Area Questions and Answers

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.

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

A square's diagonal and side satisfy \( d = a\sqrt{2} \), so the area follows from the diagonal alone.

\[ A = a^{2} = \frac{d^{2}}{2} \]

\[ A = \frac{(5.2)^{2}}{2} = \frac{27.04}{2} = 13.52 \text{ cm}^{2} \]

The diagonal cuts the square into two congruent right triangles, each with legs \( a \) and hypotenuse \( d \) - which is where the halving comes from. Forgetting it gives 27.04.

Answer: C

Work out all four perimeters and compare.

Square, area 36: side \( \sqrt{36} = 6 \), so

\[ P = 4 \times 6 = 24 \text{ cm} \]

Equilateral triangle, side 9:

\[ P = 3 \times 9 = 27 \text{ cm} \]

Rectangle, length 10 and area 40: breadth \( \tfrac{40}{10} = 4 \), so

\[ P = 2(10 + 4) = 28 \text{ cm} \]

Circle, radius 4:

\[ C = 2\pi (4) = 25.13 \text{ cm} \]

The rectangle's 28 cm is the largest. It is also the least compact shape here, which is the general pattern: for a given area, the more elongated the figure, the longer its boundary - and the circle is always the most economical.

Answer: D

Take the radii as 3 and 5, so the areas are proportional to their squares.

\[ A_{1} : A_{2} = 9 : 25 \]

The shortfall is measured against the second field:

\[ \frac{25 - 9}{25} \times 100 = \frac{16}{25} \times 100 = 64\% \]

Dividing by 9 instead would answer a different question - by how much the second exceeds the first, which is 177.8%.

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: B

Start from the pond, whose area is known outright.

\[ A_{\text{pond}} = 8^{2} = 64 \text{ m}^{2} \]

\[ A_{\text{field}} = 8 \times 64 = 512 \text{ m}^{2} \]

With \( l = 2w \):

\[ 2w^{2} = 512 \quad\Longrightarrow\quad w^{2} = 256 \quad\Longrightarrow\quad w = 16 \text{ m} \]

\[ l = 32 \text{ m} \]

The pond is one eighth of the field, so the field is eight times the pond - dividing instead of multiplying is the trap, and 16 m (the width) is offered for anyone who stops one step early.

Answer: D

Halve each diameter to get the radii: 1 m, 1 m and 0.5 m.

\[ A_{\text{plank}} = 12 \times 6 = 72 \text{ m}^{2} \]

\[ A_{\text{holes}} = \pi(1)^{2} + \pi(1)^{2} + \pi(0.5)^{2} = \pi + \pi + \frac{\pi}{4} = \frac{9\pi}{4} \]

\[ A_{\text{remaining}} = 72 - \frac{9\pi}{4} = \frac{288 - 9\pi}{4} \text{ m}^{2} \]

Numerically that is about 64.93 m². Using the diameters directly in \( \pi r^{2} \) would quadruple the holes' area.

Answer: D

The two roads cross, so their shared square must not be counted twice.

\[ \text{road along the length} = 80 \times 10 = 800 \text{ m}^{2} \]

\[ \text{road along the breadth} = 60 \times 10 = 600 \text{ m}^{2} \]

\[ \text{overlap} = 10 \times 10 = 100 \text{ m}^{2} \]

\[ \text{total} = 800 + 600 - 100 = 1300 \text{ m}^{2} \]

\[ \text{cost} = 1300 \times 3 = \text{Rs. } 3900 \]

Forgetting to subtract the crossing gives Rs. 4200 - the intersection belongs to both roads but is paved only once.