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

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Exercise 14.1 ,

Exercise-14.2,

Objective Type Questions,

Mental Maths,

Multiple Choice Questions ,(MCQ)

HOTS

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

Question 1.
Find the perimeter of each of the following figures:

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.

(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?

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?

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?

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.

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?

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?

Perimeter of the regular pentagon = 5 × Length of a side
⇒ Length of one (each) side
$\begin{array}{l}{=\frac{\text { Perimeter of the regular pentagon }}{5}} \\ {=\frac{100}{5} \mathrm{cm}=20 \mathrm{cm}}\end{array}$

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?

(a) Perimeter of the square = 4 × Length of aside
⇒ Length of a side
$\frac{\text { Perimeter of the square }}{4}=\frac{30}{4} \mathrm{cm}$
= 7.5 cm

(b) Perimeter of the equilateral triangle = 3 × Length of a side
⇒ Length of a side
$\frac{\text { Perimeter of the equilateral triangle }}{3}$
$\frac{30}{3} \mathrm{cm}=10 \mathrm{cm}$

(c) Perimeter of the regular hexagon = 6 × Length of a side
⇒ Length of a side
$\begin{array}{l}{=\frac{\text { Perimeter of the regular hexagon }}{6}} \\ {=\frac{30}{6} \mathrm{cm}=5 \mathrm{cm}}\end{array}$

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.

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?

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?

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:

(iii) Fill-filled squares = 10
∴ Total area = Area covered by full squares
= 10 × 1 sq. unit = 10 sq. units

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