## ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 18 Mensuration Objective Type Questions

**Mental Maths**

Question 1.

Fill in the blanks:

(i) Area of a parallelogram = base × …….

(ii) Area of a trapezium = \(\frac{1}{2}\) × ……….. × distance between parallel sides.

(iii) Area of a rhombus = \(\frac{1}{2}\) × product of ……..

(iv) Area is measured in ……….. units.

(v) Volume of a solid is the measurement of ………… occupied by it.

(vi) Volume is measured in ………… units.

(vii) The volume of a unit cube is ……….

(viii) 1 litre = …………… cm^{3}

(ix) 1 m^{3} = ………… litres

(x) Volume of a cuboid = ……….. × height.

(xi) Cylinders in which line segment joining the centres of the circular faces is perpendicular to the base are called ……….

(xii) Volume of a cylinder = area of base × ………..

(xiii) Area of four walls = perimeter of floor × …….

(xiv) Lateral surface area of a cube = 4 × (…………)^{2}

(xv) Total surface area of a cylinder of radius r and height h is ………..

Solution:

(i) Area of a parallelogram = base × height.

(ii) Area of a trapezium = \(\frac{1}{2}\) × (sum of parallel sides)

× distance between parallel sides.

(iii) Area of a rhombus = \(\frac{1}{2}\) × product of its diagonals.

(iv) Area is measured in square units.

(v) Volume of a solid is the measurement of the space occupied by it.

(vi) Volume is measured in cubic units.

(vii) The volume of a unit cube is 1 cubic unit.

(viii) l litre = 1000 cm^{3}

(ix) 1 m^{3} = 1000 litres.

(x) Volume of a cuboid = length × breadth × height.

OR

Volume of a cuboid = area of the base × height.

(xi) Cylinders in which line segment joining the centres of the circular faces

is perpendicular to the base are called right circular cylinders.

(xii) Volume of a cylinder = area of base × height.

(xiii) Area of four walls = perimeter of floor × height of the room.

(xiv) Lateral surface area of a cube = 4 × (edge)^{2}

(xv) Total surface area of a cylinder of radius r and height h is 2πr(h + r).

Question 2.

State which of the following statements are true (T) or false (F):

(i) Perimeter of a rectangle is the sum of lengths of its four sides.

(ii) Area of a quadrilateral can be found by splitting it into two triangles.

(iiii) Perimeter of a circle of radius r = πr^{2}.

(iv) Volume of a cube = 6 × (side)^{2}

(v) 1 m^{3} = 100000 cm^{3}

(vi) Total surface area of a cuboid

= 2 (lb + bh + hl)

(vii) There is no difference between volume and capacity.

(viii)Total surface area of a cylinder = lateral surface area + area of two circular ends.

(ix) Surface area of a cube = 4 × (side)^{2}

(x) Lateral surface area of a cuboid = perimeter of base × height.

Solution:

(i) Perimeter of a rectangle is the sum of lengths of its four sides. True

(ii) Area of a quadrilateral can be found by splitting it into two triangles. True

(iii) Perimeter of a circle of radius r = πr^{2}. False

Correct :

It is area of a circle perimeter is 2πr.

(iv) Volume of a cube = 6 × (side)^{2} False Correct :

It is surface area not volume, volume is (side)^{3}.

(v) 1 m^{3} = 100000 cm^{3} False

Correct:

1 m^{3} = 1000000 cm^{3}

(vi) Total surface area of a cuboid

= 2 (lb + bh +hl) True

(vii) There is no difference between volume and capacity. False

Correct :

Volume refers to the amount of space occupied by an object

whereas capacity refers to the quantity that a container holds.

(viii)Total surface area of a cylinder = lateral surface area

+ area of two circular ends. True

(ix) Surface area of a cube = 4 × (side)^{2} False Correct :

It is 6 × (side)^{2}

(x) Lateral surface area of a cuboid = perimeter of base × height. True

**Multiple Choice Questions**

**Choose the correct answer from the given four options (3 to 17):**

Question 3.

Area of a triangle is 30 cm^{2}. If its base is 10 cm, then its height is

(a) 5 cm

(b) 6 cm

(c) 7 cm

(d) 8 cm

Solution:

Area of a triangle = 30 cm^{2}

Base = 10 cm

Area = \(\frac{1}{2}\) × Base × Height

Height = \(\frac{A \times 2}{B}=\frac{30 \times 2}{10}\) =6 cm (b)

Question 4.

If the perimeter of a square is 80 cm, then its area is

(a) 800 cm^{2}

(b) 600 cm^{2}

(c) 400 cm^{2}

(d) 200 cm^{2}

Solution:

Perimeter of a square = 80 cm

Perimeter of square = 4(Side)

∴ Side = \(\frac{\text { Perimeter }}{4}\)

∴ Side = \(\frac{80}{4}\) = 20 cm

Area = (side)^{2} = (20)^{2} = 400 cm^{2} (c)

Question 5.

Area of a parallelogram is 48 cm^{2}. If its height is 6 cm then its base is

(a) 8 cm

(b) 4 cm

(c) 16 cm

(d) none of these

Solution:

Area of parallelogram = 48 cm^{2}

Height = 6 cm

Area of ||gm = Base × Height

Base = \(\frac{\mathrm{A}}{h}=\frac{48}{6}\) = 8 cm (a)

Question 6.

If d is the diameter of a circle, then its area is

(a) πd^{2}

(b) \(\frac{\pi d^{2}}{2}\)

(c) \(\frac{\pi d^{2}}{4}\)

(d) 2πd^{2}

Solution:

d is the diameter a circle

∴ Radius = r =\(\frac{d}{2}\)

Area = πr^{2} = π\(\left(\frac{d}{2}\right)^{2}\) = π\(\frac{d^{2}}{4}\) (c)

Question 7.

If the area of a trapezium is 64 cm^{2} and the distance between parallel sides is 8 cm, then sum of its parallel sides is

(a) 8 cm

(b) 4 cm

(c) 32 cm

(d) 16 cm

Solution:

Area of trapezium = 64 cm^{2}

Distance between parallelogram (h) = 8 cm

Area of trapezium = \(\frac{1}{2}\) × (Sum of ||gm sides) × h

Sum of parallel lines = \(\frac{\mathrm{A} \times 2}{h}\)

= \(\frac{64 \times 2}{8}\) = 16 cm (d)

Question 8.

Area of a rhombus whose diagonals are 8 cm and 6 cm is

(a) 48 cm^{2}

(b) 24 cm^{2}

(c) 12 cm^{2}

(d) 96 cm^{2}

Solution:

Area of rhombus =\(\frac{d_{1} \times d_{2}}{2}=\frac{8 \times 6}{2}\)

= \(\frac{48}{2}\) = 24 cm^{2} (b)

Question 9.

If the lengths of diagonals of a rhombus is doubled, then area of rhombus will be

(a) doubled

(b) tripled

(c) four times

(d) remains same

Solution:

Area of rhombus_{1} = \(\frac{1}{2}\) × d_{1} × d_{2}

Now the diagonals are doubled

Area of rhombus = \(\frac{1}{2}\) × 2d_{1} × 2d_{2} = 2d_{1}d_{2}

with doubled diagonals.

Lengths of diagonals are doubled, then the area will be four times. (c)

Question 10.

If the length of a diagonal of a quadrilateral is 10 cm and lengths of the perpendiculars on it from opposite vertices are 4 cm and 6 cm, then area of quadrilateral is

(a) 100 cm^{2}

(b) 200 cm^{2}

(c) 50 cm^{2}

(d) none of these

Solution:

Length of a diagonal of a quadrilateral = 10 cm

and lengths of perpendicular on it from the

opposite vertices = 4 cm and 6 cm

∴ Area = \(\frac{1}{2}\) (4 + 6) × 10 cm^{2}

= \(\frac{1}{2}\) × 10 × 10 = 50 cm^{2} (c)

Question 11.

Area of a rhombus is 90 cm^{2}. If the length of one diagonal is 10 cm then the length of other diagonal is

(a) 18 cm

(b) 9 cm

(c) 36 cm

(d) 4.5 cm

Solution:

Area of a rhombus = 90 cm^{2}

Length of one diagonal = 10 cm

Area of rhombus = \(\frac{1}{2}\) × d_{1} × d_{2}

∴ Length of second diagonal = \(\frac{A \times 2}{d_{1}}\)

= \(\frac{90 \times 2}{10}\) = 18cm (a)

Question 12.

If the volume of a cube is 729 cm^{3}, then its surface area is

(a) 486 cm^{2}

(b) 324 cm^{2}

(c) 162 cm^{2}

(d) none of these

Solution:

Volume of a cube = 729 cm^{3}

V = (Side)^{3}

∴ Side (edge) = \(\sqrt[3]{729}=\sqrt[3]{9 \times 9 \times 9}\) = 9 cm

Then surface area = 6 × (side)^{2}

= 6 × (9)^{2} = 6 × 81 cm^{2}

= 486 cm^{2} (a)

Question 13.

If the lateral surface area of a cube is 100 cm^{2}, then its volume is

(a) 25 cm^{3}

(b) 125 cm^{3}

(c) 625 cm^{3}

(d) none of these

Solution:

Lateral surface area of a cube = 4(Edge)^{2} = 100 cm^{2}

∴ 4 × (edge)^{2} = 100

⇒ (edge)^{2} = \(\frac{100}{4}\) = 25 = (5)^{2}

∴ Edge of cube = 5 cm

Volume = (edge)^{3 }= (5)^{3} = 125 cm^{3} (b)

Question 14.

If the length of side of a cube is doubled, then the ratio of volumes of new cube and original cube is

(a) 1 : 2

(b) 2 : 1

(c) 4 : 1

(d) 8 : 1

Solution:

Let original side of a cube = x

Then volume = x^{3}

If edge is doubled i.e. edge = 2x

Then volume = (2x)^{3} = 8x^{3}

∴ Ratio between new cube and original cube

= 8x^{3} : x = 8 : 1 (d)

Question 15.

If the dimensions of a rectangular room are 10m × 12m × 9m, then the cost of painting its four walls at the rate of ₹8 per m^{2} is

(a) ₹3186

(b) ₹3618

(c) ₹3168

(d) none of these

Solution:

Dimensions of a room = 10m× 12 × 9m

Area of 4 walls = 2(l + b)h

= 2(10 + 12) × 9 = 2 × 22 × 9 m^{2}

= 396 cm^{2}

Cost of painting = ₹8 per m^{2}

∴ Total cost = 396 × 8 = ₹3168 (c)

Question 16.

Volume of a cylinder is 1848 cm^{2}. If the diameter of its base is 14 cm, then the height of the cylinder is

(a) 12 cm

(b) 6 cm

(c) 3 cm

(d) none of these

Solution:

Volume of a cylinder = 1848 cm^{2}

Diameter of base = 14 cm

∴ Radius (r) = \(\frac{14}{2}\) = 7 cm

V = πr^{2}h

∴ Height = \(\frac{\mathrm{V}}{\pi r^{2}}\)

= \(\frac{1848 \times 7}{22 \times 7 \times 7}\) = 12 cm

Question 17.

If the radius of a cylinder is doubled and height is halved, then new volume is

(a) same

(b) 2 times

(c) 4 times

(d) 8 times

Solution:

Let radius = r

and height = h

Then volume = πr^{2}h

If radius is doubled i. e. 2 r and height is halved

i.e. \(\frac{h}{2}\), then

Volume = π(2r)^{2} × \(\frac{h}{2}\)

= π × 4r^{2} × \(\frac{h}{2}\) = 2πr^{2}h

∴ Its volume is doubled (2 times) (b)

**Value Based Questions**

Question 1.

Pulkit painted four walls and roof of a rectangular room of size 10m × 12m × 12m. He got ₹10 per m^{2} for his work. How much money he earned? He always give one fourth of his income to an orphanage. Find how much money he gave to orphanage? What values are being promoted?

Solution:

Dimensions of a room = 10m × 12m × 10m

∴ Area of 4-walls = 2(l + b) × h

= 2(10 + 12) × 10 m^{2}

= 2 × 22 × 10 = 440 m^{3}

and area of cielings = l x b = 10 × 12 = 120 m^{2}

Total area = 440 + 120 = 560 m^{2}

Rate of painting = ₹10 per m^{2}

Total changes for painting = ₹560 × 10 = ₹5600

Money gave to an orphanage = \(\frac{1}{4}\) of ₹5600 = ₹1400

Remaining money = ₹5600 – 1400 = ₹4200

Amount given to an orphanage is a good and noble deed.

He help the poor and needy.

Question 2.

In a slogan writing competition in a school, Rama wrote the slogan ‘Truth pays, never betrays’ on a trapezium shaped cardboard. If the lengths of parallel sides of trapezium are 60 cm and 80 cm and the distance between them is 50 cm, find the area of trapezium. What are the advantages of speaking truth?

Solution:

Parallel sides of a trapezium = 60 cm and 80 cm

and distance between then = 50 cm

∴ Area of trapezium = \(\frac{1}{2}\) (sum of parallel sides) × height

= \(\frac{1}{2}\)(60 + 80) × 50

= \(\frac{1}{2}\) × 140 × 50 cm^{3}

= 3500 cm^{3}

Rama wrote on it a slogan: Truth pays, never betrays’

Always speek the truth. It pays in the long run on speaking truth,

people will believe you, as truth is like a God.

**Higher Order Thinking Skills (Hots)**

Question 1.

The length of a room is 50% more than its breadth. The cost of carpeting the room at the rate of ₹38.50 m^{2} is ₹924 and the cost of papering the walls at ₹3.30 m^{2} is ₹214.50. If the room has one door of dimensions 1 m × 2 m and two windows each of dimensions 1 m × 1.5 m, find the dimensions of the room.

Solution:

Length of a room is 50% more than its breadth

Let breadth (b) = xm

Then length (l) = x + \(\frac{150}{100}=\frac{3}{2} x \mathrm{m}\)

Cost of carpeting the room at the rate of ₹38.50 = ₹924

Area of floor = ₹ \(\frac{924}{38.50}\)

= \(\frac{924 \times 100}{3850}\) = 24 m^{2}

∴ l × b = 24 m^{2} …(i)

Cost of papering the walls at the rate of ₹33.30 per m^{2} = ₹3214.50

∴ Area of paper \(\frac{214.50}{3.30}=\frac{21450}{330}=65 \mathrm{m}^{2}\)

Area of one door of dimension 1 m × 2 m = 2 m^{2}

and area of two windows of size

= 1 × 1.5 m = 1 × 1.5 × 2 = 3 m^{2}

∴ Area of 4-walls = 65 + 2 + 3 = 70 m^{2}

Now, l × b = 24 ⇒ \(\frac{3}{2}\)x × x = 24

⇒ x^{2} = \(\frac{24 \times 2}{3}\) = 16 = (4)^{2}

∴ x = 4

∴ Length = 4 × \(\frac{3}{2}\) = 6 m

and breadth = x = 4 m

∴ Length = 6 m, breadth = 4 m and height = 3.5 m

Question 2.

The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used up when writing 310 words on an average. How many words would use up a bottle of ink containing one-fifth of a litre? Answer correct to the nearest 100 words.

Solution:

Height of cylindrical shaped barrel (h) = 7 cm

Diameter = 5 mm

∴ Radius (r) = \(\frac{5}{2}\) mm

One-fifth of litre = 200 ml

∴ In 200 ml, words will be written

Question 3.

A cylindrical jar is 20 cm high with internal diameter 7 cm. An iron cube of edge 5 cm is immersed in the jar completely in the water which was originally 12 cm high. Find the rise in the level of water.

Solution:

Height of cylindrical jar = 20 cm

and diameter = 7 cm

Question 4.

Squares each of side 6 cm are cut off from the four comers of a sheet of tin measuring 42 cm by 30 cm. The remaining portion of the tin sheet is made into an open box by folding up the flaps. Find the capacity of the box.

Solution:

From a sheet, squares of 6 cm sides are cut and cutout

Length of sheet = 32 cm

and breadth = 30 cm

Remaining sheet is folded into a box whose length

= 42 – 6 × 2 = 42 – 12 = 30 cm

Breadth = 30 – 6 × 2 = 30 – 12 = 18 cm

and height = 6 cm

Capacity of the box = 30 × 18 × 6 = 3240 cm^{3}