Mensuration Questions and Answers

Answer: B

The value of \(3y - 15\) is 81.

The area of a square can be found from its diagonal \(d\) alone, since \(d^{2} = 2a^{2}\):

\[ \begin{aligned} y &= \frac{d^{2}}{2} = \frac{64}{2} = 32\ \text{cm}^{2} \\ 3y - 15 &= 96 - 15 = 81 \end{aligned} \]

96 is \(3y\) before subtracting 15. Treating the diagonal as the side would give an area of 64 and a far larger value, which matches none of the options.

Answer: B

The distance between the parallel sides is 8 cm.

The perimeter is the sum of all four sides, so the two parallel sides add to \(52 - 20 = 32\) cm. Then use the area formula for a trapezium.

\[ \begin{aligned} \text{Area} &= \tfrac{1}{2} \times (\text{sum of parallel sides}) \times h \\ 128 &= \tfrac{1}{2} \times 32 \times h \\ h &= \frac{128}{16} = 8 \text{ cm} \end{aligned} \]

Using the whole perimeter, 52 cm, in place of the parallel sides gives a height of about 4.9 cm, and forgetting the half gives 4 cm; neither is offered, which confirms 8 cm.

Answer: C

The person takes 30 seconds.

The other side is \(48 \div 6 = 8\) m, and the diagonal follows from Pythagoras' theorem.

\[ \begin{aligned} \text{diagonal} &= \sqrt{6^{2} + 8^{2}} = \sqrt{100} = 10 \text{ m} \\ \text{time} &= \frac{10}{20} \text{ min} = 30 \text{ s} \end{aligned} \]

Walking along two sides (14 m) would take 42 seconds, and using one side of 8 m would give 24 seconds, so the other options do not match the diagonal route.

Answer: D

The cost is ₹ 24,360.

The pool is open at the top, so plaster the floor and the four walls.

\[ \begin{aligned} \text{Floor} &= 20 \times 15 = 300\ \text{m}^2 \\ \text{Walls} &= 2(20 + 15) \times 4 = 280\ \text{m}^2 \\ \text{Total} &= 580\ \text{m}^2 \end{aligned} \]

\[ 580 \times 42 = 24360 \]

The other options correspond to areas of 590, 600 and 578 m², none of which is the floor plus the four walls. Plastering the top as well as the floor would give 880 m², far more than any option.

Answer: A

The radius of C₃ is 52 m.

A perimeter of \(2\pi r\) gives the radii: \(2\pi r_1 = 40\pi\), so \(r_1 = 20\) m; \(2\pi r_2 = 96\pi\), so \(r_2 = 48\) m.

\[ \begin{aligned} \pi r_3^2 &= \pi(20^2 + 48^2) \\ r_3^2 &= 400 + 2304 = 2704 \\ r_3 &= 52\ \text{m} \end{aligned} \]

Adding the radii (68 m) instead of the areas is the trap. 20, 48, 52 is 4 times the triple 5, 12, 13, which is a quick check.

Answer: C

The squares can be arranged in 5 ways.

Any arrangement of all 48 unit squares covers 48 cm². Reading the question as asking for rectangles, we need pairs of whole-number sides whose product is 48:

\[ \begin{aligned} 48 &= 1 \times 48 = 2 \times 24 = 3 \times 16 \\ &= 4 \times 12 = 6 \times 8 \end{aligned} \]

That gives five different rectangles; turning one round (8 by 6 instead of 6 by 8) gives the same shape. 48 has exactly these five factor pairs, so 4 and 2 leave some out, and 6 counts one more than exist.

Answer: D

The correct statement is that the shape of a figure determines the perimeter.

Perimeter depends on the shape and size of a figure, so different shapes made from the same pieces can have different perimeters. Twelve unit squares arranged as a 3 by 4 rectangle have perimeter 14 units, but in a 2 by 6 rectangle the perimeter is 16 units, though the area is 12 square units in both.

That same example disproves the first statement: equal areas do not mean equal perimeters. Conversely a 1 by 5 and a 3 by 3 rectangle both have perimeter 12 but areas 5 and 9, so equal perimeters do not mean equal areas. And the units differ: perimeter is a length (cm), area a square measure (cm²).

Answer: C

The rectangle encloses 128 cm².

The wire's length stays the same; it is the perimeter of both shapes.

\[ \begin{aligned} \text{side} &= \sqrt{144} = 12\ \text{cm} \\ \text{wire} &= 4 \times 12 = 48\ \text{cm} \\ 2(16 + b) &= 48,\ b = 8\ \text{cm} \\ \text{area} &= 16 \times 8 = 128\ \text{cm}^{2} \end{aligned} \]

The option 48 cm² is the wire's length in centimetres, not an area, and 124 and 96 do not come from a 16 cm by 8 cm rectangle. As expected, the rectangle encloses less than the square of the same perimeter.

Answer: B

The length of SR is 32 cm.

Let \(PQ = 3x\) and \(SR = 2x\). The area of a trapezium is half the sum of the parallel sides times the distance between them.

\[ \begin{aligned} \tfrac{1}{2}(3x + 2x) \times 12 &= 480 \\ 30x &= 480 \\ x &= 16 \end{aligned} \]

So \(SR = 2x = 32\) cm and \(PQ = 48\) cm. Check: \(\frac{1}{2}(48 + 32) \times 12 = 480\).

48 cm is PQ, the longer side. With SR = 24 cm the sides would be 36 and 24 and the area only 360 cm²; with SR = 36 cm the area would be 540 cm².

Answer: C

The biggest area comes from 25 m × 25 m.

Every option uses the full 100 m of wire, since each perimeter is \(2(l + b) = 100\). The areas are:

\[ \begin{aligned} 15 \times 35 &= 525\ \text{m}^{2} \\ 30 \times 20 &= 600\ \text{m}^{2} \\ 25 \times 25 &= 625\ \text{m}^{2} \\ 5 \times 45 &= 225\ \text{m}^{2} \end{aligned} \]

For a fixed perimeter, the rectangle of largest area is the square; the more unequal the sides, the smaller the area, as the 5 m by 45 m strip shows.