ML Aggarwal Mensuration Exe-14.2 Class 6 ICSE Maths Solutions

ML Aggarwal Mensuration Exe-14.2 Class 6 ICSE Maths Solutions. We Provide Step by Step Answer of  Exe-14.2 Questions for Mensuration as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-6.

ML Aggarwal Mensuration Exe-14.2 Class 6 ICSE Maths Solutions

Board ICSE
Publications Avichal Publishig Company (APC)
Subject Maths
Class 6th
Chapter-14 Mensuration
Writer ML Aggarwal
Book Name Understanding
Topics Solution of Exe-14.2 Questions
Edition 2023-2024

Mensuration Exe-14.2

ML Aggarwal Class 6 ICSE Maths Solutions

Page-294

Question 1. Find the area of the region enclosed by the following figures by counting squares:

Question 1. Find the area of the region enclosed by the following figures by counting squares:

Answer:

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 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?

Answer:

(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 largest area and rectangle (ii) smallest area.

Question 3. 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

Answer:

(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 4. Find the areas of the squares whose sides are:

(i) 8 cm
(ii) 14 m
(iii) 2 m 50 cm

Answer:

(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 cm = 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


Mensuration Exe-14.2

ML Aggarwal Class 6 ICSE Maths Solutions

Page-295

Question 5. 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?

Answer:

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 6. 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?

Answer:

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 = ₹7.5/100
∴ Cost of tiling 100000 sq. m = ₹ (7.5/100) x 100000 = ₹7500

Question 7. 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.

Answer:

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

Question 7. 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.

∴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 8. In the given figure, find the area of the path (shown shaded) which is 2 m wide all around.

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

Answer:

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 9. 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?

Answer:

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.

Answer:

(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 = (1/60) x 25202 cm = 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 = (1/60) x 14400 cm2 = 240
Hence, 240 tiles are required.

Question 11. The area of a rectangular plot is 340 sq. m. If its breadth is 17 m, find its length and the perimeter.

Answer:

Area of plot A = 340m2
Length (l) = ?
Area = l × b
⇒ 340 = l × 17
⇒ 340/17
⇒ l = 20
Length = 20 m
Perimeter = 2 (l + b) = 2 (20 + 17) = 2 (37) = 74 m

Question 12. 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.

Answer:

Area of plot = 144 m2
Length (l) = 16 m.
Breadth (b) = ?
Area = l × b
⇒ 144 = 16 × b
⇒ 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 13. The perimeter of a square is equal to that of a rectangle of length 17 m and breadth 11 m. Find the area of the square.

Answer:

Length = 17 m

Breadth = 11 m

Perimeter of rectangle = 2(l+b) = 2(17+11)

= 56 m

Perimeter of rectangle = Perimeter of square

= 56 m

Perimeter of square = 56 m

4 × side = 56

Side = 56/4 = 14m

Area of square = side × side = 14 × 14

= 196m sq.

Question 14. A rectangle has same area as that of a square of side 10m. If the breadth of the rectangle is 8 m, find the perimeter of the rectangle.

Answer:

Area of square

= Side × Side

= 10 × 10

= 100m²

According to given conditions;

Area of rectangle = Area of square

length × breadth = 100m²

length × 8 = 100m²

length = 100 ÷ 8

length = 12.5 m

Perimeter of rectangle

= 2 (length + breadth)

= 2 (12.5 + 8)

= 2 × 20.5

= 41m

Question 15. Split the following shapes into rectangles and find their areas. (The measures are given in centimetres).

Question 15. Split the following shapes into rectangles and find their areas. (The measures are given in centimetres).

Answer:

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

Question 15. Split the following shapes into rectangles and find their areas. (The measures are given in centimetres).

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

Question 15. Split the following shapes into rectangles and find their areas. (The measures are given in centimetres).

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

—  : End of ML Aggarwal Mensuration Exe-14.2 Class 6 ICSE Maths Solutions :–

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