Area Questions and Answers

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.

Answer: D

Area varies as the square of the radius.

\[ \frac{A'}{A} = (1.01)^{2} = 1.0201 \]

\[ \text{increase} = 0.0201 = 2.01\% \]

The exact rule is \( 2x + \tfrac{x^{2}}{100} = 2 + 0.01 \). For small changes the area rises by roughly twice the percentage change in the radius, with the tiny square term as the correction.

Answer: A

Square tiles must fit a whole number of times along both edges, so the tile's side must divide both dimensions. Fewest tiles means largest tiles, so take the highest common factor.

\[ 5.25 \text{ m} = 525 \text{ cm}, \qquad 5.10 \text{ m} = 510 \text{ cm} \]

\[ 525 = 3 \times 5^{2} \times 7, \qquad 510 = 2 \times 3 \times 5 \times 17 \]

\[ \text{HCF} = 3 \times 5 = 15 \text{ cm} \]

\[ n = \frac{525}{15} \times \frac{510}{15} = 35 \times 34 = 1190 \]

Converting to centimetres first is essential - the highest common factor of 5.25 and 5.10 is not a whole number of metres.

Answer: B

Work from the square to the rectangle.

\[ \text{side of square} = \frac{48}{4} = 12 \text{ cm} \quad\Longrightarrow\quad A = 144 \text{ cm}^{2} \]

\[ A_{\text{rectangle}} = 144 - 4 = 140 \text{ cm}^{2} \]

\[ \text{breadth} = \frac{140}{14} = 10 \text{ cm} \]

\[ \text{perimeter} = 2(14 + 10) = 48 \text{ cm} \]

The rectangle's perimeter coincidentally matches the square's, even though its area is smaller - equal perimeters do not imply equal areas, and the square is always the larger for a given perimeter.

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: C

A square's diagonal and side satisfy \( d = a\sqrt{2} \), so

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

\[ A = \frac{20^{2}}{2} = \frac{400}{2} = 200 \text{ m}^{2} \]

Squaring the diagonal without halving gives 400, offered as the last option; the diagonal cuts the square into two right triangles, and the halving reflects that.

Answer: B

Find the long side from the area, then the diagonal.

\[ \text{long side} = \frac{1200}{30} = 40 \text{ m} \]

\[ \text{diagonal} = \sqrt{40^{2} + 30^{2}} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \text{ m} \]

\[ \text{fencing} = 40 + 30 + 50 = 120 \text{ m} \]

\[ \text{cost} = 120 \times 10 = \text{Rs. } 1200 \]

The 30–40–50 triangle is the 3–4–5 triple scaled tenfold, so the diagonal comes out whole.

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

A diagonal splits the quadrilateral into two triangles that share it as their base, with the offsets as their heights.

\[ A = \frac{1}{2} d (h_{1} + h_{2}) \]

\[ = \frac{1}{2} \times 20 \times (9 + 6) = 10 \times 15 = 150 \text{ cm}^{2} \]

The offsets are the perpendiculars dropped onto the diagonal from the two opposite corners, which is why they add before the halving.