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 :–

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

(iv) Full-filled squares = 4
Half-filled squares = 4
Area covered by full squares = 4 × 1 sq. unit = 4 sq. units
Area covered by half squares = 4 × $\frac{1}{2}$ sq. units = 2 sq. units
∴ Total area = 4 sq. units + 2 sq. units = 6 sq. units

(v) Full-fulled squares = 2 Half-filled squares = 4
Area covered by full squarees = 2 × 1 sq. unit = 2 sq. units
Area covered by half squares = 4 × $\frac{1}{2}$ sq. unit = 2 sq. units
∴ Total area = 2 sq. units + 2 sq. units = 4 sq. units.

(vi) Full-filled squares = 3
Half-filled squares = 6
Area covered by full squares = 3 × 1 sq. unit = 3 sq. units
Area covered by half squares = 6 × $\frac{1}{2}$ sq. unit = 3 sq. units
∴ Total area = 3 sq. units + 3 sq. units = 6 sq. units

Question 2.
Find the area of the following closed figures by counting squares:

Question 3.
Find the areas of the rectangles whose lengths and breadths are:
(i) 9 m and 6 m
(ii) 17 m and 3 m
(iii) 14 m and 4 m
Which one has the largest area and which one has the smallest area?

(i) Area of the rectangle = Length × Breadth = 9 m × 6 m = 54 sq. m
(ii) Area of the rectangle = Length × Breadth = 17 m × 3 m = 51 sq. m
(iii) Area of the rectangle = Length × Breadth = 14 m × 4 m = 56 sq. m
The rectangle (iii) has the lartgest area and rectangle (ii) smallest area.

Question 4.
Find the areas of the rectangles whose two adjacent sides are:
(i) 14 cm and 23 cm
(ii) 3 km and 4 km
(iii) 2 m and 90 cm

(i) 14 cm and 23 cm
Area of the rectangle = Length × Breadth
= 14 cm × 23 cm
= 322 sq. cm

(ii) 3 km and 4 km
Area of the rectangle = Length × Breadth
= 3 km × 4 km
= 12 sq. km

(iii) 2 m and 90 cm
90 cm = 0.9 m
Area of the rectangle = l × b
= 2 m × 0.9 m
= 1.8 sq. m

Question 5.
Find the areas of the squares whose sides are:
(i) 8 cm
(ii) 14 m
(iii) 2 m 50 cm

(i) 8 cm
Area of the square = Side × Side = 8 cm × 8 cm = 64 sq. cm
(ii) 14 m
Area of the square = Side × Side = 14 m × 14 m = 196 sq. m
(iii) 2 m 50 cm
$50 \mathrm{cm}=\frac{50}{100}=0.5$ m
∴ Side = 2.5 m
Area of the square = Side × Side = 2.5 m × 2.5 m = 6.25 sq. m

Question 6.
A room is 4 m long and 3 m 25 cm wide. How many square metres of carpet is needed to cover the floor of the room?

Length of the room = 4 m
Breadth of the room = 3 m 25 cm = 3.25 m
∴ Area of the room = Length × Breadth = 4 × 3.25 sq. m = 13 sq. m
Hence, 13 square metres of carpet is needed to cover the floor of the room.

Question 7.
What is the cost of tiling a rectangular field 500 m long and 200 m wide at the rate of ₹7.5 per hundred square metres?

Length of the rectangular field = 500 m
Breadth of the rectangular field = 200 m
∴ Area of the rectangular field = Length × Breadth
= 500 m × 200 m  = 10000 sq. m
∵ Cost of tiling 100 sq. m = ₹7.5
∴ Cost of tiling 1 sq. m = ₹$\frac{7.5}{100}$
∴ Cost of tiling 100000 sq. m = ₹$\frac{7.5}{100} \times 100000$ = ₹7500

Question 8.
A floor is 5 m long and 4 m wide. A square carpet of sides 3 m is laid on the floor. Find the area of the floor that is not carpeted.

Length of the floor = 5 m
Breadth of the floor = 4 m

∴Area of the floor = Length × Breadth = 5 m × 4 m  = 20 sq m
Area of the square carpet = Side × Side = 3 m × 3 m  = 9 sq m
∴ Area of the floor that is not carpeted = 20 sq m – 9 sq m = 11 sqm

Question 9.
In the given figure, find the area of the path (shown shaded) which is 2 m wide all around.

Length of field = 100 m
Breadth of field = 60 m
Area of the field = L × B = 100 × 60 = 6000m2
Length of field exclude path = 100 – (2 + 2) = 96 m
Breadth of field exclude path = 60 – (2 + 2) = 56
Area of field exclude path = L × B = 96 m × 56 m = 5376 m2
Area of path = Area of field – Area of field exclude path = 6000 – 5376 = 624 sq. m

Question 10.
Four square flower beds of side 1 m 50 cm are dug on a rectangular piece of land 8 m long and 6 m 50 cm wide. What is the area of the remaining part of the land?

Area of 1 square flower bed = a= 1.5 × 1.5 = 2.25 m2
Area of 4 square flower bed = 4 × 2.25 = 9 m2
Length of land = 8m
Breadth of land = 6.5 m
Area of land = l × b = 8 × 6.5 = 52 m2
Area of remaining part of bed = 52 – 9 = 43 sq. m

Question 11.
How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to cover a rectangular region whose length and breadth are respectively :
(i) 70 cm and 36 cm
(ii) 144 cm and 1 m.

(i) Length (l) of tile = 12 cm
breadth (b) of tile = 5 cm
Area of tile = 1 × b = 12 cm × 5 cm = 60 cm2
Length (l) of rectangular region = 70 cm
breadth (b) of rectangular region = 36 cm
Area of rectangular region = l × b = 70 cm × 36 cm = 2520 cm2
If 60 cm2 area is covered then tile required = 1
If 2520 cm2 area is covered then tile required is = $\frac{1}{60 \mathrm{cm}} \times 2520 \mathrm{cm}^{2}=42$
Hence, 42 tiles are required.
(ii) Area of the tile = 60 cm2 (as in (i) above)
Length (l) of rectangular region = 1 m = 100 cm
breadth (b) of rectangular region = 144 cm
Area of rectangular region = 100 cm × 144 cm = 14400 cm2
If 60 cm2 are is covered then tile required = 1
If 14400 cm2 is covered then tile required = $\frac{1}{60 \mathrm{cm}} \times 14400 \mathrm{cm}^{2}=240$
Hence, 240 tiles are required.

Question 12.
The area of a rectangular plot is 340 sq. m. If its breadth is 17 m, find its length and the perimeter.

Area of plot A = 340m2
Length (l) = ?
Area = l × b
⇒ 340 = l × 17
⇒ $\frac{340}{17}$
⇒ l = 20
Length = 20 m
Perimeter = 2 (l + b) = 2 (20 + 17) = 2 (37) = 74 m

Question 13.
If the area of a rectangular plot is 144 sq. m and its length is 16 m. Find the breadth of the plot and the cost of fencing it at the rate of ?6 per metre.

Area of plot = 144 m2
Length (l) = 16 m.
Area = l × b
⇒ 144 = 16 × b
⇒ $\frac{144}{16}$ = b
⇒ b = 9 m
Cost of fencing is ₹6 per metre
Perimeter of field = 2(l + b) = 2 (16 + 9) = 50 m
Cost of fencing = 50 × 6 = ₹300

Question 14.
Split the following shapes into rectangles and find their areas. (The measures are given in centimetres).

(a) The given figure is split into 2 rectangles.

Length of Part I = 12 cm
Breadth of Part I = 2 cm
Area of Part I = Length × Breadth = (12 × 2) cm2 = 24 cm2
Length of Part II = 8 cm
Breadth of Part II = 2 cm
Area of Part II = l × b = (8 × 2) cm2 = 16 cm2
∴ Total area = Area of Part I + Area of Part II = (24 + 16) cm2 = 40 cm2

(b) The given figure is divided into 5 parts

Here length of all the rectangles = 7 cm
and breadth of all the rectangles = 7 cm
Area = l × b = 7 × 7 = 49 cm2
Total rectangles = 5
∴ Total area = 5 × 49 cm2 = 245 cm2

(c) The given figure is divided into 2 rectangles

Length of Ist Part = 4 cm
Breadth of IInd Part = 1 cm
Area = l × b = 4 × 1 = 4 cm2
Length of IInd Part = 5 cm
Breadth of IInd Part = 1 cm
Area = l × b = 5 × 1 = 5 cm2
∴ Total area = 4 cm2 + 5 cm2 = 9 cm2

### Objective Type Questions

Mental Maths, Mensuration Class-6 ML Aggarwal ICSE Mathematics Chapter-14 Solutions
Question 1.
Fill in the blanks:
(i) The perimeter of a closed plane figure is the length of its ……….
(ii) The unit of measurement of perimeter is same as that of ……….
(iii) If the side of a rhombus is 7 cm then its perimeter is ……….
(iv) The area of a closed plane figure is measured in ……….

(i) The perimeter of a closed plane figure is the length of its boundary.
(ii) The unit of measurement of perimeter is same as that of length.
(iii) If the side of a rhombus is 7 cm then its perimeter is 4 × 7 cm = 28 cm.
(iv) The area of a closed plane figure is measured in sq. units.

#### Question 2.

State whether the following statements are true (T) or false (F):
(i) Centimetre is the unit of area.
(ii) The sum of lengths of a polygon is called its area.
(iii) If the sides of a rectangle are given in centimetres, then its perimeter is measured in square centimetres.
(iv) If the side of a square is doubled, then its perimeter is also doubled.
(v) If the side of a square is doubled, then its area is also doubled.
(vi) To find the cost of constructing a road, we find its area.
(vii) To find the cost of fencing a field, we find its perimeter.

(i) Centimetre is the unit of area. False
(ii) The sum of lengths of a polygon is called its area. False
(iii) If the sides of a rectangle are given in centimetres,
then its perimeter is measured in square centimetres. False
(iv) If the side of a square is doubled, then its perimeter is also doubled. True
(v) If the side of a square is doubled, then its area is also doubled. False
(vi) To find the cost of constructing a road, we find its area. True
(vii) To find the cost of fencing a field, we find its perimeter. True

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