## ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 14 Mensuration Ex 14.2

Question 1.

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

Solution:

(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:

Solution:

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?

Solution:

(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

Solution:

(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

Solution:

(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?

Solution:

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?

Solution:

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.

Solution:

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.

Solution:

Length of field = 100 m

Breadth of field = 60 m

Area of the field = L × B = 100 × 60 = 6000m^{2}

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 m^{2}

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?

Solution:

Area of 1 square flower bed = a^{2 }= 1.5 × 1.5 = 2.25 m^{2}

Area of 4 square flower bed = 4 × 2.25 = 9 m^{2}

Length of land = 8m

Breadth of land = 6.5 m

Area of land = l × b = 8 × 6.5 = 52 m^{2}

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.

Solution:

(i) Length (l) of tile = 12 cm

breadth (b) of tile = 5 cm

Area of tile = 1 × b = 12 cm × 5 cm = 60 cm^{2}

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 cm^{2}

If 60 cm^{2} area is covered then tile required = 1

If 2520 cm^{2} 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 cm^{2} (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 cm^{2}

If 60 cm^{2} are is covered then tile required = 1

If 14400 cm^{2} 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.

Solution:

Area of plot A = 340m^{2}

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.

Solution:

Area of plot = 144 m^{2}

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).

Solution:

(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) cm^{2} = 24 cm^{2}

Length of Part II = 8 cm

Breadth of Part II = 2 cm

Area of Part II = l × b = (8 × 2) cm^{2} = 16 cm^{2}

∴ Total area = Area of Part I + Area of Part II = (24 + 16) cm^{2} = 40 cm^{2}

(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 cm^{2}

Total rectangles = 5

∴ Total area = 5 × 49 cm^{2} = 245 cm^{2}

(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 cm^{2}

Length of IInd Part = 5 cm

Breadth of IInd Part = 1 cm

Area = l × b = 5 × 1 = 5 cm^{2}

∴ Total area = 4 cm^{2} + 5 cm^{2} = 9 cm^{2}