Mensuration Questions and Answers

Answer: D

Statement I is false but Statement II is true.

Area grows with the square of the side. Doubling a side of length a gives

\[ (2a)^2 = 4a^2 \]

so the area becomes four times as large, not twice. Statement I is false.

The perimeter is \(4a\), a straight-line (linear) function of a: doubling the side doubles the perimeter. Statement II is true. The options calling Statement I true confuse the perimeter's behaviour with the area's.

Answer: A

About \(6.67 \times 10^{6}\) litres fall into the sea each minute.

The water flowing past in one minute is a block as wide and deep as the river and as long as the distance the water moves in a minute. 2 km/hr is 2000 m in 60 minutes:

\[ \begin{aligned} \text{length} &= 2000/60 = 33.33\ \text{m} \\ V &= 4 \times 50 \times 33.33 \\ &= 6666.7\ \text{m}^{3} \end{aligned} \]

Since 1 m³ = 1000 L, that is 6,666,667 L, or \(6.67 \times 10^{6}\) L. The other options do not come from these dimensions; forgetting the conversion to litres would give 6666.7, not any listed figure.

Answer: A

The area of the circle is 154 square centimetre.

The rectangle's perimeter is \(2(14 + 8) = 44\) cm, which equals the circle's circumference:

\[ \begin{aligned} 2 \pi r &= 44 \\ r &= \frac{44 \times 7}{2 \times 22} = 7 \\ \pi r^{2} &= \frac{22}{7} \times 49 = 154 \end{aligned} \]

308 is twice the area, 77 is half of it, and 112 is the area of the rectangle (14 × 8), not of the circle.

Answer: C

The longest side is 23 units.

A perimeter of 70 means length + breadth = 35, and the area gives length × breadth = 276. So the sides are the roots of

\[ t^{2} - 35t + 276 = 0 \Rightarrow (t - 23)(t - 12) = 0 \]

The sides are 23 and 12, and 23 × 12 = 276 checks. 12 is the shorter side. 18 would need a breadth of 17, giving an area of 306, and 36 cannot be a side at all because the two sides add to only 35.

Answer: A

The total area grazed is \(\pi P^{2}\).

Both cows are tied to the same pole, so each grazes a circle centred on that pole: one of radius P and one of radius Q. Since P is greater than Q, the smaller circle lies entirely inside the larger one, and the total area grazed is just the area of the larger circle, \(\pi P^{2}\).

\(\pi P^{2} + \pi Q^{2}\) counts the inner circle twice. \(\pi P^{2} - \pi Q^{2}\) is the ring grazed only by the cow with the longer rope, and \(\pi Q^{2}\) is the area the smaller cow alone can reach.

Answer: C

About 30% is saved.

For a square of side \(a\), walking along two edges covers \(2a\) and walking the diagonal covers \(\sqrt{2}a \approx 1.414a\):

\[ \begin{aligned} \text{saving} &= \frac{2a - 1.414a}{2a} \\ &= \frac{0.586}{2} \approx 29.3\% \end{aligned} \]

which rounds to 30%. The other options are too small: 20%, 15% or 12% would need the diagonal to be 1.6, 1.7 or 1.76 times the side, but it is only 1.414 times.

Answer: A

12 cubic metres is about 424 cubic feet.

One metre is 3.2808 feet, so one cubic metre is

\[ \begin{aligned} 3.2808^{3} &\approx 35.31\ \text{ft}^{3} \\ 12 \times 35.31 &\approx 423.8\ \text{ft}^{3} \end{aligned} \]

which rounds to 424. 464 and higher would need about 38.7 cubic feet per cubic metre or more, overstating the conversion; multiplying by 3.28 only once, as for a length, would give just 39.4.

Answer: D

The total area is 924 cm².

Each blade sweeps a sector of radius 21 cm through 120°, one third of a full circle:

\[ \begin{aligned} \text{one sector} &= \frac{120}{360} \times \frac{22}{7} \times 21^{2} \\ &= \frac{1}{3} \times 1386 = 462\ \text{cm}^{2} \end{aligned} \]

The two wipers do not overlap, so the total is \(2 \times 462 = 924\) cm². The trap is to stop at one wiper (462) or to forget that a 120° sweep is only a third of the circle; 824, 1024 and 536 do not arise from the correct sector formula.

Answer: A

The expenditure is Rs. 1650.

From the circumference, \(2\pi R = 88\), so \(R = 14\) m. The strip runs inside the boundary, so the inner radius is \(14 - 3 = 11\) m, and the strip is the ring between them:

\[ \begin{aligned} \text{area} &= \tfrac{22}{7}(14^{2} - 11^{2}) \\ &= \tfrac{22}{7} \times 75 = \tfrac{1650}{7}\ \text{m}^{2} \end{aligned} \]

At Rs. 7 per square metre the cost is \(\tfrac{1650}{7} \times 7 = 1650\). Taking the strip outside the circle (radius 17 m) gives a larger area and a cost of 2046, close to the 2050 offered, which is the trap.

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.