Mensuration Questions and Answers

Answer: C

\[ V = a^{3} = 1^{3} = 1 \text{ cm}^{3} \]

A cube of unit edge has unit volume; its total surface area, a different quantity, would be \( 6a^{2} = 6 \) cm².

Answer: C

The diagonals of a rhombus bisect each other at right angles, cutting it into four congruent right triangles. That gives

\[ A = \frac{1}{2} d_{1} d_{2} \]

Each of the four triangles has area \( \tfrac{1}{2} \cdot \tfrac{d_{1}}{2} \cdot \tfrac{d_{2}}{2} = \tfrac{d_{1}d_{2}}{8} \), and four of them make \( \tfrac{d_{1}d_{2}}{2} \).

The same half-product formula works for any quadrilateral whose diagonals are perpendicular, including the square and the kite.

Answer: D

Convert to degrees first.

\[ \frac{2\pi}{3} = \frac{2 \times 180^{\circ}}{3} = 120^{\circ} \]

\[ \sin 120^{\circ} = \sin(180^{\circ} - 120^{\circ}) = \sin 60^{\circ} = \frac{\sqrt{3}}{2} = 0.866 \]

Sine never exceeds 1, so any option above it can be discarded immediately.

Answer: B

Convert the radian measure to degrees, using \( \pi \text{ radians} = 180^{\circ} \).

\[ 2.3\pi = 2.3 \times 180^{\circ} = 414^{\circ} \]

Subtracting a full turn:

\[ 414^{\circ} - 360^{\circ} = 54^{\circ} \]

\[ \cos 54^{\circ} = 0.587 \]

Cosine never exceeds 1, so the options above 1 can be discarded before any calculation.

Answer: D

The immersed cube displaces its own volume of water, which spreads over the vessel's base.

\[ V_{\text{cube}} = 15^{3} = 3375 \text{ cm}^{3} \]

\[ \text{base area} = 20 \times 15 = 300 \text{ cm}^{2} \]

\[ \text{rise} = \frac{3375}{300} = 11.25 \text{ cm} \]

The cube's own 15 cm edge is not the rise: the water spreads over a base twice the cube's footprint, so it climbs less far.

Answer: A

An increase of \( 9.09\% \) is a factor of \( \tfrac{1}{11} \), since \( \tfrac{1}{11} = 0.0909 \). The radius therefore becomes \( \tfrac{12}{11}r \).

Surface area varies as the square of the radius:

\[ \frac{S'}{S} = \left(\frac{12}{11}\right)^{2} = \frac{144}{121} = 1.1901 \]

\[ \text{increase} = 19.01\% \approx 19\% \]

Recognising 9.09% as \( \tfrac{1}{11} \) keeps the arithmetic exact; the decimal route gives 18.998% and invites a rounding error.

Answer: A

Correct answer: 240 sq. cm

\[\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}\]

\[= \frac{1}{2} \times (15 + 25) \times 12 = \frac{1}{2} \times 40 \times 12 = 240 \text{ sq. cm}\]

The height must be the perpendicular distance between the parallel sides, not the length of a slanting side.

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: 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.

Answer: D

In \( C = 2\pi r \), the letter \( r \) is the radius - the distance from the centre to any point on the circle.

The expression \( 2\pi r \) is itself the circumference, so \( r \) cannot be that; area and perimeter are likewise the results of formulas, not the input to this one.