Area Questions and Answers

Answer: A

The parallel sides are 24 cm and 40 cm.

Let the parallel sides be \(3k\) and \(5k\). The area of a trapezium is half the sum of the parallel sides times the height:

\[ \begin{aligned} \tfrac{1}{2}(3k + 5k) \times 12 &= 384 \\ 48k &= 384 \\ k &= 8 \end{aligned} \]

So the sides are \(3 \times 8 = 24\) cm and \(5 \times 8 = 40\) cm. The other pairs are not in the ratio 3:5 (12 and 18 is 2:3, 15 and 20 is 3:4), or do not give an area of 384.

Answer: C

The radius of the inscribed circle is 3 cm.

Since \(8^{2} + 15^{2} = 64 + 225 = 289 = 17^{2}\), the triangle is right-angled. Its area is \(\frac{1}{2} \times 8 \times 15 = 60\) and its semi-perimeter is \((8 + 15 + 17)/2 = 20\):

\[ r = \frac{\text{Area}}{s} = \frac{60}{20} = 3\ \text{cm} \]

For a right triangle there is also the shortcut \(r = (a + b - c)/2 = (8 + 15 - 17)/2 = 3\). 4 and 5 cm and \(2\sqrt{2}\) cm do not satisfy either formula.

Answer: B

The area of the quadrant is 38.5 cm².

From the circumference, find the radius, then take a quarter of the circle's area (with \(\pi = 22/7\)):

\[ \begin{aligned} 2\pi r &= 44 \Rightarrow r = 7 \\ \tfrac{1}{4}\pi r^{2} &= \tfrac{1}{4} \times 154 = 38.5 \end{aligned} \]

77 cm² is a semicircle and 19.25 cm² an eighth of the circle, so both use the wrong fraction.

Answer: B

The shaded area is \(102.73\ \text{cm}^{2}\).

An angle in a semicircle is a right angle, so with AC as diameter, the angle at B is 90°.

\[ \begin{aligned} AC &= \sqrt{12^2 + 5^2} = 13 \\ r &= 6.5 \\ \text{circle} &= \pi (6.5)^2 \approx 132.73 \\ \text{triangle} &= \tfrac{1}{2} \times 12 \times 5 = 30 \end{aligned} \]

The shaded region is the circle minus the triangle: \(132.73 - 30 = 102.73\ \text{cm}^{2}\), using \(\pi \approx 3.1416\).

All four options end in .73, so the question turns only on the whole number: the circle's 132.73 minus the triangle's 30.

Answer: D

Each revolution advances the wheel by one circumference.

\[ C = 2\pi r = 2 \times \frac{22}{7} \times 14 = 88 \text{ cm} \]

\[ n = \frac{1056}{88} = 12 \]

A radius that is a multiple of 7 makes the \( \tfrac{22}{7} \) cancel exactly, which is why the circumference lands on a whole 88 cm.

Answer: B

A square's diagonal and side are related by \( d = a\sqrt{2} \), so the area can be written directly in terms of the diagonal:

\[ A = a^{2} = \frac{d^{2}}{2} \]

The factor \( \tfrac{1}{2} \) is common to both squares and cancels:

\[ \frac{A_{1}}{A_{2}} = \frac{d_{1}^{2}}{d_{2}^{2}} = \frac{2^{2}}{5^{2}} = \frac{4}{25} \]

So the ratio is \( 4 : 25 \). Any two similar figures have areas in the square of their linear ratio, whichever length is used to compare them.

Answer: C

A rhombus's diagonals bisect each other at right angles, so half of each diagonal and a side form a right triangle.

\[ \text{half of the known diagonal} = \frac{24}{2} = 12 \text{ cm} \]

\[ \text{half of the other} = \sqrt{13^{2} - 12^{2}} = \sqrt{169 - 144} = 5 \text{ cm} \]

\[ \text{other diagonal} = 10 \text{ cm} \]

\[ \text{area} = \frac{1}{2} d_{1} d_{2} = \frac{1}{2} \times 24 \times 10 = 120 \text{ cm}^{2} \]

5–12–13 is a Pythagorean triple, which keeps the arithmetic exact. Forgetting the \( \tfrac{1}{2} \) gives 240, the second option.

Answer: D

The fencing runs round the perimeter, so divide the cost by the rate.

\[ \text{perimeter} = \frac{5300}{26.50} = 200 \text{ m} \]

\[ 2\big((b + 20) + b\big) = 200 \quad\Longrightarrow\quad 2b + 20 = 100 \]

\[ b = 40 \text{ m}, \qquad l = 60 \text{ m} \]

The length is 60 m, which appears nowhere among the choices, so the answer is none of these. The 40 m on offer is the breadth - a genuine intermediate value, but not what was asked.

Answer: D

Let the original square have side \( a \), so its diagonal is \( a\sqrt{2} \).

The second square is built on that diagonal as its side:

\[ A_{1} = a^{2}, \qquad A_{2} = \left(a\sqrt{2}\right)^{2} = 2a^{2} \]

\[ \frac{A_{1}}{A_{2}} = \frac{a^{2}}{2a^{2}} = \frac{1}{2} \quad\Longrightarrow\quad 1 : 2 \]

The square on the diagonal is always exactly double - the diagonal is longer by \( \sqrt{2} \), and areas scale by the square of that.

Answer: A

Multiply the two factors and compare with 1.

\[ \frac{A'}{A} = (1 + 0.25)(1 - 0.20) = 1.25 \times 0.80 = 1.00 \]

The area is therefore unchanged.

The percentages look unequal but exactly offset each other, because \( \tfrac{5}{4} \times \tfrac{4}{5} = 1 \). Adding them (\( +25 - 20 = +5 \)) would suggest a 5% rise, which is the distractor.