Area Questions and Answers

Answer: B

Start from the pond, whose area is known outright.

\[ A_{\text{pond}} = 8^{2} = 64 \text{ m}^{2} \]

\[ A_{\text{field}} = 8 \times 64 = 512 \text{ m}^{2} \]

With \( l = 2w \):

\[ 2w^{2} = 512 \quad\Longrightarrow\quad w^{2} = 256 \quad\Longrightarrow\quad w = 16 \text{ m} \]

\[ l = 32 \text{ m} \]

The pond is one eighth of the field, so the field is eight times the pond - dividing instead of multiplying is the trap, and 16 m (the width) is offered for anyone who stops one step early.

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.

Answer: A

\[ l = 3b \quad\Longrightarrow\quad 2(3b + b) = 96 \quad\Longrightarrow\quad 8b = 96 \]

\[ b = 12 \text{ m}, \qquad l = 36 \text{ m} \]

\[ A = 36 \times 12 = 432 \text{ m}^{2} \]

The perimeter contains four "breadths" worth of length in total (\( 2 \times 3b + 2 \times b \)), which is what makes the division by 8 so clean.

Answer: A

The path lies outside, so each dimension grows by twice its width.

\[ \text{outer} = (75 + 5) \times (55 + 5) = 80 \times 60 = 4800 \text{ m}^{2} \]

\[ \text{field} = 75 \times 55 = 4125 \text{ m}^{2} \]

\[ \text{path} = 4800 - 4125 = 675 \text{ m}^{2} \]

\[ \text{cost} = 675 \times 2 = \text{Rs. } 1350 \]

The 2.5 m appears on both sides of each dimension, so 5 m is added in total to each - not 2.5 m.

Answer: D

The two roads cross, so their shared square must not be counted twice.

\[ \text{road along the length} = 80 \times 10 = 800 \text{ m}^{2} \]

\[ \text{road along the breadth} = 60 \times 10 = 600 \text{ m}^{2} \]

\[ \text{overlap} = 10 \times 10 = 100 \text{ m}^{2} \]

\[ \text{total} = 800 + 600 - 100 = 1300 \text{ m}^{2} \]

\[ \text{cost} = 1300 \times 3 = \text{Rs. } 3900 \]

Forgetting to subtract the crossing gives Rs. 4200 - the intersection belongs to both roads but is paved only once.

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: B

Both dimensions grow, so multiply the two factors.

\[ \frac{A'}{A} = (1 + 0.10)(1 + 0.25) = 1.10 \times 1.25 = 1.375 \]

\[ \text{increase} = 1.375 - 1 = 0.375 = 37.5\% \]

Adding the percentages gives 35%; the extra 2.5% is the cross term \( 0.10 \times 0.25 \), the part of the rectangle that gains in both directions at once.

Answer: D

The diagonals of a rhombus bisect each other at right angles, so half of each and a side form a right triangle.

\[ \text{half of the known diagonal} = 10 \text{ m} \]

\[ \text{half of the other} = \sqrt{26^{2} - 10^{2}} = \sqrt{676 - 100} = \sqrt{576} = 24 \text{ m} \]

\[ \text{other diagonal} = 48 \text{ m} \]

\[ A = \frac{1}{2} \times 20 \times 48 = 480 \text{ m}^{2} \]

10–24–26 is the 5–12–13 triple doubled.

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: B

All four sides of a rhombus are equal, so start from the perimeter.

\[ \text{side} = \frac{68}{4} = 17 \text{ cm} \]

\[ \text{half of the given diagonal} = 8 \text{ cm} \]

\[ \text{half of the other} = \sqrt{17^{2} - 8^{2}} = \sqrt{289 - 64} = \sqrt{225} = 15 \text{ cm} \]

\[ \text{other diagonal} = 30 \text{ cm} \]

\[ A = \frac{1}{2} \times 16 \times 30 = 240 \text{ cm}^{2} \]

8–15–17 is another Pythagorean triple, so the second diagonal comes out whole.