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.
Breadth (b) = ?
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

### Multiple Choice Questions

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

Choose the correct answer from, the given four options (3 to 15):
Question 3.
If the perimeter of a square is 50 cm, then its side is
(a) 200 cm
(b) 150 cm
(c) 25 cm
(d) 12.5 cm

Perimeter of a square = 50 cm
⇒ 4 × length of a side = 50 cm
⇒ length of a side = $\frac{50}{4}$ cm = 12.5 cm (d)

Question 4.
The area of the rectangle with length 25 cm and breadth 12 cm is
(a) 300 sq. m
(b) 74 cm
(c) 300 sq. cm
(d) 74 sq. cm

Length = 25 cm
Breadth = 12 cm
Area = Length × Breadth
= 25 cm × 12 cm
= 300 sq. cm (c)

Question 5.
If the perimeter of a square is 36 cm, then its area is
(a) 6 sq. cm
(b) 9 sq. cm
(c) 18 sq. cm
(d) 81 sq. cm

Perimeter of a square = 36 cm
⇒ 4 × length of a side = 36
⇒ length of a side = $\frac{36}{4}$ = 9 cm
Area of a square = (Length of a side)= (9 cm)2 = 81 cm2 (d)

Question 6.
If the area of rectangular plot is 180 sq. m and its length is 15 m, then its breadth is
(a) 12 m
(b) 12 cm
(c) 60 m
(d) 9 m

Length = 15 m
Area of a rectangle = 180 sq. m
Length × Breadth = 180 sq. m
Breadth = $\frac{180}{15}$ = 12 m (a)

Question 7.
If the length and the breadth of a rectangle are doubled, then its perimeter
(a) remains the same
(b) doubles
(c) becomes four times
(d) becomes half

Let the length of rectangle = l
Let the breadth of rectangle = b
Perimeter = 2(l + b)
If length and breadth are doubles then
Length = 2l
Perimeter = 2(2l + 2b) = 2 × 2(l +b)
Hence perimeter becomes doubles. (b)

Question 8.
If the length and the breadth of a rectangular are doubled then its area
(a) remains same
(b) becomes half
(c) doubles
(d) becomes four times.

If length = x
and breadth = y
Then area of rectangle = x × y = xy
And If length and breadth are doubled
i. e. length = 2x and breadth = 2y
Then area of rectangle becomes = 2x × 2y = 4 xy
Hence, It shows that the area of rectangle becomes four times. (d)

Question 9.
If the sides of a square are halved, then its area
(a) remains same
(b) becomes half
(c) becomes one-fourth
(d) doubles

Let us assume side of a square = x cm
∴ Area = (x)2 = x2 cm2
and if we half the side
∴ New side of a square = $\frac{x}{2}$ cm
∴ Area = $\frac{x^{2}}{4}$ cm2
Hence, It shows if side of square are halved,
then its area become one-fourth. (c)

Question 10.
A square-shaped park ABCD of side 100 m has two equal flower beds of size 10 m x 5 m as show in the given figure. The perimeter of the remaining park is

(a) 340 m
(b) 370 m
(c) 400 m
(d) 430 m

Perimeter of square = 4 × Side = 4 × 100 m = 400 m (c)

Question 11.
In the given figure, a square of side 1 cm is joined to a square of side 3 cm. The perimeter of the new figure is

(a) 13 cm
(b) 14 cm
(c) 15 cm
(d) 16 cm

The given figure is

Perimeter = AB + BC + CD + DE + EF + FG + GA
As we know all the sides of square are equal
∴ AB = BC = CD = DA = 3 cm
Also, ED = DG = GF = FE = 1 cm
But in perimeter we need AG
AG = AD – GD = 3 cm – 1 cm = 2 cm
Hence perimeter = (3 + 3 + 3 + 1 + 1 + 1 + 2) cm = 14 cm (b)

Question 12.
Two regular hexagons of perimeter 30 cm each are joined as shown in the given figure. The perimeter of the new figure is

(a) 65 cm
(b) 60 cm
(c) 55 cm
(d) 50 cm

As per given figure,

Perimeter = 30 cm
∴ Side = $\frac{30}{6}$ = 5 cm
Hence, remaining perimeter
= AB + BC + CD + DE + EF + FG + GH + HI + IJ + JA
= (5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5) cm
= 50 cm (d)

Question 13.
If the area of a square is numerically equal to its perimeter, then the length of each side is
(a) 1 unit
(b) 2 units
(c) 3 units
(d) 4 units

Let the side of the square be s
∴ Area = (side)2 = s2
And perimeter = 4 × length of the side = 4 s
But, we have, area = perimeter
∴ s2 = 4s ⇒ s = 4
∴ The length of each side is 4 units (d)

Question 14.
If a ribbon of length 10 m is stitched around a rectangular table cloth making 2 rounds along its boundary, then the perimeter of the table cloth is
(a) 20 m
(b) 10 m
(c) 5 m
(d) 2.5 m

Length of the Ribbon = 10m
As per the question,
the ribbon of length is stitched around a rectangular table
cloth-making 2 rounds along with its bounding
∴ Perimeter of the table cloth

Question 15.
A picture is 60 cm wide and 1.8 m long. The ratio of its width to its perimeter in lowest form is
(a) 1 : 2
(b) 1 : 3
(c) 1:6
(d) 1 : 8

Width of picture = 60 cm
Length = 1.8 m = 180 cm
Perimeter = 2(l + b) = 2(60 + 180) = 2(240) = 480 cm
Ratio = $\frac{60}{480}=\frac{1}{8}=1 : 8$ (d)

### Higher Order Thinking Skills

HOTS, Mensuration Class-6 ML Aggarwal ICSE Mathematics Chapter-14 Solutions

Question 1.
How many envelopes of size 25 cm x 15 cm can be made from a rectangular sheet of size 4 m x 1.2 m?

Size of envelope = 25 cm × 15 cm
Area of envelope = 25 × 15 cm2 = 375 cm2
Size of rectangular sheet = 4 m × 1.2 m
Area of rectangular sheet = 400 cm × 120 cm = 48000 cm2

Question 2.
The perimeter of a rectangle is 36 cm. What will be length and breadth (in natural number) of that rectangle whose area is
(i) maximum?
(ii) minimum?

(i) When area is maximum Then, l = 9 cm, b = 9 cm
(ii) When area is minimum Then, l = 17 cm b = 1 cm

### Check Your Progress

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

Question 1.
Look at the given figure and fill in the following blanks:

(i) The contribution in estimation of given area due to completely covered squares is
(ii) The contribution in estimation of given area due to more than half covered squares is
(iii) The contribution in estimation of given area due to exactly half covered square, is
(iv) The contribution in estimation of given area due to less than half covered squares is
(v) The total estimated area is

(i) The contribution in estimation of given area
due to completely covered squares is 2 sq. units.
(ii) The contribution in estimation of given area
due to more than half covered squares is 2 sq. units.
(iii) The contribution in estimation of given area
due to exactly half covered square is 3 sq. units.
(iv) The contribution in estimation of given area
due to less than half covered squares is 0 sq. units.
(v) The total estimated area is 7 sq. units.

Question 2.
The perimeter of a square ABCD is twice the perimeter of ∆PQR. Find the area of the square ABCD.

Perimeter of ∆PQR = 6 cm + 5 cm + 7 cm = 18 cm
Perimeter of square = 2 × 18 = 36 cm
Side of square = ?
4 × side = 36
⇒ Side = $\frac{36}{4}$ = 9
Area of square = (side)2 = (9)= 9 × 9 = 81 cm2

Question 3.
A wire is in the shape of a square of side 10 cm. If the wire is rebent into a rectangle of length 12 cm, find its breadth. Which encloses more area, the square or the rectangle and by how much?

Side of square = 10 cm
Perimeter of square = 4 × 10 cm = 40 cm
According to question,
Perimeter of rectangular = 2(l + b)
⇒ 40 cm = 2 (12 cm + b)
⇒ 40 cm = 24 cm + 2b
⇒ 2b = 40 cm – 24 cm
⇒ 2b = 16 cm
⇒ b = $\frac{16}{2}$ cm = 8 cm
Area of square = (a)= (10 cm)2 = 100 cm2
Area of rectangular = (l × b) = 12 × 8 = 96 cm2
Area of square is more and it is = 100 cm2 – 96 cm2 = 4 cm2

Question 4.
A rectangular room is 9 m long and 6 m wide. Find the cost of covering the floor with carpet 2 m wide at ₹35 per metre.

Length of room = 9m
Width of room = 6m
Area of room = l × b = 9 × 6 = 54 m2
Width of carpet = 2m
Area of carpet = 54 m2
Length = ?
Length of carpet = $\frac{54}{2}$ = 27 m
Cost of covering = ₹35 × 27 m = ₹945

Question 5.
If the cost of fencing a square plot at the rate of ₹ 2.50 per metre is ₹ 200, then find the length of each side of the field.

Total cost of fencing a square plot = ₹200
Rate of fencing = ₹2.50
∴ Perimeter of square = $\frac{\text { Total cost }}{\text { Rate }}$ = $\frac{200}{2.5}$ = 80m
Since, we know that,
Perimeter of square = 4a
⇒ 80 m = 4a
⇒ a = 20 m
∴ Length of a square plot = 20 m

Question 6.
If the cost of fencing a rectangular park at the rate of ₹7.50 per metre is ₹600 and the length of the park is 24 m, find the breadth of the park.

Cost of fencing a rectangular park = ₹600
Rate of fencing = ₹7.50
Perimeter of a park = $\frac{\text { Total cost }}{\text { Rate }}$ = $\frac{600}{7.5}$ = 80m
Length of the park = 24 cm
Let breadth of the park = b
∴ Perimeter of a square = 80 m
⇒ 2(l + b) = 80 m
⇒ 2(24 + b) = 80 m
⇒ 24 + b = 80 m
⇒ b = 40 – 24 m
∴ b = 16 m

Question 7.
By splitting the following figures into rectangle, find their areas (The measures are given in centimetres).

(a) Area of the figure,
= (3 × 1 + 3 × 1 + 3 × 1) sq m
= (3 + 3 + 3) sq m
= 9 sq m

(b) Area of the figure,
= (3 × 3 + 1 × 2 + 3 × 3 + 4 × 2) sq cm
= (9 + 2 + 9 + 8) sq cm
= 28 sq cm

-: End of Mensuration Class-6 ML Aggarwal Solutions  :–

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