ML Aggarwal Mensuration MCQs Class 8 ICSE Ch-18 Maths Solutions

ML Aggarwal Mensuration MCQs Class 8 ICSE Ch-18 Maths Solutions. We Provide Step by Step Answer of  MCQs Questions for Mensuration as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-8.

ML Aggarwal Mensuration MCQs Class 8 ICSE Maths Solutions

Board ICSE
Publications Avichal Publishig Company (APC)
Subject Maths
Class 8th
Chapter-18 Mensuration
Writer ML Aggarwal
Book Name Understanding
Topics Solution of MCQs
Edition 2023-2024

Mensuration MCQs

ML Aggarwal Class 8 ICSE Maths Solutions

Page-313

Mental Maths

Question 1. Fill in the blanks:

(i) Area of a parallelogram = base × …….
(ii) Area of a trapezium = 1/2 × ……….. × distance between parallel sides.
(iii) Area of a rhombus = 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 = …………… cm3
(ix) 1 m3 = ………… 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 ………..

Answer:

(i) Area of a parallelogram = base × height.
(ii) Area of a trapezium = 1/2 × (sum of parallel sides) × distance between parallel sides.
(iii) Area of a rhombus = 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 cm3
(ix) 1 m3 = 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 = πr2.
(iv) Volume of a cube = 6 × (side)2
(v) 1 m3 = 100000 cm3
(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.

Answer:

(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 = πr2. 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 m3 = 100000 cm3 False
Correct:
1 m3 = 1000000 cm3
(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


Mensuration MCQs

ML Aggarwal Class 8 ICSE Maths Solutions

Page-314

MCQs, Mensuration Class-8 ML Aggarwal ICSE Mathematics Solutions Chapter-18

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

Question 3. Area of a triangle is 30 cm2. If its base is 10 cm, then its height is

(a) 5 cm
(b) 6 cm
(c) 7 cm
(d) 8 cm

Answer:

Area of a triangle = 30 cm2
Base = 10 cm
Area = 1/2 × Base × Height
Height = (A x 2)/B = (30 x 2)10 =6 cm (b)

Question 4. If the perimeter of a square is 80 cm, then its area is

(a) 800 cm2
(b) 600 cm2
(c) 400 cm2
(d) 200 cm2

Answer:

Perimeter of a square = 80 cm
Perimeter of square = 4(Side)
∴ Side = Perimeter/4
∴ Side = 80/2 = 20 cm
Area = (side)2 = (20)2 = 400 cm2 (c)

(ML Aggarwal Mensuration MCQs Class 8 ICSE Maths)

Question 5. Area of a parallelogram is 48 cm2. If its height is 6 cm then its base is

(a) 8 cm
(b) 4 cm
(c) 16 cm
(d) none of these

Answer:

Area of parallelogram = 48 cm2
Height = 6 cm
Area of ||gm = Base × Height
Base =A/h = 48/6 = 8 cm (a)

Question 6. If d is the diameter of a circle, then its area is

(a) πd2
(b) πd2/2
(c) πd2/4
(d) 2πd2

Answer:

d is the diameter a circle
∴ Radius = r =d/2
Area = πr2 = π(d/2)² = πd²/4 (c)

Question 7. If the area of a trapezium is 64 cm2 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

Answer:

Area of trapezium = 64 cm2
Distance between parallelogram (h) = 8 cm
Area of trapezium = 1/2 × (Sum of ||gm sides) × h
Sum of parallel lines = (A x 2)/h
= (64 x 2)/8 = 16 cm (d)

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

(a) 48 cm2
(b) 24 cm2
(c) 12 cm2
(d) 96 cm2

Answer:

Area of rhombus = (d1 x d2)/2 = (8 x 6)/2
= 48/2 = 24 cm2 (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

Answer:

Area of rhombus1 = 1/2 × d1 × d2
Now the diagonals are doubled
Area of rhombus = 1/2 × 2d1 × 2d2 = 2d1d2
with doubled diagonals.
Lengths of diagonals are doubled, then the area will be four times. (c)

(ML Aggarwal Mensuration MCQs Class 8 ICSE Maths)

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 cm2
(b) 200 cm2
(c) 50 cm2
(d) none of these

Answer:

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 = 1/2 (4 + 6) × 10 cm2
= 1/2 × 10 × 10 = 50 cm2 (c)

Question 11. Area of a rhombus is 90 cm2. 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

Answer:

Area of a rhombus = 90 cm2
Length of one diagonal = 10 cm
Area of rhombus = 1/2 × d1 × d2
∴ Length of second diagonal = (A x 2)d1
= (90 x 2)/10  = 18cm (a)


Mensuration MCQs

ML Aggarwal Class 8 ICSE Maths Solutions

Page-315

Question 12. If the volume of a cube is 729 cm3, then its surface area is

(a) 486 cm2
(b) 324 cm2
(c) 162 cm2
(d) none of these

Answer:

Volume of a cube = 729 cm3
V = (Side)3
∴ Side (edge) = ml class 8 chapter 18 imh 40 = 9 cm
Then surface area = 6 × (side)2
= 6 × (9)2 = 6 × 81 cm2
= 486 cm2 (a)

Question 13. If the lateral surface area of a cube is 100 cm2, then its volume is

(a) 25 cm3
(b) 125 cm3
(c) 625 cm3
(d) none of these

Answer:

Lateral surface area of a cube = 4(Edge)2 = 100 cm2
∴ 4 × (edge)2 = 100
⇒ (edge)2 = 100/4 = 25 = (5)2
∴ Edge of cube = 5 cm
Volume = (edge)= (5)3 = 125 cm3 (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

Answer:

Let original side of a cube = x
Then volume = x3
If edge is doubled i.e. edge = 2x
Then volume = (2x)3 = 8x3
∴ Ratio between new cube and original cube
= 8x3 : 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 m2 is

(a) ₹3186
(b) ₹3618
(c) ₹3168
(d) none of these

Answer:

Dimensions of a room = 10m× 12 × 9m
Area of 4 walls = 2(l + b)h
= 2(10 + 12) × 9 = 2 × 22 × 9 m2
= 396 cm2
Cost of painting = ₹8 per m2
∴ Total cost = 396 × 8 = ₹3168 (c)

Question 16. Volume of a cylinder is 1848 cm2. 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

Answer:

Volume of a cylinder = 1848 cm2
Diameter of base = 14 cm
∴ Radius (r) = 14/2 = 7 cm
V = πr2h
∴ Height = V/πr2
= (1848 x 7)/ (22 x 7 x 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

Answer:

Let radius = r
and height = h
Then volume = πr2h
If radius is doubled i. e. 2 r and height is halved
i.e. h/2, then
Volume = π(2r)2 × h/2
= π × 4r2 × h/2 = 2πr2h
∴ Its volume is doubled (2 times) (b)


Mensuration MCQs

ML Aggarwal Class 8 ICSE Maths Solutions

Page-315

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 m2 is ₹924 and the cost of papering the walls at ₹3.30 m2 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.

Answer:

Length of a room is 50% more than its breadth
Let breadth (b) = xm
Then length (l) = x + (150/100) = (3/2)xm
Cost of carpeting the room at the rate of ₹38.50 = ₹924
Area of floor = ₹ 924/38.50
= (924 x 100)/3850 = 24 m2
∴ l × b = 24 m2 …(i)
Cost of papering the walls at the rate of ₹33.30 per m2 = ₹3214.50
∴ Area of paper (214.50)/3.30 = (21450/330) = 65m²
Area of one door of dimension 1 m × 2 m = 2 m2
and area of two windows of size
= 1 × 1.5 m = 1 × 1.5 × 2 = 3 m2
∴ Area of 4-walls = 65 + 2 + 3 = 70 m2
Now, l × b = 24 ⇒ 3/2x × x = 24
⇒ x2 = (24 x 2)/3  = 16 = (4)2
∴ x = 4
∴ Length = 4 × 3/2 = 6 m
and breadth = x = 4 m

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 m2 is ₹924 and the cost of papering the walls at ₹3.30 m2 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.
∴ 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.

Answer:

Height of cylindrical shaped barrel (h) = 7 cm
Diameter = 5 mm
∴ Radius (r) = 5/2 mm
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.
One-fifth of litre = 200 ml
∴ In 200 ml, words will be written
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.

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.

Answer:

Height of cylindrical jar = 20 cm
and diameter = 7 cm
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.

(ML Aggarwal Mensuration MCQs Class 8 ICSE Maths)

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.

Answer:

From a sheet, squares of 6 cm sides are cut and cutout
Length of sheet = 32 cm
and breadth = 30 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.
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 cm3

— End of Mensuration  MCQs Class 8 ICSE Maths Solutions :–

Return to : ML Aggarwal Maths Solutions for ICSE Class -8

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