Area Questions and Answers

Answer: B

The perimeter of a semicircle includes the straight diameter as well as the curved arc.

\[ P = \pi r + 2r = r(\pi + 2) \]

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

\[ P = 6.3 \times \frac{36}{7} = 0.9 \times 36 = 32.4 \text{ cm} \]

The arc alone measures \( \pi r = 19.8 \) cm; leaving out the 12.6 cm diameter is the standard omission.

Answer: B

A diagonal splits the quadrilateral into two triangles that share it as their base, with the offsets as their heights.

\[ A = \frac{1}{2} d (h_{1} + h_{2}) \]

\[ = \frac{1}{2} \times 20 \times (9 + 6) = 10 \times 15 = 150 \text{ cm}^{2} \]

The offsets are the perpendiculars dropped onto the diagonal from the two opposite corners, which is why they add before the halving.

Answer: C

Let the longer diagonal be \( d \); the shorter is then \( 0.6d \).

A rhombus's diagonals bisect each other at right angles, so its area is half their product:

\[ A_{\text{rhombus}} = \frac{1}{2} \times d \times 0.6d = 0.3d^{2} \]

The square is built on the longer diagonal as its side:

\[ A_{\text{square}} = d^{2} \]

\[ \frac{A_{\text{square}}}{A_{\text{rhombus}}} = \frac{d^{2}}{0.3d^{2}} = \frac{10}{3} \quad\Longrightarrow\quad 10 : 3 \]

The unknown \( d \) cancels, so the ratio holds for any rhombus with that diagonal proportion.

Answer: B

Both dimensions shrink, so multiply the two surviving fractions.

\[ \frac{A'}{A} = (1 - 0.20)(1 - 0.10) = 0.80 \times 0.90 = 0.72 \]

\[ \text{decrease} = 1 - 0.72 = 0.28 = 28\% \]

Adding the two losses gives 30%, which overstates the shrinkage by the product \( 0.20 \times 0.10 = 2\% \) - the part counted twice.

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

Answer: D

The altitude of an equilateral triangle splits it into two 30–60–90 triangles, giving

\[ h = \frac{\sqrt{3}}{2}a \]

\[ h = \frac{\sqrt{3}}{2} \times 2\sqrt{3} = \frac{2 \times 3}{2} = 3 \text{ cm} \]

Equivalently, by Pythagoras: \( \sqrt{(2\sqrt{3})^{2} - (\sqrt{3})^{2}} = \sqrt{12 - 3} = 3 \).

Answer: B

Bending conserves the length of the wire, so the circle's circumference becomes the rectangle's perimeter.

\[ C = 2\pi r = 2 \times \frac{22}{7} \times 21 = 132 \text{ cm} \]

With sides \( 6x \) and \( 5x \):

\[ 2(6x + 5x) = 132 \quad\Longrightarrow\quad 22x = 132 \quad\Longrightarrow\quad x = 6 \]

\[ \text{sides} = 36 \text{ cm and } 30 \text{ cm} \]

\[ \text{area} = 36 \times 30 = 1080 \text{ cm}^{2} \]

It is the perimeter that is preserved, not the area: the circle enclosed \( \pi r^{2} = 1386 \) cm², noticeably more than the rectangle. A circle always encloses the most area for a given perimeter.

Answer: B

Find the rectangle's sides from its perimeter first.

\[ 2\big((b + 12) + b\big) = 200 \quad\Longrightarrow\quad 2b + 12 = 100 \]

\[ b = 44 \text{ cm}, \qquad l = 56 \text{ cm} \]

\[ A = 56 \times 44 = 2464 \text{ cm}^{2} \]

Now set the circle's area equal to it, with \( \pi = \tfrac{22}{7} \):

\[ \frac{22}{7}r^{2} = 2464 \quad\Longrightarrow\quad r^{2} = \frac{2464 \times 7}{22} = 784 \]

\[ r = 28 \text{ cm} \quad\Longrightarrow\quad d = 56 \text{ cm} \]

The question asks for the diameter, so the radius of 28 cm must be doubled - and 28 is offered as a distractor for exactly that reason.

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: 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.