Area Questions and Answers

Answer: A

The parallel sides are 24 cm and 40 cm.

Let the parallel sides be \(3k\) and \(5k\). The area of a trapezium is half the sum of the parallel sides times the height:

\[ \begin{aligned} \tfrac{1}{2}(3k + 5k) \times 12 &= 384 \\ 48k &= 384 \\ k &= 8 \end{aligned} \]

So the sides are \(3 \times 8 = 24\) cm and \(5 \times 8 = 40\) cm. The other pairs are not in the ratio 3:5 (12 and 18 is 2:3, 15 and 20 is 3:4), or do not give an area of 384.

Answer: C

The radius of the inscribed circle is 3 cm.

Since \(8^{2} + 15^{2} = 64 + 225 = 289 = 17^{2}\), the triangle is right-angled. Its area is \(\frac{1}{2} \times 8 \times 15 = 60\) and its semi-perimeter is \((8 + 15 + 17)/2 = 20\):

\[ r = \frac{\text{Area}}{s} = \frac{60}{20} = 3\ \text{cm} \]

For a right triangle there is also the shortcut \(r = (a + b - c)/2 = (8 + 15 - 17)/2 = 3\). 4 and 5 cm and \(2\sqrt{2}\) cm do not satisfy either formula.

Answer: B

The area of the quadrant is 38.5 cm².

From the circumference, find the radius, then take a quarter of the circle's area (with \(\pi = 22/7\)):

\[ \begin{aligned} 2\pi r &= 44 \Rightarrow r = 7 \\ \tfrac{1}{4}\pi r^{2} &= \tfrac{1}{4} \times 154 = 38.5 \end{aligned} \]

77 cm² is a semicircle and 19.25 cm² an eighth of the circle, so both use the wrong fraction.

Answer: B

The shaded area is \(102.73\ \text{cm}^{2}\).

An angle in a semicircle is a right angle, so with AC as diameter, the angle at B is 90°.

\[ \begin{aligned} AC &= \sqrt{12^2 + 5^2} = 13 \\ r &= 6.5 \\ \text{circle} &= \pi (6.5)^2 \approx 132.73 \\ \text{triangle} &= \tfrac{1}{2} \times 12 \times 5 = 30 \end{aligned} \]

The shaded region is the circle minus the triangle: \(132.73 - 30 = 102.73\ \text{cm}^{2}\), using \(\pi \approx 3.1416\).

All four options end in .73, so the question turns only on the whole number: the circle's 132.73 minus the triangle's 30.

Answer: A

The area of the plot is \(1500\sqrt{3}\ \text{m}^{2}\).

Dividing the perimeter of 300 m in the ratio 3 : 5 : 7 (15 parts of 20 m each) gives sides of 60 m, 100 m and 140 m. The semi-perimeter is s = 150 m. By Heron's formula:

\[ \begin{aligned} A &= \sqrt{s(s - a)(s - b)(s - c)} \\ &= \sqrt{150 \times 90 \times 50 \times 10} \\ &= \sqrt{6750000} = 1500\sqrt{3}\ \text{m}^{2} \end{aligned} \]

since \(6750000 = 1500^{2} \times 3\). This is about 2,598 m². The options with 1300 or \(\sqrt{5}\) do not come from these sides; a quick check is that the angle opposite the 140 m side is 120°, giving \(\frac{1}{2}(60)(100)\sin 120^{\circ} = 1500\sqrt{3}\).

Answer: C

The fencing would cost INR 45,450.

The mowing cost gives the area, the area gives the side of the square, and the side gives the perimeter:

\[ \begin{aligned} \text{area} &= 4375 / 700 = 6.25 \text{ ha} \\ &= 62500 \text{ m}^2 \\ \text{side} &= \sqrt{62500} = 250 \text{ m} \\ \text{perimeter} &= 4 \times 250 = 1000 \text{ m} \end{aligned} \]

\[ \text{cost} = 1000 \times 45.45 = 45450 \]

The step most often missed is the conversion of 1 hectare to 10,000 square metres. 43,750 is just ten times the mowing cost, and 10,000 and 25,000 are not the price of 1,000 metres at INR 45·45.

Answer: B

The remaining area is 126 m².

The largest square in a circle has its diagonal equal to the diameter. From the cost, the square's area is \(882 \div 4 = 220.5\) m².

\[ \begin{aligned} d^2 &= 2 \times 220.5 = 441 \Rightarrow d = 21 \\ \text{circle} &= \tfrac{22}{7} \times 10.5^2 = 346.5 \\ \text{rest} &= 346.5 - 220.5 = 126 \end{aligned} \]

So 126 m² of the circular field remains. The other options do not follow from these figures with \(\pi = 22/7\).

Answer: D

The ratio of the bases is 15:8.

The area of a triangle is \(\frac{1}{2} \times \text{base} \times \text{height}\), so the ratio of bases is the ratio of areas divided by the ratio of heights.

\[ \begin{aligned} \frac{b_1}{b_2} &= \frac{A_1 / A_2}{h_1 / h_2} \\ &= \frac{3/2}{4/5} = \frac{15}{8} \end{aligned} \]

8:15 inverts the answer, and 7:8 comes from wrongly subtracting or adding the ratio terms instead of dividing.

Answer: D

The answer is None of these: the ratio works out to \(3\sqrt{3} : \pi\).

For an equilateral triangle, height \(h = \frac{\sqrt{3}}{2}a\), so \(a = 6\) cm, and the inradius is one third of the height, \(r = \sqrt{3}\) cm.

\[ \begin{aligned} \text{triangle} &= \frac{\sqrt{3}}{4} \times 6^2 = 9\sqrt{3} \\ \text{circle} &= \pi r^2 = 3\pi \\ \text{ratio} &= 9\sqrt{3} : 3\pi = 3\sqrt{3} : \pi \end{aligned} \]

Option A, \(3\sqrt{3} : 4\), has 4 where \(\pi\) should be, and B inverts it. Neither matches, and C is unrelated.

Answer: D

The area is \(96\ \text{cm}^{2}\).

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

\[ \begin{aligned} \text{half of } d_2 &= \sqrt{10^2 - 6^2} = 8 \\ d_2 &= 16\ \text{cm} \\ \text{area} &= \frac{12 \times 16}{2} = 96\ \text{cm}^{2} \end{aligned} \]

192 forgets to halve the product of the diagonals, and 48 halves it twice.