Area Questions and Answers

Answer: D

The compound wall covers one side, so the wire spans the opposite side and both ends. Let the fenced-opposite side be \( L \) and each end \( W \).

\[ L + 2W = 30, \qquad L \times W = 100 \]

Substituting \( L = 30 - 2W \):

\[ (30 - 2W)W = 100 \quad\Longrightarrow\quad 2W^{2} - 30W + 100 = 0 \]

\[ W^{2} - 15W + 50 = 0 \quad\Longrightarrow\quad (W - 5)(W - 10) = 0 \]

\( W = 5 \) gives \( L = 20 \); \( W = 10 \) gives \( L = 10 \). Only 20 m × 5 m appears among the options.

Checking: \( 20 + 5 + 5 = 30 \) m of wire and \( 20 \times 5 = 100 \) m².

Answer: B

Walking entirely around a semicircular garden means covering the curved arc and the straight diameter.

\[ \text{perimeter} = \pi r + 2r = r(\pi + 2) \]

With \( \pi = \tfrac{22}{7} \), so \( \pi + 2 = \tfrac{36}{7} \):

\[ r \times \frac{36}{7} = 324 \quad\Longrightarrow\quad r = \frac{324 \times 7}{36} = 63 \text{ m} \]

\[ A = \frac{1}{2}\pi r^{2} = \frac{1}{2} \times \frac{22}{7} \times 3969 = 6237 \text{ m}^{2} \]

Omitting the straight edge would make the perimeter \( \pi r \) alone and give a radius of 103 m - the diameter is a real part of the boundary he walks.

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.