Area Questions and Answers

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: B

Convert each perimeter to a side, then add the areas.

\[ a_{1} = \frac{48}{4} = 12 \text{ cm}, \qquad a_{2} = \frac{20}{4} = 5 \text{ cm} \]

\[ A = 12^{2} + 5^{2} = 144 + 25 = 169 \text{ cm}^{2} \]

\[ a = \sqrt{169} = 13 \text{ cm} \]

\[ \text{diagonal} = a\sqrt{2} = 13\sqrt{2} \text{ cm} \]

The perimeter of the combined square is 52 cm - not the sum 68 cm of the two originals, since areas add but perimeters do not.

Answer: C

\[ 2(l + b) = 600 \quad\Longrightarrow\quad l + b = 300 \]

\[ l = 300 - 100 = 200 \text{ m} \]

Halving the perimeter first is essential - subtracting the breadth from the full 600 m would give 500 m.

Answer: D

Turn the two facts into a sum and a product.

\[ 2(l + b) = 50 \quad\Longrightarrow\quad l + b = 25 \]

\[ lb = 150 \]

So \( l \) and \( b \) are the roots of

\[ x^{2} - 25x + 150 = 0 \quad\Longrightarrow\quad (x - 15)(x - 10) = 0 \]

\[ l = 15 \text{ m}, \qquad b = 10 \text{ m} \]

Checking the options directly also works: only 15 and 10 give both a sum of 25 and a product of 150.

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: B

Walking diagonally means covering the rectangle's diagonal, which forms a right triangle with the two sides.

\[ b = \sqrt{20^{2} - 16^{2}} = \sqrt{400 - 256} = \sqrt{144} = 12 \text{ m} \]

This is the 3–4–5 triple scaled by four: 12, 16, 20.

Answer: C

Use the ratio to write both sides with one unknown.

\[ 3k \times 2k = 3750 \quad\Longrightarrow\quad 6k^{2} = 3750 \quad\Longrightarrow\quad k^{2} = 625 \]

\[ k = 25 \quad\Longrightarrow\quad \text{sides} = 75 \text{ m and } 50 \text{ m} \]

\[ \text{perimeter} = 2(75 + 50) = 250 \text{ m} \]

\[ \text{cost} = 250 \times \text{Rs. } 0.50 = \text{Rs. } 125 \]

50 paise is half a rupee, so the cost is half the perimeter in rupees - treating the rate as Rs. 50 would inflate the answer a hundredfold.

Answer: A

Convert the mixed numbers to decimals and divide the areas.

\[ \text{stone} = 2.5 \times 2 = 5 \text{ m}^{2} \]

\[ \text{courtyard} = 30 \times 16.5 = 495 \text{ m}^{2} \]

\[ n = \frac{495}{5} = 99 \]

The division is exact, so the stones tile the courtyard with no cutting - which is what makes the volume comparison legitimate here.

Answer: B

The 0.25 m border runs along all four sides, so each dimension loses twice that.

\[ \text{tiled area} = (4 - 0.5)(9 - 0.5) = 3.5 \times 8.5 = 29.75 \text{ m}^{2} \]

\[ \text{tile} = 0.25 \times 0.25 = 0.0625 \text{ m}^{2} \]

\[ n = \frac{29.75}{0.0625} = 476 \]

Subtracting only 0.25 m from each dimension instead of 0.5 is the usual slip - the space is left on both opposite sides.

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.