# Mensuration Class-6 ML Aggarwal ICSE Maths Solutions

**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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**Objective Type Questions, **

**Multiple Choice Questions ,(MCQ)**

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

Question 1.

Find the perimeter of each of the following figures:

#### Answer

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.

#### Answer

(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?

#### Answer

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?

#### Answer

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?

#### Answer

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.

#### Answer

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?

#### Answer

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?

#### Answer

Perimeter of the regular pentagon = 5 × Length of a side

⇒ Length of one (each) side

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?

#### Answer

(a) Perimeter of the square = 4 × Length of aside

⇒ Length of a side

=

= 7.5 cm

(b) Perimeter of the equilateral triangle = 3 × Length of a side

⇒ Length of a side

=

=

(c) Perimeter of the regular hexagon = 6 × Length of a side

⇒ Length of a side

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.

#### Answer

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?

#### Answer

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?

#### Answer

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:

#### Answer

(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 × 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 × 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 × 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:

#### Answer

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?

#### 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 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

#### 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 5.

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

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?

#### 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 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?

#### 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 = ₹

∴ Cost of tiling 100000 sq. m = ₹ = ₹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.

#### Answer

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.

#### Answer

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?

#### Answer

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.

#### Answer

(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 =

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 =

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.

#### Answer

Area of plot A = 340m^{2}

Length (l) = ?

Area = l × b

⇒ 340 = l × 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.

#### Answer

Area of plot = 144 m^{2}

Length (l) = 16 m.

Breadth (b) = ?

Area = l × b

⇒ 144 = 16 × b

⇒ = 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).

#### Answer

(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}

**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 ……….

#### Answer

(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.

#### Answer

(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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