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: 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.