Mensuration Class 8 Extra Questions with Answers

These Class 8 Maths Extra Questions Chapter 11 Mensuration will help students prepare well for the exams.

Class 8 Maths Chapter 11 Extra Questions Mensuration

Class 8 Maths Mensuration Extra Questions

Question 1.
The below shape is formed by joining cubes of side 1 cm. Calculate the volume of the shape formed.

Solution:
l = 1 + 1 + 1 + 1 = 4 cm
b = 1 + 1 + 1 = 3 cm
h = 1 + 1 = 2 cm
Volume = l × b × h
= 4 × 3 × 2 = 24 cm³

Question 2.
A rectangular sheet of length l metres and breadth h metres is rolled to form a cylinder of height h metres and diameter d metres.
(The opposite edges coincide exactly without any overlap).

Solution:

Required length
= circumference of circle
= 2πr = πd

Mensuration Extra Questions Class 8

Question 1.
Sarah has a square-shaped corkboard on her wall with a side length of 12cm. If each of her photos measures 4 cm by 6 cm, how many maxiumum photos can Sarah pin on the corkboard without overlapping ?
Solution:
Side of carboard = 12 cm
Area of cardboard = s × s
= 12 × 12 cm²
Measurements of each photo = 4 cm,6 cm
Area of each photo = 4 cm × 6 cm
∴ Number of maximum photos can pin
on the cardboard = \(\frac{12 \times 12}{4 \times 6}\) = 6

Question 2.
The perimeter of a rhombus is 68 cm and the length of one diagonal is 16 cm. Find the area of the rhombus.
Solution:

Perimeter of Rhombus = 68 cm
All sides are equal
∴ 4 × side = 68 cm
side = \(\frac{68}{4}\) cm = 17 cm
One diagonal
length d₁ = 16 cm
Let BD = 16 cm
Then OB = OD
= \(\frac{B D}{2}\) = \(\frac{16}{2}\) cm = 8 cm
In ΔAOB, ∠AOB = 90°
∴ ΔAOD is a right angled triangle
By Pathagoras theorem OA² = AD² – OD²
= 17² – 8²
= 289 – 64 = 225
OA = \(\sqrt{225}\) cm = 15 cm
Now AC = 2 × OA= d₂ = 2 × 15 cm = 30 cm
∴ Area of Rhombus = d₁d₂
= \(\frac{1}{2}\) × 16 × 30 cm²
= 240 cm²

Question 3.
A child builds a big cube using 27 smaller cubes, each measuring 3 cm on each side.

He wants to paint the entire surface of the big cube, and it costs Rs. 5 to paint every square centimeter. How much will it cost the child to paint the whole structure ?
Solution:
Side of each small cube = 3 cm
A big cube is formed by 27 smaller cubes.
∴ Side of big cube = \(\frac{27}{3}\) cm = 9 cm
Big cube has six sides
∴ Total surface area of big cube = 6 × 9 × 9 cm² = 486 cm²
Cost of painting per 1 cm² = ₹ 5
∴ Cost of painting to whole cube = ₹ 486 × 5 = ₹ 2430

Extra Questions of Mensuration Class 8

Question 1.
The length and breadth of a rectangular garden are (4x + 3y) meters and (5x + 2y) meters respectively. A rectangular fountain with dimensions (x + 2y) meters in length and (x – y) meters in width is located at the center of the garden.
a) What is the area of the garden not occupied by the fountain ?
b) How much would it cost to fence this garden at a rate of Rs. (z + 5) per
Solution:
Length of the rectangular garden (l) = (4x + 3y) m
Breadth (b) = (5x + 2y) m
Area of garden = l × b
= (4x + 3y) (5x + 2y) m²
= 4x(5x + 2y) + 3y(5x + 2y) m²
– 20x² + 8xy + 15xy + 6y²
= 20x² + 23xy + 6y² m² ________ (1)
Length of the fountain (l) = (x + 2y) m
Breadth (b) = (x – y) m
Area of fountain = l × b
= (x + 2y) (x – y)
= x² – xy + 2xy – 2y²
= x² + xy – 2y² ________ (2)

a) Area of garden not occupied by fountain
= (1) – (2)
= 20x² + 23xy + 6y² – (x² + xy – 2y²)
= (20x² + 23xy + 6y² – x² – xy + 2y²)m²
= (19x² + 22xy + 8y²) m²

b) Perimeter of the rectangle = 2(l + b)
= 2(4x + 3y + 5x + 2y)
= 2(9x + 5y) m
= (18x + 10 y) m
Cost of fencing per meter = ₹ (z + 5)
Cost of fencing for garden
= ₹ (18x + 10y)(z + 5)
= ₹ 18xz + 90x + 10yz + 50y
= ₹ (18xz + 90x + 50y + 10yz)

Mensuration Class 8 Extra Questions

Question 1.
Find the area of quadrilateral PQRS shown in Fig.

(OR)
Find the area of quadrilateral PQRS.

Solution:
In this case, d = 5.5 cm, h₁ = 2.5cm, h₂ = 1.5 cm,
Area = \(\frac{1}{2}\) d ( h₁ + h₂)
= \(\frac{1}{2}\) × 5.5 × (2.5 + 1.5) cm²
= \(\frac{1}{2}\) × 5.5 × 4 cm² = 11 cm²

Question 2.
Find the area of a rhombus whose diagonals are of lengths 10 cm and 8.2 cm.
Solution:
Area of the rhombus = \(\frac{1}{2}\)d₁ d₂
where d₁, d₂ are lengths of diagonals.
= \(\frac{1}{2}\) × 10 × 8.2 cm² = 41 cm².

Question 1.
The area of a trapezium shaped field is 480 m², the distance between two parallel sides is 15 m and one of the parallel side is 20 m. Find the other parallel side.
Solution:
One of the parallel sides of the trapezium is a = 20 m, let another parallel side be b, height h = 15 m.
The given area of trapezium = 480 m².
Area of a trapezium = \(\frac{1}{2}\) h (a + b)
So 480 = \(\frac{1}{2}\) × 15 × (20 + b) or
\(\frac{480 \times 2}{15}\) = 20 + b
or 64 = 20 + b or b = 44 m
Hence the other parallel side of the trapezium is 44 m.

Question 2.
The area of a rhombus is 240 cm² and one of the diagonals is 16 cm. Find the other diagonal.
Solution:
Let length of one diagonal d₁ = 16 cm and length of the other diagonal = d₂
Area.of the rhombus = \(\frac{1}{2}\) d₁. d₂ = 240
So, \(\frac{1}{2}\)16 . d₂ = 240 2
Therefore, d₂ = 30 cm
Hence the length of the second diagonal is 30 cm.

Question 3.
There is a hexagon MNOPQR of side 5 cm (Fig). Aman and Ridhima divided it in two different ways (Fig).

Find the area of this hexagon using both ways.
Solution:
Aman’s method : Since it is a hexagon so NQ divides the hexagon into two congruent trapeziums. You can verify it by paper folding (Fig).

Now area of trapezium MNQR
= 4 × \(\frac{(11 + 5)}{2}\) = 2 × 16 = 32 cm².
So the area of hexagon MNOPQR
= 2 × 32 = 64 cm².
Ridhima’s method : Δ MNO and Δ RPQ are congruent triangles with altitude 3 cm (Fig).

You can verify this by cutting off these two triangles and placing them on one another.
Area of Δ MNO = \(\frac{1}{2}\) × 8 × 3 = 12 cm²
= Area of Δ RPQ
Area of rectangle MOPR = 8 × 5 = 40 cm².
Now, area of hexagon MNOPQR
= 40 + 12 + 12 = 64 cm².

Question 4.
An aquarium is in the form of a cuboid whose external measures are 80 cm × 30 cm × 40 cm. The base, side faces and back face are to be covered with a coloured paper. Find the area of the paper needed ?

Solution:
The length of the aquarium = l = 80 cm
Width of the aquarium = b = 30 cm
Height of the aquarium = h = 40 cm
Area of the base = l × b = 80 × 30
= 2400 cm²
Area of the side face = b × h = 30 × 40
= 1200 cm²
Area of the back face = l × h = 80 × 40
= 3200 cm²
Required area = Area of the base + area of the back face + (2 × area of a side face)
= 2400 + 3200 + (2 × 1200) = 8000 cm²
Hence the area of the coloured paper required is 8000 cm².

Question 5.
The internal measures of a cuboidal room are 12 m × 8 m × 4 m. Find the total cost of whitewashing all four walls of a room, if the cost of white washing is ₹ 5 per m². What will be the cost of white washing if the ceiling of the room is also whitewashed.
Solution:
Let the length of the room = l = 12 m
Width of the room = b = 8 m
Height of the room = h = 4 m
Area of the four walls of the room
= Perimeter of the base × Height of the room
= 2 (l + b) × h = 2 (12 + 8) × 4
= 2 × 20 × 4 = 160 m².
Cost of white washing per m² = ₹ 5
Hence the total cost of white washing four walls of the room = ₹ (160 × 5) = ₹ 800
Area of ceiling is 12 × 8 = 96 m²
Cost of white washing the ceiling = ₹ (96 × 5) = ₹ 480
So the total cost of white washing = ₹ (800 + 480) = ₹ 1280

Question 6.
In a building there are 24 cylindrical pillars. The radius of each pillar is 28 cm and height is 4 m. Find the total cost of painting the curved surface area of all pillars at the rate of ₹ 8 per m².

Solution:
Radius of cylindrical pillar, r = 28 cm = 0.28 m
height, h = 4 m
curved surface area of a cylinder = 2πrh
curved surface area of a pillar = 2 × \(\frac{22}{7}\) × 0.28 × 4 = 7.04 m²
curved surface area of 24 such pillar = 7.04 × 24 = 168.96 m²
cost of painting an area of 1 m² = ₹ 8
Therefore, cost of painting 1689.6 m² = 168.96 × 8 = ₹ 1351.68

Question 7.
Find the height of a cylinder whose radius is 7 cm and the total surface area is 968 cm².

Solution:
Let height of the cylinder = h, radius = r = 7cm
Total surface area = 2πr (h + r)
i.e., 2 × \(\frac{22}{7}\) × 7 × (7 + h) = 968 h = 15 cm
Hence, the height of the cylinder is 15cm.

Question 8.
Find the height of a cuboid whose volume is 275 cm³ and base area is 25 cm².
Solution:
Volume of a cuboid = Base area × Height
Hence height of the cuboid = \(\frac{\text { Volume of cuboid }}{\text { Base area }}\) = \(\frac{275}{25}\) = 11 cm
Height of the cuboid is 11 cm.

Question 9.
A godown is in the form of a cuboid of measures 60 m × 40 m × 30 m. How many cuboidal boxes can be stored in it if the volume of one box is 0.8 m³ ?
Solution:
Volume of one box = 0.8 m³
Volume of godown = 60 × 40 × 30
= 72000 m³
Number of boxes that can be stored in Volume of the godown = \(\frac{\text { Volume of the godown }}{\text { Volume of one box }}\)
= \(\frac{60 \times 40 \times 30}{0.8}\) = 90,000
Hence the number of cuboidal boxes that can be stored in the godown is 90,000

Question 10.
A rectangular paper of width 14 cm is rolled along its width and a cylinder of radius 20 cm is formed. Find the volume of the cylinder (Fig). (Take \(\frac{22}{7}\) for π)
Solution:
A cylinder is formed by rolling a rectangle about its width. Hence the width of the paper becomes height and radius of the cylinder is 20 cm.

Height of the cylinder = h = 14 cm
Radius = r = 20 cm
Volume of the cylinder = V = π r² h
= \(\frac{22}{7}\) × 20 × 20 × 14 = 17600 cm³
Hence, the volume of the cylinder is 17600 cm³.

Question 11.
A rectangular piece of paper 11 cm × 4 cm is folded without overlapping to make a cylinder of height 4 cm. Find the volume of the cylinder.
Solution:
Length of the paper becomes the perimeter of the base of the cylinder and width becomes height.
Let radius of the cylinder = r and height = h
Perimeter of the base of the cylinder = 2πr = 11
or 2 × \(\frac{22}{7}\) × r = 11
Therefore, r = \(\frac{7}{4}\) cm
Volume of the cylinder = V = πr²h
= \(\frac{22}{7}\) × \(\frac{7}{4}\) × \(\frac{7}{4}\) × 4 cm³ = 38.5 cm³.
Hence the volume of the cylinder is 38.5 cm³.