# Mensuration Class-6 ML Aggarwal ICSE Maths Solutions

**Mensuration Class-6 ML Aggarwal** ICSE Mathematics Chapter-14 Solutions. We provide step by step Solutions of Exercise / lesson-14 Basic **Mensuration**** **ICSE Class-6th ML Aggarwal Mathematics.

Our Solutions contain all type Questions with Exe- 14.1, Exe-14.2, Objective Type Questions (includes: Mental Maths, Multiple Choice Questions , HOT ) and Check Your Progress to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-6 Mathematics.

**Mensuration Class-6 ML Aggarwal** ICSE Mathematics Chapter-14 Solutions

**–: Select Topic :–**** **

**Objective Type Questions, **

**Multiple Choice Questions ,(MCQ)**

**Ex 14.1,** **Mensuration Class-6 ML Aggarwal** ICSE Mathematics Solutions

Question 1.

Find the perimeter of each of the following figures:

#### Answer

Perimeter = Sum of all the sides.

(i) Perimeter = 5 cm + 3 cm + 2 cm + 7 cm = 17 cm

(ii) Perimeters = 31 cm + 38 cm + 48 cm + 38 cm = 155 cm

(iii) Perimeter = 19 cm + 19 cm + 19 cm + 19 cm = 76 cm

(iv) Perimeter = 7 cm + 7 cm + 7 cm + 7 cm + 7 cm = 35 cm

Question 2.

Find the perimeter of each of the following shapes:

(i) A triangle of sides 3 cm, 4 cm and 6 cm.

(ii) A triangle of sides 7 cm, 5.4 cm and 10.2 cm.

(iii) An equilateral triangle of side 11 cm.

(iv) An isosceles triangle with equal sides 10 cm each and third side 7 cm.

#### Answer

(i) Perimeter of the triangle with sides 3 cm, 4 cm and 6 cm

= 3 cm + 4 cm + 6 cm

= 13 cm

(ii) Perimeter of the triangle with sides 7 cm, 5.4 cm, 10.2 cm

= 7 cm + 5.4 cm + 10.2 cm

= 22.6 cm

(iii) Perimeter of an equilateral triangle

= 3 × length of a side

= 3 × 11 cm

= 33 cm

(iv) Perimeter of isosceles triangle

= 10 cm + 10 cm + 7 cm

= 27 cm

Question 3.

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

#### Answer

Length of the tape required

= Perimeter of the rectangular box

= 2 × (Length + Breadth)

= 2 × (40 cm + 10 cm)

= 2 × (50 cm)

= 100 cm

= 1 m

Question 4.

Table-Top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?

#### Answer

Perimeter of the table-top

= 2 × (Length + Breadth)

= 2 × (2 m 25 cm + 1 m 50 cm)

= 2 × (2.25 m + 1.50 m)

= 2 × (3.75 m)

= 7.5 m

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?

#### Answer

Perimeter of the rectangle

= 2 × (Length + Breadth)

= 2 × (0.7 km + 0.5 km)

= 2 × (1.2 km)

= 2.4 km

Length of the wire needed

= 4 × Perimeter of the rectangle

= 4 × (2.4 km)

= 9.6 km

Question 6.

Find the perimeter of a regular hexagon with each side measuring 7.5 m.

#### Answer

The perimeter of a regular hexagon

= 6 × Length of a side

= 6 × 7.5 m

= 45 m

Question 7.

The lengths of two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is the length of its third side?

#### Answer

Perimeter of a triangle = 12 cm + 14 cm + l

⇒ 36 cm = 12 cm + 14 cm + l

⇒ 36 cm = 26 cm + l

⇒ l = 36 cm – 26 cm

⇒ l = 10 cm

Question 8.

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

#### Answer

Perimeter of the regular pentagon = 5 × Length of a side

⇒ Length of one (each) side

Question 9.

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?

#### Answer

(a) Perimeter of the square = 4 × Length of aside

⇒ Length of a side

=

= 7.5 cm

(b) Perimeter of the equilateral triangle = 3 × Length of a side

⇒ Length of a side

=

=

(c) Perimeter of the regular hexagon = 6 × Length of a side

⇒ Length of a side

Question 10.

Find the cost of fencing a rectangular park of length 225 m and breadth 115 m at the rate of ₹13 per metre.

#### Answer

Perimeter of the rectangular park

= 2 × (Length + Breadth)

= 2 × (225 m + 115 m)

= 2 × (340 m)

= 680 m

∴ Cost of fencing the rectangular park

at the rate of ₹13 per metre = ₹13 × 680 m = ₹8840

Question 11.

Meera went to a rectangular park 140 m long and 90 m wide. She took 5 complete rounds on its boundary. What is the distance covered by her?

#### Answer

Length (l) = 140 m

Width (b) = 90 m

∴ Perimeter of park = 2(l + b)

= 2 (140 + 90)

= 2(230)

= 460 m

She takes 5 complete round,

therefore distance covered by her = 5 × 460m = 2300 m

Question 12.

Pinky runs 8 times around a rectangular park with length 80 m and breadth 55 m while Pankaj runs 7 times around a square park of side 75 cm. Who covers more distance and by how much?

#### Answer

Perimeter of rectangular park = 2 × (Length + Breadth)

= 2 × (80 + 55)

= 270 m

Pinky runs 8 times = 8 × 270 = 2160 m

And, perimeter of square park = 4 × Length of a side

= 4 × 75

= 300

Pankaj runs 7 times = 7 × 300 = 2100 m

∴ Pinky covers more distance i.e., 2160 – 2100 = 60 m

**Mensuration Class-6 ML Aggarwal** ICSE Mathematics Solutions **Ex 14.2**

Question 1.

Find the area of the region enclosed by the following figures by counting squares:

#### Answer

(iii) Fill-filled squares = 10

∴ Total area = Area covered by full squares

= 10 × 1 sq. unit = 10 sq. units

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