Area Questions and Answers

Answer: C

The area is 2.

Simplify the function first. \(\sqrt{e^{x}} = e^{x/2}\), so

\[ \begin{aligned} f(x) &= 2\ln(e^{x/2}) = 2 \cdot \frac{x}{2} = x \\ \int_{0}^{2} x\, dx &= \frac{2^{2}}{2} = 2 \end{aligned} \]

The region is a right triangle with base 2 and height 2, area half of 4. Forgetting the factor 2 in front of the logarithm gives f(x) = x/2 and an area of 1; dropping the square root gives f(x) = 2x and an area of 4. 1/2 matches neither.

Answer: B

The ratio is 1 : 1.

In case M each circle has diameter 1/2 cm, and in case N each has diameter 1/3 cm. The total shaded area is

\[ \begin{aligned} M &: 4 \times \pi \left(\tfrac{1}{4}\right)^{2} = \frac{\pi}{4} \\ N &: 9 \times \pi \left(\tfrac{1}{6}\right)^{2} = \frac{\pi}{4} \end{aligned} \]

Both shade \(\pi/4\) of the 1 cm² sheet, so both leave \(1 - \pi/4\) unshaded, and the ratio is 1 : 1. In general, n² touching circles in an n by n grid always cover \(\pi/4\) of the square. The options 2 : 3 and 3 : 2 come from comparing the numbers of circles instead of their areas.

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.

Answer: A

Cutting 5 m from the length removes a strip 5 m long and \( b \) wide.

\[ 5b = 200 \quad\Longrightarrow\quad b = 40 \text{ m} \]

\[ l = \frac{2400}{40} = 60 \text{ m} \]

\[ \text{perimeter} = 2(60 + 40) = 200 \text{ m} \]

The lost strip has the full breadth of the park, which is what makes the breadth recoverable from the area change alone.

Answer: B

\[ P = r(\pi + 2) = r \times \frac{36}{7} \]

\[ 144 = \frac{36r}{7} \quad\Longrightarrow\quad r = \frac{144 \times 7}{36} = 28 \text{ cm} \]

Using \( \pi r \) alone would give \( r = 45.8 \) cm - the diameter is part of a semicircle's boundary and must be included.