Area Questions and Answers

Answer: B

Walking entirely around a semicircular garden means covering the curved arc and the straight diameter.

\[ \text{perimeter} = \pi r + 2r = r(\pi + 2) \]

With \( \pi = \tfrac{22}{7} \), so \( \pi + 2 = \tfrac{36}{7} \):

\[ r \times \frac{36}{7} = 324 \quad\Longrightarrow\quad r = \frac{324 \times 7}{36} = 63 \text{ m} \]

\[ A = \frac{1}{2}\pi r^{2} = \frac{1}{2} \times \frac{22}{7} \times 3969 = 6237 \text{ m}^{2} \]

Omitting the straight edge would make the perimeter \( \pi r \) alone and give a radius of 103 m - the diameter is a real part of the boundary he walks.

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: C

Find the original breadth, then the new one.

\[ b = \frac{840}{48} = 17.5 \text{ cm} \]

\[ A_{\text{new}} = 840 \times 2\tfrac{6}{7} = 840 \times \frac{20}{7} = 2400 \text{ cm}^{2} \]

The length is unchanged, so

\[ b_{\text{new}} = \frac{2400}{48} = 50 \text{ cm} \]

\[ \text{perimeter} = 2(48 + 50) = 196 \text{ cm} = 1.96 \text{ m} \]

The answer is offered in metres, so the 196 cm must be divided by 100 - the step the distractors are spaced around.

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: C

The triangle formed by joining the midpoints of a triangle's sides is the medial triangle. Each of its sides is half the parallel side of the original, by the midpoint theorem, so its perimeter is exactly half.

\[ P_{\triangle CDE} = 3 + 8 + 4 = 15 \]

\( \triangle CDE \) is the medial triangle of \( \triangle PQR \):

\[ P_{\triangle PQR} = 2 \times 15 = 30 \]

and \( \triangle PQR \) is the medial triangle of \( \triangle LMN \):

\[ P_{\triangle LMN} = 2 \times 30 = 60 \]

Each step outward doubles the perimeter, so two steps multiply it by four.

Answer: D

Check the sides first - \( 6^{2} + 8^{2} = 36 + 64 = 100 = 10^{2} \), so this is a right-angled triangle with legs 6 km and 8 km.

\[ A = \frac{1}{2} \times 6 \times 8 = 24 \text{ km}^{2} \]

\[ \text{cost} = 24 \times 8 = \text{Rs. } 192 \]

Heron's formula gives the same result: with \( s = 12 \), \( \sqrt{12 \times 4 \times 2 \times 6} = \sqrt{576} = 24 \). Spotting the Pythagorean triple simply saves the work.

Answer: C

Two parallelograms on the same base and between the same parallels have the same height, and area is base times height:

\[ A = \text{base} \times \text{height} \]

Both the base and the height are shared, so the two areas are equal and the ratio is \( 1 : 1 \).

The condition "between the same parallels" is doing the real work here: a rhombus with the same side but a different angle has area \( a^{2}\sin\theta \), which is less than the square's \( a^{2} \) unless \( \theta = 90^{\circ} \).

Answer: B

The area must stay fixed, so the two factors have to multiply to 1.

\[ 1.25 \times (1 - d) = 1 \quad\Longrightarrow\quad 1 - d = \frac{1}{1.25} = 0.8 \]

\[ d = 0.2 = 20\% \]

A 25% rise is undone by a 20%, not a 25%, fall, because the reduction applies to the larger figure. The general rule is

\[ d = \frac{x}{100 + x} \times 100 = \frac{25}{125} \times 100 = 20\% \]

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: D

Perimeter = Distance travelled in 8 minutes,
=> Perimeter = 12000/60 * 8 = 1600 meter. [because Distance = Speed * Time]
As per question length is 3x and width is 2x
We know perimeter of rectangle is 2(L+B)
So, 2(3x+2x) = 1600
=> x = 160
So Length = 160*3 = 480 meter
and Width = 160*2 = 320 meter
Finally, Area = length * breadth
=480*320

=153600