Area Questions and Answers

Answer: C

Area varies as the square of the radius, so the ratio is squared.

\[ \frac{A_{1}}{A_{2}} = \frac{\pi r_{1}^{2}}{\pi r_{2}^{2}} = \left(\frac{1}{3}\right)^{2} = \frac{1}{9} \]

So the areas are in the ratio \( 1 : 9 \). Carrying the linear ratio straight through gives \( 1 : 3 \) - the circumferences, not the areas.

Answer: D

Take the radii as 3 and 5, so the areas are proportional to their squares.

\[ A_{1} : A_{2} = 9 : 25 \]

The shortfall is measured against the second field:

\[ \frac{25 - 9}{25} \times 100 = \frac{16}{25} \times 100 = 64\% \]

Dividing by 9 instead would answer a different question - by how much the second exceeds the first, which is 177.8%.

Answer: C

The calf grazes a circle, so the extra ground is the ring between the old and new radii.

\[ \Delta A = \pi\left(R^{2} - r^{2}\right) = \frac{22}{7}\left(23^{2} - 12^{2}\right) \]

\[ = \frac{22}{7}(529 - 144) = \frac{22}{7} \times 385 \]

\[ = 22 \times 55 = 1210 \text{ m}^{2} \]

The difference of the squares is needed, not the square of the difference: \( \pi(23-12)^{2} = 380 \) m² is a quite different and much smaller area.

Answer: D

All lengths in a square scale together: if the diagonal is three times larger, so is the side, and so is the perimeter.

\[ d = a\sqrt{2} \quad\Longrightarrow\quad \frac{d_{1}}{d_{2}} = 3 \;\Longrightarrow\; \frac{a_{1}}{a_{2}} = 3 \]

\[ \frac{P_{1}}{P_{2}} = \frac{4a_{1}}{4a_{2}} = 3 \quad\Longrightarrow\quad 3 : 1 \]

Only the areas would be squared, in the ratio 9 : 1. Perimeter, side and diagonal are all first powers of length and share one ratio.

Answer: B

The perimeter of a semicircle includes the straight diameter as well as the curved arc.

\[ P = \pi r + 2r = r(\pi + 2) \]

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

\[ P = 6.3 \times \frac{36}{7} = 0.9 \times 36 = 32.4 \text{ cm} \]

The arc alone measures \( \pi r = 19.8 \) cm; leaving out the 12.6 cm diameter is the standard omission.

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.

Answer: C

The walk lies outside the path, so it is the ring between radius 13 m and radius 15 m.

\[ A = \pi\left(15^{2} - 13^{2}\right) = \frac{22}{7}(225 - 169) = \frac{22}{7} \times 56 \]

\[ = 22 \times 8 = 176 \text{ m}^{2} \]

At 25 paise, that is a quarter of a rupee per square metre:

\[ \text{cost} = 176 \times 0.25 = \text{Rs. } 44 \]

Reading 25p as Rs. 25 would inflate the answer a hundredfold.

Answer: A

Each semicircle sits on a side of the square as its diameter, so each has diameter 20 cm and radius 10 cm.

\[ A_{\text{square}} = 20^{2} = 400 \text{ cm}^{2} \]

\[ A_{\text{one semicircle}} = \frac{1}{2}\pi (10)^{2} = 50\pi \text{ cm}^{2} \]

Two of them together make a full circle of radius 10:

\[ A_{\text{cut away}} = 100\pi \text{ cm}^{2} \]

\[ A_{\text{remaining}} = (400 - 100\pi) \text{ cm}^{2} \approx 85.8 \text{ cm}^{2} \]

The semicircles are on opposite sides, so they do not overlap and their areas simply add.

Answer: C

A sector is a triangle-like wedge whose "base" is its arc and whose "height" is the radius:

\[ A = \frac{1}{2} \times r \times \text{arc length} \]

\[ = \frac{1}{2} \times 5 \times 3.5 = 8.75 \text{ cm}^{2} \]

This form needs no angle at all. It follows from \( A = \tfrac{\theta}{2\pi}\pi r^{2} \) and \( \text{arc} = r\theta \), which combine to give exactly \( \tfrac{1}{2}r \times \text{arc} \).

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.