Area Questions and Answers

Answer: A

The length is 18 cm.

Let the length be \(l\) and the breadth \(b\). The perimeter is five times the breadth:

\[ \begin{aligned} 2(l + b) &= 5b \Rightarrow l = 1.5b \\ l \times b &= 1.5b^{2} = 216 \\ b^{2} &= 144,\ b = 12 \end{aligned} \]

So \(l = 18\) cm. Check: the perimeter is \(2(18 + 12) = 60\) cm, and 60 : 12 = 5 : 1. 24 cm would need a breadth of 9 cm, which gives a perimeter ratio of 66 : 9, not 5 : 1.

Answer: B

The diameter is 28m.

Use \(\text{area} = \pi r^{2}\) with \(\pi = \frac{22}{7}\):

\[ \begin{aligned} \frac{22}{7} r^{2} &= 616 \\ r^{2} &= 196,\ r = 14\ \text{m} \\ d &= 2r = 28\ \text{m} \end{aligned} \]

14 m is the radius, which is the usual slip; 16, 18 and 20 m do not give an area of 616 m².

Taking \(\pi = \frac{22}{7}\) is the cue: 616 is \(22 \times 28\), so \(r^{2} = 28 \times 7 = 196\), a perfect square.

Answer: C

The other diagonal is 12 cm.

The area of a rhombus is half the product of its diagonals.

\[ \begin{aligned} \tfrac{1}{2} \times 16 \times d &= 96 \\ 8d &= 96 \\ d &= 12\ \text{cm} \end{aligned} \]

Forgetting the half gives 6 cm, which is not offered; 14, 18 and 20 cm would give areas of 112, 144 and 160 square centimetres.

Answer: C

The area is 616 square centimetres.

First find the radius from the circumference \(2\pi r\), taking \(\pi = \frac{22}{7}\).

\[ \begin{aligned} 2 \times \tfrac{22}{7} \times r &= 88 \\ r &= 14\ \text{cm} \\ \text{area} &= \tfrac{22}{7} \times 14^{2} \\ &= 616\ \text{cm}^{2} \end{aligned} \]

352 is \(\frac{22}{7} \times 112\), a slip in squaring 14, and 600 and 432 do not come from any correct radius.

Answer: B

Work in a single unit - centimetres - before dividing.

\[ \text{floor} = 800 \times 400 = 320{,}000 \text{ cm}^{2} \]

\[ \text{tile} = 25 \times 20 = 500 \text{ cm}^{2} \]

\[ n = \frac{320{,}000}{500} = 640 \]

Dividing the floor in square metres (32) by the tile in square centimetres (500) is the classic slip and produces the nonsensical 6.4 offered as the last option.

Answer: D

Only the four walls are painted, and the door and window are not.

\[ \text{four walls} = 2(l + b)h = 2(6 + 5) \times 4 = 88 \text{ m}^{2} \]

\[ \text{door} = 2.5 \times 1.2 = 3 \text{ m}^{2}, \qquad \text{window} = 1 \times 1 = 1 \text{ m}^{2} \]

\[ \text{paintable area} = 88 - 3 - 1 = 84 \text{ m}^{2} \]

\[ \text{cost} = 84 \times 2 = \text{Rs. } 168 \]

Rs. 84 is the area mistaken for the cost, and Rs. 176 would come from forgetting to deduct the openings.

Answer: C

Let the longer diagonal be \( d \); the shorter is then \( 0.6d \).

A rhombus's diagonals bisect each other at right angles, so its area is half their product:

\[ A_{\text{rhombus}} = \frac{1}{2} \times d \times 0.6d = 0.3d^{2} \]

The square is built on the longer diagonal as its side:

\[ A_{\text{square}} = d^{2} \]

\[ \frac{A_{\text{square}}}{A_{\text{rhombus}}} = \frac{d^{2}}{0.3d^{2}} = \frac{10}{3} \quad\Longrightarrow\quad 10 : 3 \]

The unknown \( d \) cancels, so the ratio holds for any rhombus with that diagonal proportion.

Answer: D

Halve each diameter to get the radii: 1 m, 1 m and 0.5 m.

\[ A_{\text{plank}} = 12 \times 6 = 72 \text{ m}^{2} \]

\[ A_{\text{holes}} = \pi(1)^{2} + \pi(1)^{2} + \pi(0.5)^{2} = \pi + \pi + \frac{\pi}{4} = \frac{9\pi}{4} \]

\[ A_{\text{remaining}} = 72 - \frac{9\pi}{4} = \frac{288 - 9\pi}{4} \text{ m}^{2} \]

Numerically that is about 64.93 m². Using the diameters directly in \( \pi r^{2} \) would quadruple the holes' area.

Answer: D

The road runs outside the farm, so it forms a border of width 5 m around a rectangle \( l \times b \).

\[ \text{road area} = (l + 10)(b + 10) - lb = 10l + 10b + 100 \]

\[ 600 = 10(l + b) + 100 \quad\Longrightarrow\quad l + b = 50 \]

This fixes only the sum of the two dimensions. The pairs \( 25 \times 25 \), \( 20 \times 30 \) and \( 10 \times 40 \) all satisfy it and give different ratios, so the ratio cannot be determined.

A second independent fact - the farm's area or perimeter - would be needed to pin the shape down.

Answer: C

The area of an equilateral triangle is \( \dfrac{\sqrt{3}}{4}a^{2} \), so it varies as the square of the side.

\[ \frac{A'}{A} = \left(\frac{1.2a}{a}\right)^{2} = 1.44 \]

\[ \text{increase} = 1.44 - 1 = 0.44 = 44\% \]

The constant \( \tfrac{\sqrt{3}}{4} \) cancels, so this 44% holds for any shape scaled by 20% - square, circle or triangle alike. The exact rule is \( 2x + \tfrac{x^{2}}{100} = 40 + 4 \).