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Worksheet Solutions: Mensuration

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

Answer the following Questions

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.

Answer the following Questions

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.

Answer the following Questions

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 cmof juice can be poured in a cuboidal can whose dimensions are 15cm × 10cm × 25cm. How many cubical packs of 25cmvolume 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.

The document Worksheet Solutions: Mensuration is a part of the Class 8 Course Mathematics (Maths) Class 8.
All you need of Class 8 at this link: Class 8

FAQs on Worksheet Solutions: Mensuration

1. How do I calculate the surface area of a cube and cuboid in mensuration?
Ans. Surface area of a cube equals 6a², where 'a' is the side length. For a cuboid, it's 2(lw + lh + wh), where l, w, and h are length, width, and height. These formulas measure the total area covering all outer faces, essential for solving CBSE Class 8 mensuration problems involving three-dimensional shapes.
2. What's the difference between curved surface area and total surface area for cylinders?
Ans. Curved surface area covers only the cylindrical sides: 2πrh, where r is radius and h is height. Total surface area includes curved surface plus both circular bases: 2πr(r + h). Understanding this distinction prevents calculation errors in worksheet solutions and helps students identify which measurement a problem requires.
3. How do I find the volume of different shapes like cubes, cuboids, and cylinders?
Ans. Volume of a cube is a³; for a cuboid, it's length × width × height; for a cylinder, it's πr²h. Volume measures space inside three-dimensional objects, calculated in cubic units. These formulas form the foundation of mensuration problems and are critical for CBSE Class 8 assessments involving capacity and storage calculations.
4. Why do I keep getting the wrong answer when calculating areas of composite shapes?
Ans. Composite shapes require breaking them into simpler parts-rectangles, triangles, circles-calculating each area separately, then adding or subtracting. Common mistakes include forgetting to subtract overlapping regions or missing inner cutouts. Visualising shapes through diagrams and checking units prevents errors in mensuration worksheet solutions.
5. What formulas do I need to memorise for the lateral surface area of pyramids and cones?
Ans. Lateral surface area of a cone is πrl, where r is base radius and l is slant height. For pyramids, it's ½ × perimeter × slant height. These measurements exclude the base, essential for distinguishing from total surface area in CBSE mensuration problems involving pointed three-dimensional solids and architectural calculations.
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