Area Questions and Answers

Answer: A

Convert to a single unit before dividing.

\[ \text{courtyard} = 2500 \times 1600 = 4{,}000{,}000 \text{ cm}^{2} \]

\[ \text{brick} = 20 \times 10 = 200 \text{ cm}^{2} \]

\[ n = \frac{4{,}000{,}000}{200} = 20{,}000 \]

A square metre holds 10,000 cm², so the 400 m² courtyard is four million square centimetres - the factor most often lost is that square metre conversion.

Answer: A

The track's length is

\[ C = 2\pi r = 2 \times \frac{22}{7} \times 70 = 440 \text{ m} \]

Runners moving in opposite directions meet once every time their combined distance equals one lap:

\[ v_{\text{John}} + v_{\text{Jan}} = \frac{440}{66} = \frac{20}{3} \text{ m/s} \]

\[ v_{\text{Joni}} + v_{\text{Jan}} = \frac{440}{110} = 4 \text{ m/s} \]

Subtracting eliminates Janardan's speed:

\[ v_{\text{John}} - v_{\text{Joni}} = \frac{20}{3} - 4 = \frac{8}{3} \text{ m/s} \]

John and Joni run the same way, so one laps the other when this difference covers a full circuit:

\[ t = \frac{440}{8/3} = 165 \text{ seconds} \]

Same direction means subtract the speeds; opposite directions means add them. Getting that the wrong way round here would give 66 seconds.

Answer: A

The diagonal, the given side and the unknown side form a right triangle.

\[ \text{other side} = \sqrt{5^{2} - 4^{2}} = \sqrt{25 - 16} = 3 \text{ m} \]

\[ A = 4 \times 3 = 12 \text{ m}^{2} \]

3–4–5 is the smallest Pythagorean triple. Multiplying the given side by the diagonal would give 20 - the diagonal is not a side of the rectangle.

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: C

\[ A = \frac{1}{2} d_{1} d_{2} = \frac{1}{2} \times 15 \times 20 = 150 \text{ cm}^{2} \]

The half-product formula works because the perpendicular diagonals cut the rhombus into four right triangles, each of area \( \tfrac{1}{2} \times \tfrac{d_{1}}{2} \times \tfrac{d_{2}}{2} \). Omitting the half gives 300, the second option.

Answer: A

Write the sides as multiples of a common factor and use the perimeter.

\[ 5k + 12k + 13k = 300 \quad\Longrightarrow\quad 30k = 300 \quad\Longrightarrow\quad k = 10 \]

\[ \text{sides} = 50,\; 120,\; 130 \text{ m} \]

Since \( 5^{2} + 12^{2} = 13^{2} \), the triangle is right-angled with the two shorter sides as legs:

\[ A = \frac{1}{2} \times 50 \times 120 = 3000 \text{ m}^{2} \]

Recognising the 5–12–13 triple avoids Heron's formula entirely.

Answer: B

The path lies within the plot's borders, so it eats 4 m in from each side and the untouched middle shrinks by 8 m in both directions.

\[ A_{\text{plot}} = 20 \times 15 = 300 \text{ m}^{2} \]

\[ A_{\text{inner}} = (20 - 8)(15 - 8) = 12 \times 7 = 84 \text{ m}^{2} \]

\[ A_{\text{path}} = 300 - 84 = 216 \text{ m}^{2} \]

\[ \text{cost} = 216 \times 6 = \text{Rs. } 1296 \]

Subtracting only 4 m from each dimension rather than 8 is the standard error - the path runs along both opposite sides.

Answer: A

At the same speed, the time saved is proportional to the distance saved. Running around the park covers the circumference; crossing it covers the diameter.

\[ \text{distance saved} = 2\pi r - 2r = 2r(\pi - 1) \]

Sixty minutes at 60 m/min is 3600 m:

\[ 2r(\pi - 1) = 3600 \]

With \( \pi = \tfrac{22}{7} \), so that \( \pi - 1 = \tfrac{15}{7} \):

\[ 2r \times \frac{15}{7} = 3600 \quad\Longrightarrow\quad 2r = \frac{3600 \times 7}{15} = 1680 \text{ m} \]

So the diameter is 1680 m and the radius 840 m. Note that \( 2r \) - the diameter itself - falls out of the algebra directly, with no need to find the radius first.

Answer: B

Only three sides are fenced, and the 20 ft side is left open.

\[ \text{other side} = \frac{680}{20} = 34 \text{ ft} \]

The open side is one of the pair of 20 ft sides, so the fencing covers the other 20 ft side and both 34 ft sides:

\[ 20 + 34 + 34 = 88 \text{ ft} \]

Fencing all four sides would need the full perimeter of 108 ft; leaving out one 20 ft side saves exactly 20 ft.

Answer: C

Find the side, then the length of wire once the gates are deducted.

\[ \text{side} = \sqrt{3136} = 56 \text{ m} \quad\Longrightarrow\quad \text{perimeter} = 224 \text{ m} \]

Two gates of 1 m each are left open:

\[ 224 - 2 = 222 \text{ m per round} \]

\[ \text{total wire} = 222 \times 3 = 666 \text{ m} \]

\[ \text{cost} = 666 \times 1.50 = \text{Rs. } 999 \]

The gates must be subtracted before multiplying by the three rounds, since every round skips them.