CBSE Class 8 Mathematics Mensuration Assignment Set 08

VIII - Mathematics Assignment No-04 - Mensuration

Q1. Find the volume and total surface area of a cuboid whose length = 5 m, breadth = 3 m and height = 2.5 m.
Answer: Length (\(l\)) = 5 m
Breadth (\(b\)) = 3 m
Height (\(h\)) = 2.5 m
Volume of cuboid = \(l \times b \times h\)
Volume = \(5 \times 3 \times 2.5 = 37.5\text{ m}^3\)
Total Surface Area (T.S.A.) = \(2(lb + bh + hl)\)
T.S.A. = \(2(5 \times 3 + 3 \times 2.5 + 2.5 \times 5)\)
T.S.A. = \(2(15 + 7.5 + 12.5) = 2(35) = 70\text{ m}^2\)
In simple words: The volume is 37.5 cubic meters and the total surface area is 70 square meters.

Exam Tip: Remember to write the correct units for volume (cubic units) and surface area (square units).

Q2. The T.S.A. of a cube is 384 sq. cm. Find its volume.
Answer: Total Surface Area of a cube = \(6a^2\), where \(a\) is the side length
\(6a^2 = 384\)
\(a^2 = \frac{384}{6} = 64\)
\(a = \sqrt{64} = 8\text{ cm}\)
Volume of cube = \(a^3\)
Volume = \(8^3 = 8 \times 8 \times 8 = 512\text{ cm}^3\)
In simple words: First find the side length from the surface area, then cube it to get the volume.

Exam Tip: Always calculate the side length (\(a\)) first before finding the volume of a cube.

Q3. A brick measures \(24\text{ cm} \times 12\text{ cm} \times 8\text{ cm}\). How many bricks will be required to construct a wall measuring 24 m long, 8 m high and 60 cm thick?
Answer: Dimensions of the wall:
Length (\(L\)) = \(24\text{ m} = 2400\text{ cm}\)
Height (\(H\)) = \(8\text{ m} = 800\text{ cm}\)
Thickness (\(W\)) = \(60\text{ cm}\)
Volume of the wall = \(L \times W \times H = 2400 \times 60 \times 800\text{ cm}^3\)
Dimensions of one brick:
Length (\(l\)) = 24 cm, breadth (\(b\)) = 12 cm, height (\(h\)) = 8 cm
Volume of one brick = \(l \times b \times h = 24 \times 12 \times 8\text{ cm}^3\)
Number of bricks = \(\frac{\text{Volume of the wall}}{\text{Volume of one brick}}\)
Number of bricks = \(\frac{2400 \times 60 \times 800}{24 \times 12 \times 8}\)
Number of bricks = \(\frac{2400}{24} \times \frac{60}{12} \times \frac{800}{8} = 100 \times 5 \times 100 = 50,000\)
In simple words: Divide the total volume of the wall by the volume of a single brick after converting all units to centimeters.

Exam Tip: Always convert all dimensions to the same unit (centimeters here) before calculating volumes.

Q4. Two cubes each of side 12 cm are joined end to end. Find the S.A. of the resulting cuboid.
Answer: When two cubes of side 12 cm are joined end to end, the dimensions of the resulting cuboid are:
Length (\(l\)) = \(12 + 12 = 24\text{ cm}\)
Breadth (\(b\)) = 12 cm
Height (\(h\)) = 12 cm
Surface Area of the cuboid = \(2(lb + bh + hl)\)
Surface Area = \(2(24 \times 12 + 12 \times 12 + 12 \times 24)\)
Surface Area = \(2(288 + 144 + 288)\)
Surface Area = \(2(720) = 1440\text{ cm}^2\)
In simple words: Joining two cubes doubles the length while breadth and height remain the same.

Exam Tip: Be careful to only change the dimension along the axis where the cubes are joined.

Q5. A cuboid measures \(80\text{ cm} \times 75\text{ cm} \times 36\text{ cm}\). It is melted and a cube is formed. Find the edge of the cube.
Answer: Volume of the cuboid = \(80 \times 75 \times 36\text{ cm}^3\)
Volume = \(216,000\text{ cm}^3\)
Since the cuboid is melted to form a cube, their volumes are equal.
Let the edge of the cube be \(a\).
Volume of the cube = \(a^3 = 216,000\text{ cm}^3\)
\(a = \sqrt[3]{216,000} = 60\text{ cm}\)
In simple words: The volume remains unchanged when melting a shape to form another. Find the cube root of the total volume to get the edge.

Exam Tip: Equate the volumes when an object is melted and recast into a new shape.

Q6. Find the diagonal of a cuboid 25 cm long, 20 cm wide and 15 cm high.
Answer: Length (\(l\)) = 25 cm
Breadth (\(b\)) = 20 cm
Height (\(h\)) = 15 cm
Diagonal of cuboid = \(\sqrt{l^2 + b^2 + h^2}\)
Diagonal = \(\sqrt{25^2 + 20^2 + 15^2}\)
Diagonal = \(\sqrt{625 + 400 + 225}\)
Diagonal = \(\sqrt{1250}\)
Diagonal = \(25\sqrt{2} \approx 25 \times 1.414 = 35.35\text{ cm}\)
In simple words: Use the 3D diagonal formula by taking the square root of the sum of squares of length, breadth, and height.

Q7. The diagonal of a cube is \(6\sqrt{3}\text{ cm}\). Find its volume and S.A.
Answer: Formula for the diagonal of a cube = \(a\sqrt{3}\), where \(a\) is the side length.
\(a\sqrt{3} = 6\sqrt{3}\)
\(a = 6\text{ cm}\)
Volume of the cube = \(a^3 = 6^3 = 216\text{ cm}^3\)
Surface Area of the cube = \(6a^2 = 6 \times 6^2 = 6 \times 36 = 216\text{ cm}^2\)
In simple words: Equate the cube diagonal formula to the given value to find the side length, then compute volume and surface area.

Exam Tip: The diagonal of a cube is always \(\sqrt{3}\) times its side.

Q8. A granary is in the shape of a cuboid of size \(8\text{ m} \times 6\text{ m} \times 3\text{ m}\). If a bag of grain occupies a space of \(0.65\text{ m}^3\), how many bags can be stored in the granary?
Answer: Volume of the granary = \(8 \times 6 \times 3 = 144\text{ m}^3\)
Space occupied by 1 bag = \(0.65\text{ m}^3\)
Number of bags = \(\frac{\text{Volume of granary}}{\text{Volume occupied by 1 bag}}\)
Number of bags = \(\frac{144}{0.65} = \frac{14400}{65} = 221.53\)
Since the number of complete bags must be an integer, 221 bags can be stored.
In simple words: Divide the total volume of the granary by the volume of one grain bag.

Exam Tip: For practical items like bags or boxes, round down to the nearest whole number if the result is a decimal.

Q9. A reservoir is in the form of a rectangular parallelepiped (cuboid). Its length is 20 m. If 18 kl of water is removed from the reservoir, the water level goes down by 15 cm. Find the width of the reservoir (\(1\text{ kl} = 1\text{ m}^3\)).
Answer: Volume of water removed = \(18\text{ kl} = 18\text{ m}^3\)
Let the width of the reservoir be \(b\) meters.
Length (\(l\)) = 20 m
Drop in height (\(h\)) = \(15\text{ cm} = 0.15\text{ m}\)
Volume of water removed = Length \(\times\) Width \(\times\) Drop in height
\(18 = 20 \times b \times 0.15\)
\(18 = 3 \times b\)
\(b = \frac{18}{3} = 6\text{ m}\)
In simple words: The volume of the removed water equals the volume of the cuboid formed by the length, width, and the drop in water level.

Exam Tip: Ensure all measurements are converted to meters before applying the volume formula.

Q10. A closed iron box 15 m long, 10 m wide and 6 m deep is to be made. Find the cost of iron sheet used at the rate of Rs. 50 per metre if the sheet is 2 m wide.
Answer: Dimensions of the iron box:
Length (\(l\)) = 15 m
Breadth (\(b\)) = 10 m
Depth/Height (\(h\)) = 6 m
Total surface area of the closed box = \(2(lb + bh + hl)\)
Surface Area = \(2(15 \times 10 + 10 \times 6 + 6 \times 15)\)
Surface Area = \(2(150 + 60 + 90)\)
Surface Area = \(2(300) = 600\text{ m}^2\)
Width of the iron sheet = 2 m
Length of the iron sheet required = \(\frac{\text{Total Surface Area}}{\text{Width of sheet}}\)
Length of sheet = \(\frac{600}{2} = 300\text{ m}\)
Cost of iron sheet = Length \(\times\) Rate per meter
Cost = \(300 \times \text{Rs. } 50 = \text{Rs. } 15,000/
In simple words: Find the surface area of the box, divide by the sheet width to get the total length of sheet needed, and multiply by the cost per meter.

Exam Tip: The area of the rectangular sheet equals length \(\times\) width, so length equals area divided by width.