Mensuration Class 8 Worksheet Maths Chapter 9

Multiple Choice Questions

Q1: The area of a parallelogram whose base is 9cm and altitude is 6cm
(a) 45cm²
(b) 54cm²
(c) 48cm²
(d) 84cm²
​Ans: (b)
Sol: Area of a parallelogram = base × altitude
= 9 cm × 6 cm
= 54 cm²
Hence the correct option is (b).

Q2: The volume of a cube whose edge is 6a is
(a) 25a³
(b) 216a³
(c) 125a³
(d) None of these 
​Ans: (b)
Sol: Volume of a cube = (edge)³
= (6a)³
= 6 × 6 × 6 × a³ = 216a³
Therefore the correct option is (b).

Q3: The sum of the areas of all six faces of a cuboid is the _____ of the cuboid. 
(a) Volume 
(b) Surface area 
(c) Area 
(d) Curved surface area 
​Ans: (b)
Sol: The sum of the areas of all six faces of a cuboid gives its total surface area.
Formula: Surface area = 2(lb + bh + hl).
Therefore the correct option is (b).

Q4: The area of a Rhombus is 240 cm² and one of the diagonals is 16 cm Then other diagonal is
(a) 25cm
(b) 30cm
(c) 18cm
(d) 35cm
​Ans: (b)
Sol: Area of a rhombus = (1/2) × d₁ × d₂, where d₁ and d₂ are the diagonals.
Given area = 240 cm², d₁ = 16 cm.
So 240 = (1/2) × 16 × d₂.
Multiply both sides by 2: 480 = 16 × d₂.
d₂ = 480 / 16 = 30 cm.
Therefore the correct option is (b).

Q5: The volume of water tank is 3m³. Its capacity in litres is 
(a) 30
(b) 300
(c) 3000
(d) None of these 
​Ans: (c)
Sol: 1 m³ = 1000 litres.
So 3 m³ = 3 × 1000 = 3000 litres.
Therefore the correct option is (c).

Fill in The Blanks

Q1: 10000 m² = ________ hectare.
Ans: 1 hectare.

Q2: 1 m² = ________ cm2.
Ans: 10,000 cm²

Q3: Perimeter of a regular polygon = length of one side ×  ________.
Ans: number of sides

Q4: The distance around a circle is its  ________.
Ans: circumference

Q5: If a wire in the shape of the square is rebent into a rectangle, then the  ________  of both shapes remain same, but  ________  may vary.
Ans: perimeter, area

True(T) or False(F)

Q1: All the triangles that are equal in area are congruent.
Ans: False
Explanation: Two triangles can have the same area but different shapes and side lengths. Equal area does not guarantee identical sides or angles, so they need not be congruent.

Q2: All congruent triangles are equal in area.
Ans: True
Explanation: Congruent triangles are identical in shape and size; hence their areas are equal.

Q3: Ratio of the circumference and the diameter of a circle is more than 3.
Ans: True
Explanation: The ratio equals π, which is approximately 3.1416, so it is greater than 3.

Q4: If the area of rectangle increases from 2 cm² to 4 cm², then perimeter will remains same.
Ans: False
Explanation: Perimeter depends on the side lengths. If area changes, side lengths normally change and so does the perimeter, unless specific dimensions change in a way that keeps perimeter constant (rare).

Answer the following Questions

Q1: Find the area of a square, the length of the diagonal is 2√2 m
Ans:
Let diagonal = d = 2√2 m.
For a square, Area = (d²)/2.
d² = (2√2)² = 4 × 2 = 8.
Area = 8 / 2 = 4 m².

Q2: If the parallel sides of a parallelogram are 2cm apart and their sum is 12cm then find its area.
Ans: Distance between the parallel sides (height) = 2 cm
Sum of the parallel sides = 12 cm
Area of a parallelogram = height × sum of parallel sides
Area = 2 × 12 = 24 cm²

Q3: The length, breadth and height of a cuboid are 20cm, 15cm and 10cm respectively. Find its total surface area. 
Ans:
L = 20 cm, B = 15 cm, H = 10 cm.
Surface area = 2(lb + bh + hl).
= 2(20×15 + 15×10 + 10×20) cm².
= 2(300 + 150 + 200) cm² = 2(650) cm² = 1300 cm².

Q4: Volume of Cube is 8000cm³. Find its surface area.
Ans:
V = l³ = 8000 cm³.
So edge l = ∛8000 = 20 cm.
Surface area = 6l² = 6 × (20)² = 6 × 400 = 2400 cm².

Q5: Find the ratio of the areas of two circles whose radii is 7cm and 14cm.
Ans:
r₁ = 7 cm ⇒ A₁ = π(7)² = 49π.
r₂ = 14 cm ⇒ A₂ = π(14)² = 196π.
Ratio A₁:A₂ = 49π : 196π = 49 : 196 = 1 : 4.

Q6: Find the diameter of the circle whose circumference is 230m.
Ans:
Circumference C = 230 m.
C = 2πr ⇒ r = C / (2π) = 230 / (2π).
Using π = 22/7, r = 230 × 7 / (2 × 22) = 1610 / 44 ≈ 36.5909 m.
Diameter d = 2r ≈ 73.1818 m ≈ 73.18 m.

Q7: Find the area of the figure if the upper portion is a semicircle

Ans:
Total area = Area of semicircle + Area of rectangle.
Semicircle radius r = 7 cm (since diameter matches rectangle width 14 cm).
Area of semicircle = (1/2)πr² = (1/2)π(7)² = (1/2)π×49 = 49π/2.
Using π = 22/7: 49π/2 = 49×22/(7×2) = 77 cm².
Area of rectangle = length × breadth = 14 × 8 = 112 cm².
Total area = 77 + 112 = 189 cm².

Q8: A goat is tied to one corner of a square field of side 8m by a rope 3m long. Find the area it can graze? Also find the area the goat cannot graze.

Ans:
Side of square = 8 m ⇒ Area of square = 8 × 8 = 64 m².
Rope length = 3 m = radius r of grazing circle sector inside the square.
When tied at a corner, the goat can graze one quarter of a full circle of radius 3 m inside the square.
Area grazed = (1/4)πr² = (1/4)π(3)² = (9π)/4.
Using π = 22/7: (9π)/4 = 9×22/(7×4) ≈ 7.07 m².
Area the goat cannot graze = Area of square - Area grazed = 64 - (9π)/4 ≈ 64 - 7.07 = 56.93 m².

Q9: If x units are added to the length of the radius of a circle, what is the number of units by which the circumference of the circle is increased?
Ans:
Let original radius = r, original circumference = 2πr.
New radius = r + x, new circumference = 2π(r + x) = 2πr + 2πx.
Increase in circumference = 2πx units.

Q10: Find the area of the shaded portion if diameter of circle is 16cm  and ABCD is a square.

Ans:
Radius of circle = 16/2 = 8 cm ⇒ Area of circle = πr² = π(8)² = 64π.
Square ABCD has diagonal = 16 cm. Area of a square = (diagonal)² / 2 = 16² / 2 = 256 / 2 = 128 cm².
Area of shaded portion = Area of circle - Area of square = 64π - 128.
Using π = 22/7 gives approximate value: 64π ≈ 201.14, so shaded area ≈ 201.14 - 128 = 73.14 cm² (approx). Exact value: 64π - 128 cm².

Q11: How many cm³ of juice can be poured in a cuboidal can whose dimensions are 15cm × 10cm × 25cm. How many cubical packs of 25cm³ volume can be made?
Ans:
Volume of cuboid = Length × Breadth × Height = 15 cm × 10 cm × 25 cm = 3750 cm³.
Each cubical packet has volume = 25 cm³.
Number of such packets = 3750 ÷ 25 = 150 packets.

Q12: A rectangular piece of paper66 cm long and10cm broad is rolled along the length to form a cylinder. What is the radius of the base and calculate volume of cylinder?
Ans:
When rolled along the length, the length 66 cm becomes the circumference of the base of the cylinder.
Circumference C = 66 cm = 2πr ⇒ r = 66 / (2π) = 33 / π.
Using π = 22/7, r = 33 × 7 / 22 = 10.5 cm.
Height of cylinder h = breadth = 10 cm.
Volume of cylinder = πr²h = π × (10.5)² × 10.
(10.5)² = 110.25, so
Volume = π × 110.25 × 10 = 1102.5π cm³.
Using π = 22/7, Volume = 1102.5 × 22 / 7 = 3465 cm³.