Area Questions and Answers

Answer: A

Let the base be \( b \); the altitude is then \( 2b \).

\[ A = b \times 2b = 2b^{2} = 128 \]

\[ b^{2} = 64 \quad\Longrightarrow\quad b = 8 \text{ m} \]

A parallelogram's area is base times perpendicular height, with no factor of one half - that half belongs to the triangle.

Answer: B

a = 100         a² = 10000

a = 125         a² = 15625

                         ----------------

        10000 --------- 5625

        100 -------? => 56.25 %

Answer: D

A trapezium's area is the average of the parallel sides times the depth between them.

\[ A = \frac{1}{2}(a + b)h \]

\[ 640 = \frac{1}{2}(10 + 6)h = 8h \]

\[ h = 80 \text{ m} \]

The two widths must be averaged, not added outright: using 16 as the effective width would halve the depth to 40 m.

Answer: D

Area varies as the square of the radius.

\[ \frac{A'}{A} = (1.01)^{2} = 1.0201 \]

\[ \text{increase} = 0.0201 = 2.01\% \]

The exact rule is \( 2x + \tfrac{x^{2}}{100} = 2 + 0.01 \). For small changes the area rises by roughly twice the percentage change in the radius, with the tiny square term as the correction.

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: B

\[ A = \pi r^{2} = \frac{22}{7} \times 7^{2} = \frac{22}{7} \times 49 \]

\[ = 22 \times 7 = 154 \text{ m}^{2} \]

A radius of 7 is chosen so that the 7 in \( \tfrac{22}{7} \) cancels cleanly. The circumference, a different quantity, would be \( 2\pi r = 44 \) m.

Answer: B

Find the radius from the area, then use it for the circumference.

\[ \pi r^{2} = 616 \quad\Longrightarrow\quad r^{2} = \frac{616 \times 7}{22} = 196 \]

\[ r = 14 \text{ cm} \]

\[ C = 2\pi r = 2 \times \frac{22}{7} \times 14 = 88 \text{ cm} \]

The two formulas share \( \pi \) but not the power of \( r \), so the radius must be extracted before the circumference can be found.

Answer: B

Fencing follows the circumference, and the diameter is given directly.

\[ C = \pi d = \frac{22}{7} \times 28 = 88 \text{ m} \]

\[ \text{cost} = 88 \times 1.50 = \text{Rs. } 132 \]

Using \( 2\pi d \) - effectively treating 28 as the radius - would double the answer to Rs. 264.

Answer: A

One revolution advances the wheel by its own circumference.

\[ C = 2\pi r = 2 \times \frac{22}{7} \times 1.75 = 11 \text{ m} \]

\[ 11 \text{ km} = 11{,}000 \text{ m} \]

\[ n = \frac{11{,}000}{11} = 1000 \]

The radius 1.75 m is \( \tfrac{7}{4} \), so the 7 in \( \tfrac{22}{7} \) cancels and the circumference comes out exactly 11 m - which is what makes the final division so clean.

Answer: D

Each revolution advances the wheel by one circumference.

\[ C = 2\pi r = 2 \times \frac{22}{7} \times 14 = 88 \text{ cm} \]

\[ n = \frac{1056}{88} = 12 \]

A radius that is a multiple of 7 makes the \( \tfrac{22}{7} \) cancel exactly, which is why the circumference lands on a whole 88 cm.