Area Questions and Answers

Answer: C

Let the original length be \( L \), with breadth 20 cm.

\[ A_{\text{old}} = 20L \]

The length rises by 30% and the breadth falls by 20%:

\[ A_{\text{new}} = (1.3L)(0.8 \times 20) = (1.3L)(16) = 20.8L \]

\[ A_{\text{new}} - A_{\text{old}} = 0.8L = 100 \quad\Longrightarrow\quad L = 125 \text{ cm} \]

\[ A_{\text{new}} = 20.8 \times 125 = 2600 \text{ cm}^{2} \]

The question asks for the new area; the old one is 2500 cm², offered as the last distractor.

Answer: A

Find the barn's area, then its dimensions, then its perimeter.

\[ A_{\text{farm}} = 240^{2} = 57{,}600 \text{ m}^{2} \]

\[ A_{\text{barn}} = \frac{57{,}600}{8} = 7200 \text{ m}^{2} \]

With breadth \( b \) and length \( b + 10 \):

\[ b(b + 10) = 7200 \quad\Longrightarrow\quad b^{2} + 10b - 7200 = 0 \]

\[ b = \frac{-10 + \sqrt{100 + 28800}}{2} = \frac{-10 + 170}{2} = 80 \text{ m} \]

So the barn is 80 m by 90 m.

\[ \text{perimeter} = 2(80 + 90) = 340 \text{ m} \]

\[ \text{cost} = 340 \times 1.5 = \text{Rs. } 510 \]

The wall is priced by length, so the area of 7200 m² is only an intermediate step - pricing it by area would give Rs. 10,800.

Answer: A

Square tiles must fit a whole number of times along both edges, so the tile's side must divide both dimensions. Fewest tiles means largest tiles, so take the highest common factor.

\[ 5.25 \text{ m} = 525 \text{ cm}, \qquad 5.10 \text{ m} = 510 \text{ cm} \]

\[ 525 = 3 \times 5^{2} \times 7, \qquad 510 = 2 \times 3 \times 5 \times 17 \]

\[ \text{HCF} = 3 \times 5 = 15 \text{ cm} \]

\[ n = \frac{525}{15} \times \frac{510}{15} = 35 \times 34 = 1190 \]

Converting to centimetres first is essential - the highest common factor of 5.25 and 5.10 is not a whole number of metres.

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: C

A rhombus's diagonals bisect each other at right angles, so half of each diagonal and a side form a right triangle.

\[ \text{half of the known diagonal} = \frac{24}{2} = 12 \text{ cm} \]

\[ \text{half of the other} = \sqrt{13^{2} - 12^{2}} = \sqrt{169 - 144} = 5 \text{ cm} \]

\[ \text{other diagonal} = 10 \text{ cm} \]

\[ \text{area} = \frac{1}{2} d_{1} d_{2} = \frac{1}{2} \times 24 \times 10 = 120 \text{ cm}^{2} \]

5–12–13 is a Pythagorean triple, which keeps the arithmetic exact. Forgetting the \( \tfrac{1}{2} \) gives 240, the second option.

Answer: B

Work from the square to the rectangle.

\[ \text{side of square} = \frac{48}{4} = 12 \text{ cm} \quad\Longrightarrow\quad A = 144 \text{ cm}^{2} \]

\[ A_{\text{rectangle}} = 144 - 4 = 140 \text{ cm}^{2} \]

\[ \text{breadth} = \frac{140}{14} = 10 \text{ cm} \]

\[ \text{perimeter} = 2(14 + 10) = 48 \text{ cm} \]

The rectangle's perimeter coincidentally matches the square's, even though its area is smaller - equal perimeters do not imply equal areas, and the square is always the larger for a given perimeter.

Answer: A

Area varies as the square of the side, so take square roots to recover the length ratio.

\[ \frac{a_{1}^{2}}{a_{2}^{2}} = \frac{1}{4} \quad\Longrightarrow\quad \frac{a_{1}}{a_{2}} = \frac{1}{2} \]

Perimeter is \( 4a \), a plain multiple of the side, so it keeps that same ratio:

\[ \frac{P_{1}}{P_{2}} = \frac{4a_{1}}{4a_{2}} = \frac{1}{2} \quad\Longrightarrow\quad 1 : 2 \]

Carrying the area ratio straight through to the perimeters gives 1 : 4, the first distractor.

Answer: A

The field is a square of side 40 m. Each cow, tied at a corner with a 14 m rope, sweeps a quarter circle inside the field.

Four quarter circles of the same radius make one full circle:

\[ A_{\text{grazed}} = 4 \times \frac{1}{4}\pi r^{2} = \pi (14)^{2} = \frac{22}{7} \times 196 = 616 \text{ m}^{2} \]

\[ A_{\text{field}} = 40^{2} = 1600 \text{ m}^{2} \]

\[ A_{\text{ungrazed}} = 1600 - 616 = 984 \text{ m}^{2} \]

Since \( 14 < 20 \), the ropes are too short for neighbouring cows to reach the same patch, so no area is double-counted.

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: A

At the same speed, the time saved is proportional to the distance saved. Running around the park covers the circumference; crossing it covers the diameter.

\[ \text{distance saved} = 2\pi r - 2r = 2r(\pi - 1) \]

Sixty minutes at 60 m/min is 3600 m:

\[ 2r(\pi - 1) = 3600 \]

With \( \pi = \tfrac{22}{7} \), so that \( \pi - 1 = \tfrac{15}{7} \):

\[ 2r \times \frac{15}{7} = 3600 \quad\Longrightarrow\quad 2r = \frac{3600 \times 7}{15} = 1680 \text{ m} \]

So the diameter is 1680 m and the radius 840 m. Note that \( 2r \) - the diameter itself - falls out of the algebra directly, with no need to find the radius first.