Mensuration Questions and Answers

Answer: B

Correct answer: 154 sq. cm

Find the radius from the circumference first.

\[2\pi r = 44 \quad\Longrightarrow\quad r = \frac{44 \times 7}{2 \times 22} = 7 \text{ cm}\]

\[\text{Area} = \pi r^2 = \frac{22}{7} \times 49 = 154 \text{ sq. cm}\]

Answer: C

Drop a horizontal from the shorter fort's top to the taller one. That gives a right triangle whose legs are the ground gap and the height difference.

\[ \text{height difference} = 15 - 8 = 7 \text{ cm}, \qquad \text{ground gap} = 24 \text{ cm} \]

\[ d = \sqrt{24^{2} + 7^{2}} = \sqrt{576 + 49} = \sqrt{625} = 25 \text{ cm} \]

7-24-25 is a Pythagorean triple, which is why the answer comes out exact. Answering 24 cm ignores the height difference; 31 cm adds the numbers instead of squaring them.

Answer: C

A cuboid has three pairs of identical faces.

\[ \text{TSA} = 2(lb + bh + hl) \]

\[ = 2(16 \times 14 + 14 \times 7 + 7 \times 16) \]

\[ = 2(224 + 98 + 112) = 2 \times 434 = 868 \text{ m}^{2} \]

The volume, a different quantity, would be \( 16 \times 14 \times 7 = 1568 \) m³.

Answer: B

The longest pole runs along the space diagonal of the room.

\[ d = \sqrt{l^{2} + b^{2} + h^{2}} \]

\[ = \sqrt{12^{2} + 8^{2} + 9^{2}} = \sqrt{144 + 64 + 81} = \sqrt{289} \]

\[ = 17 \text{ m} \]

The floor diagonal alone would be \( \sqrt{144 + 64} = 14.4 \) m; the height adds the third term.

Answer: D

Let the breadth be \( b \). Then the height is \( 5b \) and the length \( 8 \times 5b = 40b \).

\[ V = b \times 5b \times 40b = 200b^{3} = 12.8 \]

\[ b^{3} = 0.064 \quad\Longrightarrow\quad b = 0.4 \text{ m} = 40 \text{ cm} \]

"Eight times as long as it is high" refers to the height \( 5b \), not the breadth - reading it as \( 8b \) would give a different wall.

Answer: C

Work in centimetres throughout.

\[ V_{\text{wall}} = 2400 \times 800 \times 60 = 115{,}200{,}000 \text{ cm}^{3} \]

Mortar fills 10%, so bricks occupy 90%:

\[ 0.9 \times 115{,}200{,}000 = 103{,}680{,}000 \text{ cm}^{3} \]

\[ V_{\text{brick}} = 24 \times 12 \times 8 = 2304 \text{ cm}^{3} \]

\[ n = \frac{103{,}680{,}000}{2304} = 45{,}000 \]

Ignoring the mortar would give 50,000 - a tenth too many.

Answer: B

Convert the volume to cubic centimetres to match the base area.

\[ 2.6 \text{ m}^{3} = 2.6 \times 1{,}000{,}000 = 2{,}600{,}000 \text{ cm}^{3} \]

\[ \text{depth} = \frac{2{,}600{,}000}{6500} = 400 \text{ cm} = 4 \text{ m} \]

A cubic metre is a million cubic centimetres, not ten thousand - the factor that decides the answer.

Answer: B

Turn the weight into a volume, then divide out the two known dimensions.

\[ 112 \text{ kg} = 112{,}000 \text{ g} \quad\Longrightarrow\quad V = \frac{112000}{25} = 4480 \text{ cm}^{3} \]

\[ \text{length} = \frac{4480}{28 \times 5} = \frac{4480}{140} = 32 \text{ cm} \]

The density is per gram, so the kilograms must be converted first.

Answer: B

Hammering preserves the volume, which spreads over the given area.

\[ V = 0.5 \text{ m}^{3} = 500{,}000 \text{ cm}^{3} \]

\[ A = 1 \text{ hectare} = 10{,}000 \text{ m}^{2} = 100{,}000{,}000 \text{ cm}^{2} \]

\[ \text{thickness} = \frac{500{,}000}{100{,}000{,}000} = 0.005 \text{ cm} \]

Both conversions matter: a cubic metre is \( 10^{6} \) cm³ and a square metre \( 10^{4} \) cm².

Answer: B

The rain forms a thin slab over the ground.

\[ 5 \text{ cm} = 0.05 \text{ m}, \qquad 1.5 \text{ hectares} = 15{,}000 \text{ m}^{2} \]

\[ V = 15{,}000 \times 0.05 = 750 \text{ m}^{3} \]

A hectare is 10,000 m², and the depth must be in metres to match.