Area Questions and Answers

Answer: B

Express the breadth in terms of the length and use the perimeter.

\[ b = 0.6l \quad\Longrightarrow\quad 2(l + 0.6l) = 800 \]

\[ 2(1.6l) = 800 \quad\Longrightarrow\quad l = 250 \text{ m} \]

\[ b = 0.6 \times 250 = 150 \text{ m} \]

\[ A = 250 \times 150 = 37{,}500 \text{ m}^{2} \]

Reading the 60% as applying to the length rather than the breadth reverses the two and halves the area to 18,750 - the first distractor.

Answer: B

\[ A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 5 = 10 \text{ m}^{2} \]

The factor of one half is what distinguishes the triangle from the rectangle on the same base and height, whose area would be 20 m².

Answer: A

Check the triangle inequality before computing anything: the sum of any two sides must exceed the third.

\[ 1 + 2 = 3 \]

This is equality, not inequality, so the three points are collinear and no triangle encloses any area.

Heron's formula confirms it. With \( s = \tfrac{1+2+3}{2} = 3 \):

\[ A = \sqrt{3(3-1)(3-2)(3-3)} = \sqrt{3 \times 2 \times 1 \times 0} = 0 \]

The zero factor \( (s - c) \) is the algebra registering the degenerate case.

Answer: D

Use the equilateral-triangle area formula and solve for the side.

\[ A = \frac{\sqrt{3}}{4}a^{2} = 4\sqrt{3} \]

\[ a^{2} = \frac{4\sqrt{3} \times 4}{\sqrt{3}} = 16 \]

\[ a = 4 \text{ cm} \]

The \( \sqrt{3} \) cancels from both sides, which is why the answer comes out a whole number.

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: D

Find the missing perpendicular side by Pythagoras.

\[ \text{height} = \sqrt{10^{2} - 8^{2}} = \sqrt{100 - 64} = \sqrt{36} = 6 \]

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

The hypotenuse is never a side of the right angle, so it cannot be used as a height - using 8 and 10 as the two legs would give 40.

Answer: A

Write the sides as multiples of a common factor and use the perimeter.

\[ 5k + 12k + 13k = 300 \quad\Longrightarrow\quad 30k = 300 \quad\Longrightarrow\quad k = 10 \]

\[ \text{sides} = 50,\; 120,\; 130 \text{ m} \]

Since \( 5^{2} + 12^{2} = 13^{2} \), the triangle is right-angled with the two shorter sides as legs:

\[ A = \frac{1}{2} \times 50 \times 120 = 3000 \text{ m}^{2} \]

Recognising the 5–12–13 triple avoids Heron's formula entirely.

Answer: A

\[ A = a^{2} = 15^{2} = 225 \text{ m}^{2} \]

The perimeter would be \( 4 \times 15 = 60 \) m - a different quantity with different units, and worth keeping distinct from the area.

Answer: C

A square's diagonal and side satisfy \( d = a\sqrt{2} \), so

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

\[ A = \frac{20^{2}}{2} = \frac{400}{2} = 200 \text{ m}^{2} \]

Squaring the diagonal without halving gives 400, offered as the last option; the diagonal cuts the square into two right triangles, and the halving reflects that.

Answer: B

From \( d = a\sqrt{2} \):

\[ a = \frac{8\sqrt{2}}{\sqrt{2}} = 8 \]

\[ \text{perimeter} = 4a = 32 \text{ m} \]

The \( \sqrt{2} \) in the given diagonal is a deliberate signal that the side is a whole number.