Area Questions and Answers

Answer: C

Express the length in terms of the breadth and substitute.

\[ l = 1.15b \quad\Longrightarrow\quad 1.15b \times b = 460 \]

\[ b^{2} = \frac{460}{1.15} = 400 \quad\Longrightarrow\quad b = 20 \text{ m} \]

The length is then \( 1.15 \times 20 = 23 \) m - offered as a distractor for anyone who solves for the wrong side. Checking: \( 23 \times 20 = 460 \) m².

Answer: B

\[ A = \pi r^{2} = \frac{22}{7} \times 7^{2} = \frac{22}{7} \times 49 \]

\[ = 22 \times 7 = 154 \text{ m}^{2} \]

A radius of 7 is chosen so that the 7 in \( \tfrac{22}{7} \) cancels cleanly. The circumference, a different quantity, would be \( 2\pi r = 44 \) m.

Answer: B

All four sides of a rhombus are equal, so start from the perimeter.

\[ \text{side} = \frac{68}{4} = 17 \text{ cm} \]

\[ \text{half of the given diagonal} = 8 \text{ cm} \]

\[ \text{half of the other} = \sqrt{17^{2} - 8^{2}} = \sqrt{289 - 64} = \sqrt{225} = 15 \text{ cm} \]

\[ \text{other diagonal} = 30 \text{ cm} \]

\[ A = \frac{1}{2} \times 16 \times 30 = 240 \text{ cm}^{2} \]

8–15–17 is another Pythagorean triple, so the second diagonal comes out whole.

Answer: A

Area varies as the square of the side, so take square roots to recover the length ratio.

\[ \frac{a_{1}^{2}}{a_{2}^{2}} = \frac{1}{4} \quad\Longrightarrow\quad \frac{a_{1}}{a_{2}} = \frac{1}{2} \]

Perimeter is \( 4a \), a plain multiple of the side, so it keeps that same ratio:

\[ \frac{P_{1}}{P_{2}} = \frac{4a_{1}}{4a_{2}} = \frac{1}{2} \quad\Longrightarrow\quad 1 : 2 \]

Carrying the area ratio straight through to the perimeters gives 1 : 4, the first distractor.

Answer: C

A sector's area is the same fraction of the circle as its angle is of a full turn.

\[ A = \frac{\theta}{360} \times \pi r^{2} \]

\[ = \frac{42}{360} \times \frac{22}{7} \times 144 \]

\[ = \frac{7}{60} \times \frac{22}{7} \times 144 = \frac{22 \times 144}{60} \]

\[ = \frac{3168}{60} = 52.8 \text{ m}^{2} \]

The whole circle covers \( \tfrac{22}{7} \times 144 = 452.6 \) m², and \( 42^{\circ} \) is just under a twelfth of it - a quick check that the answer is the right size.

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 \( 2x + \tfrac{x^{2}}{100} = 40 + 4 \).

Answer: C

Three sides are painted and one is not. The unpainted side is 9 ft, so the side opposite it is also 9 ft and is painted, along with the two equal widths.

\[ 9 + 2w = 37 \quad\Longrightarrow\quad 2w = 28 \quad\Longrightarrow\quad w = 14 \text{ ft} \]

\[ A = 9 \times 14 = 126 \text{ ft}^{2} \]

The 37 ft covers all three painted sides, not two - reading it as the two widths alone would give \( w = 18.5 \) and an area of 166.5 ft².

Answer: A

Convert to a single unit before dividing.

\[ \text{courtyard} = 2500 \times 1600 = 4{,}000{,}000 \text{ cm}^{2} \]

\[ \text{brick} = 20 \times 10 = 200 \text{ cm}^{2} \]

\[ n = \frac{4{,}000{,}000}{200} = 20{,}000 \]

A square metre holds 10,000 cm², so the 400 m² courtyard is four million square centimetres - the factor most often lost is that square metre conversion.

Answer: A

The track's length is

\[ C = 2\pi r = 2 \times \frac{22}{7} \times 70 = 440 \text{ m} \]

Runners moving in opposite directions meet once every time their combined distance equals one lap:

\[ v_{\text{John}} + v_{\text{Jan}} = \frac{440}{66} = \frac{20}{3} \text{ m/s} \]

\[ v_{\text{Joni}} + v_{\text{Jan}} = \frac{440}{110} = 4 \text{ m/s} \]

Subtracting eliminates Janardan's speed:

\[ v_{\text{John}} - v_{\text{Joni}} = \frac{20}{3} - 4 = \frac{8}{3} \text{ m/s} \]

John and Joni run the same way, so one laps the other when this difference covers a full circuit:

\[ t = \frac{440}{8/3} = 165 \text{ seconds} \]

Same direction means subtract the speeds; opposite directions means add them. Getting that the wrong way round here would give 66 seconds.