Area Questions and Answers

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 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: A

Let original length = x metres and original breadth = y metres
Original Area=xy m²
New Length =120100x=65x
New Breadth=120100y=65y
⇒New Area=65x * 65y   
⇒New Area=3625xy
Area Difference=3625xy−xy                            
=1125xy
Increase%=DifferenceActual * 100   
=11xy25 * 1xy * 100
=44%

Answer: D

A trapezium's area is the average of the parallel sides times the depth between them.

\[ A = \frac{1}{2}(a + b)h \]

\[ 640 = \frac{1}{2}(10 + 6)h = 8h \]

\[ h = 80 \text{ m} \]

The two widths must be averaged, not added outright: using 16 as the effective width would halve the depth to 40 m.

Answer: A

Two facts describe the same rectangle, so use the ratio to introduce a single unknown.

\[ l = 7k, \qquad b = 4k \]

\[ l - b = 15 \quad\Longrightarrow\quad 3k = 15 \quad\Longrightarrow\quad k = 5 \]

\[ l = 35 \text{ m}, \qquad b = 20 \text{ m} \]

\[ A = 35 \times 20 = 700 \text{ m}^{2} \]

The difference of the ratio terms, \( 7 - 4 = 3 \), is what corresponds to the given 15 m - matching the right difference to the right number is the whole problem.

Answer: A

Let the base be \( b \); the altitude is then \( 2b \).

\[ A = b \times 2b = 2b^{2} = 128 \]

\[ b^{2} = 64 \quad\Longrightarrow\quad b = 8 \text{ m} \]

A parallelogram's area is base times perpendicular height, with no factor of one half - that half belongs to the triangle.

Answer: D

Only the four walls are painted, and the door and window are not.

\[ \text{four walls} = 2(l + b)h = 2(6 + 5) \times 4 = 88 \text{ m}^{2} \]

\[ \text{door} = 2.5 \times 1.2 = 3 \text{ m}^{2}, \qquad \text{window} = 1 \times 1 = 1 \text{ m}^{2} \]

\[ \text{paintable area} = 88 - 3 - 1 = 84 \text{ m}^{2} \]

\[ \text{cost} = 84 \times 2 = \text{Rs. } 168 \]

Rs. 84 is the area mistaken for the cost, and Rs. 176 would come from forgetting to deduct the openings.

Answer: A

One revolution advances the wheel by its own circumference.

\[ C = 2\pi r = 2 \times \frac{22}{7} \times 1.75 = 11 \text{ m} \]

\[ 11 \text{ km} = 11{,}000 \text{ m} \]

\[ n = \frac{11{,}000}{11} = 1000 \]

The radius 1.75 m is \( \tfrac{7}{4} \), so the 7 in \( \tfrac{22}{7} \) cancels and the circumference comes out exactly 11 m - which is what makes the final division so clean.

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

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.