perimeter = (4 + 4 + 4 + 4 +4) cm = 20 cm

(e) Perimeter = (1 + 4 + 0.5 + 2.5 + 2.5 + 0.5+ 4) cm = 15 cm

(f) Perimeter = (1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4) = 52 cm

Question 2.

The lid of a rectangular box of sides 40 cm by 10 cm is sealed all around with tape. What is the length of the tape required?

Solution:

Length (l) of rectangular box = 40 cm

Breadth (b) of rectangular box = 10 cm

Length o f tape required = perimeter of rectangular box

= 2 ( l + b)

= 2(40 + 10) = 100 cm

Question 3.

A table – top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table – top?

Solution:

Length (l) of the top of the table = 2 m 25 cm = 2 + 0.25

= 2.25 m

Breadth (b) of the top of the table = 1 m 50 cm = 1 + 0.50 = 1.50 m

Perimeter of table – top = 2 (l + b)

= 2 × (2.25 + 1.50)

= 2 × 3.75 = 7.5 m

The perimeter of the table-top is 7.5 m

Question 4.

What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively?

Solution:

Length (l) of the photograph = 32cm

Breadth (b) of the photograph = 21 cm

Perimeter of photograph = 2 × (l + b)

= 2 × (32 + 21)

= 2 × (53)

= 106 cm

Length of wooden strip required is 106 cm.

Question 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?

Solution:

Given,

Length of the Land (l) = 0.7 km

Breadth of the land (b) = 0.5 km.

perimeter of the land = 2 × (l + b)

= 2 × (0.7 + 0.5)

= 2 × (1.2)

= 2.4 km.

The length of the wire required to fenced the land is given by 4 × 2.4 = 9.6 km.

Question 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 8 cm each and third side 6 cm.

Solution:

a) perimeter of a ∆le = (AB + BC + CA) cm

= (3 + 4 + 5) cm.

= 12 cm

b) perimeter of an equilateral triangle = 3 × side of ∆le

= (3 × 9 ) cm

= 27 cm.

c) perimeter of an isosceles triangle = 8 + 8 + 6 = 22 cm.

Question 7.

Find the perimeter of a triangle with sides measuring 10 cm, 14 cm, and 15 cm.

Solution:

Perimeter of triangle = sum of lengths of all sides of the triangle

∴ Perimeter of ∆le = 10 + 14 + 15 = 39 cm.

Question 8.

Find the perimeter of a regular hexagon with each side measuring 8m.

Solution:

Perimeter of regular hexagon = 6 × side of regular hexagon

= 6 × 8 m

∴ Perimeter of regular hexagon = 48 m

Question 9.

Find the side of the square whose perimeter is 20m.

Solution:

Perimeter of square = 4 × side 20

= 4 × side

∴ Side of the square = 5 m

Question 10.

The perimeter of a regular pentagon is 100 cm. How long is its each side?

Solution:

Perimeter of regular pentagon = 5 × side of regular pentagon.

100 = 5 × side

∴ Side of the regular pentagon = 20 cm.

Question 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?

Solution:

a) perimeter a square = 4 × side of the square

= 30 = 4 × side of the square

Side of the square = 7.5 cm.

b) perimeter an equilateral triangle = 3 × side of the ∆le

30 = 3 × side of the ∆le

Side of the equilateral ∆le = 10 cm.

c) perimeter of a regular hexagon = 6 × side of the regular hexagon

30 = 6 × side

Side of the regular hexagon = 5 cm.

Question 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?

Solution:

Given AB = 12 cm, and BC = 14 cm, CA = ?

Perimeter of triangle = sum of all sides of the triangles

36 = AB + BC + CA

36 = 26 + CA

CA = 36 – 26 = 10 cm.

CA = 10 cm.

Hence, the third side of the triangle is 10 cm.

Question 13.

Find the cost of fencing a square park of side 250 m at the rate of Rs 20 per metre.

Solution:

Length of fence required is equal to the perimeter of the square park

= 4 × side

= 4 × 250

= 1000 m

Cost for fencing 1 m of square park = Rs. 20

Cost for fencing 1000m of square park = 1000 × 20

= Rs. 20,000

Question 14.

Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of Rs 12 per metre.

Solution:

Given:-

Length of rectangular park (l) = 175 cm

Breadth of rectangular park (b) = 125 cm

Length of wire required for fencing the park = Perimeter of the park

= 2 × (l + b)

= 2 × (175 + 125)

= 2 × 300 = 600 m

Cost for fencing 1 m of the park = Rs 12.

Cost for fencing 600m of the square park = 600 × 12

= Rs. 7200

Question 15.

Sweety runs around a square park of side 75m. Bulbul runs around a rectangular park with length 60m and breadth 45m. Who covers less distance?

Solution:

The distance covered by sweety = 4 × rectangular square park

= 4 × 75 m = 300 m

The distance covered by bulbul = perimeter of rectangular park

= 2 × (60 + 45)

= 2 × 105

= 210 m

Therefore, bulbul covers the less distance than sweety.

Question 16.

What is the perimeter of each of the following figures? What do you infer from the answers?

Solution:

a) perimeter of square = 4 × 25 = 100 cm.

b) perimeter of rectangle = 2 × (10 + 40) = 2 × 50 = 100 cm

c) perimeter of rectangle = 2 × (20 + 30) = 100 cm.

d) perimeter of ∆le = 30 + 30 + 40 = 100 cm

Imfered:- The above figures have the same perimetre .

Question 17.

Avneet buys 9 square paving slabs, each with a side of \(\frac{1}{2}\) m. He lays them in the form of a square.

Solution:

a) What iis 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)]?

perimeter of cross = 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1

= 10 m

c) Which has greater perimeter?

The arrangement in the form of a cross 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?

Arrangements with a perimeter greater than 10m can not be determined.