Area Questions and Answers

Answer: C

The area is 2.

Simplify the function first. \(\sqrt{e^{x}} = e^{x/2}\), so

\[ \begin{aligned} f(x) &= 2\ln(e^{x/2}) = 2 \cdot \frac{x}{2} = x \\ \int_{0}^{2} x\, dx &= \frac{2^{2}}{2} = 2 \end{aligned} \]

The region is a right triangle with base 2 and height 2, area half of 4. Forgetting the factor 2 in front of the logarithm gives f(x) = x/2 and an area of 1; dropping the square root gives f(x) = 2x and an area of 4. 1/2 matches neither.

Answer: B

The ratio is 1 : 1.

In case M each circle has diameter 1/2 cm, and in case N each has diameter 1/3 cm. The total shaded area is

\[ \begin{aligned} M &: 4 \times \pi \left(\tfrac{1}{4}\right)^{2} = \frac{\pi}{4} \\ N &: 9 \times \pi \left(\tfrac{1}{6}\right)^{2} = \frac{\pi}{4} \end{aligned} \]

Both shade \(\pi/4\) of the 1 cm² sheet, so both leave \(1 - \pi/4\) unshaded, and the ratio is 1 : 1. In general, n² touching circles in an n by n grid always cover \(\pi/4\) of the square. The options 2 : 3 and 3 : 2 come from comparing the numbers of circles instead of their areas.

Answer: D

The area is 64 \(\mathrm{m}^{2}\).

In 15 seconds, a quarter of a minute, P covers \(56 \div 4 = 14\) m along the diagonal, and R covers \(72 \div 4 = 18\) m along two sides. With sides l and b:

\[ \begin{aligned} l + b &= 18,\quad l^{2} + b^{2} = 196 \\ 2lb &= (l + b)^{2} - (l^{2} + b^{2}) \\ &= 324 - 196 = 128 \\ lb &= 64 \end{aligned} \]

128 is twice the area, the value of 2lb before dividing by 2.

Answer: D

The new area becomes 4 times the original.

The area of a square is the square of its side:

\[ \begin{aligned} \text{Original area} &= 5 \times 5 = 25\ \text{cm}^{2} \\ \text{New area} &= 10 \times 10 = 100\ \text{cm}^{2} \\ \frac{100}{25} &= 4 \end{aligned} \]

In general, doubling the side multiplies the area by \(2^{2} = 4\). The answer 2 times is the common error of thinking area grows in the same proportion as the side, and 8 times is what happens to the volume of a cube when its side is doubled.

Answer: A

The area is 228 cm².

Since AD = BC = 13 cm, the trapezium is isosceles. Dropping perpendiculars from A and B to CD gives two right triangles with hypotenuse 13 cm and height 12 cm, so each base piece is

\[ \sqrt{13^2 - 12^2} = \sqrt{25} = 5 \text{ cm} \]

So AB = 24 - 5 - 5 = 14 cm, and

\[ \begin{aligned} \text{Area} &= \tfrac{1}{2}(14 + 24) \times 12 \\ &= 228 \text{ cm}^2 \end{aligned} \]

114 is half the correct area, and 247 comes from using 13 cm as the height.

Answer: D

The perimeter of the rectangle is 114 cm.

The triangle is right-angled, since 24² + 45² = 576 + 2025 = 2601 = 51². So its legs 24 cm and 45 cm are base and height:

\[ \begin{aligned} \text{area} &= \tfrac{1}{2} \times 24 \times 45 = 540 \\ \text{breadth} &= 540 \div 45 = 12 \\ \text{perimeter} &= 2(45 + 12) = 114 \end{aligned} \]

57 cm is the half-perimeter (45 + 12), forgetting to double. 110 cm and 55 cm come from a breadth of 10 cm, an arithmetic slip in the area.

Answer: B

The remaining area is 88 cm².

The sheet has diameter 16 cm, so radius 8 cm. Subtract the area of the circle of radius 6 cm:

\[ \begin{aligned} A &= \pi(8^{2} - 6^{2}) \\ &= \frac{22}{7} \times 28 = 88 \end{aligned} \]

The trap is to use 16 cm, the diameter, as the radius. Of the other options, 176 is twice the correct area and 44 is half of it, and 66 is three quarters; none equals the area of the ring.

Answer: C

The perimeter is 24 units.

Each small square has area 4 sq. units, so its side is 2 units. The rectangle drawn on the grid is 4 squares long and 2 squares wide:

\[ \begin{aligned} \text{length} &= 4 \times 2 = 8,\quad \text{breadth} = 2 \times 2 = 4 \\ \text{perimeter} &= 2(8 + 4) = 24\ \text{units} \end{aligned} \]

12 units is the perimeter if each square's side is taken as 1 unit, and 32 units is the rectangle's area (8 squares × 4), not its perimeter. 8 units is just its length.

Answer: B

The diagonal is 28.4 cm.

A diagonal divides the quadrilateral into two triangles with the diagonal as their common base and the two perpendiculars as their heights:

\[ \begin{aligned} \text{Area} &= \tfrac{1}{2} d (h_{1} + h_{2}) \\ 227.2 &= \tfrac{1}{2} d (7.2 + 8.8) = 8d \\ d &= 28.4\ \text{cm} \end{aligned} \]

Forgetting the factor one half gives 14.2 cm, not an option; the other values do not satisfy the area formula (for example 8 × 26.8 = 214.4).

Answer: C

The area is 128 cm².

The perimeter is the two parallel sides plus the two non-parallel sides, so the parallel sides together measure

\[ 52 - (10 + 10) = 32\ \text{cm} \]

The area of a trapezium is half the sum of the parallel sides times the distance between them:

\[ \begin{aligned} \text{Area} &= \tfrac{1}{2} \times 32 \times 8 \\ &= 128\ \text{cm}^{2} \end{aligned} \]

Forgetting the half doubles this to 256, which is not offered, and 64 is half the correct area. The individual lengths of the parallel sides are not needed, only their sum.