Area Questions and Answers

Answer: C

The area is 242 square metres. Let the breadth be b, so the length is 2b. The perimeter is 2(2b + b) = 6b = 66, so b = 11 m and the length is 22 m. The area is 22 x 11 = 242 square metres. The option 121 is the square of the breadth and 338 and 616 do not come from the given dimensions. The area is length times breadth, 22 x 11 = 242 square metres, and the perimeter is 2(22 + 11) = 66 metres, which agrees with the data. The options 338, 616 and 121 do not give a perimeter of 66 for any rectangle whose length is twice its breadth.

Answer: A

Let the breadth be b, so the length is 5b. The perimeter of a rectangle is twice the sum of the length and the breadth, so 2(5b + b) = 12b = 300 cm, and b = 25 cm. The length is then 5 x 25 = 125 cm. Check: 2(125 + 25) = 300 cm, and 125 is five times 25. So the length is 125 cm, option (A). The value 25 cm is the breadth, and 100 cm, 150 cm and 250 cm do not give a perimeter of 300 cm with a length five times the breadth.

Answer: B

The perimeter is 48 cm.

A regular hexagon has six equal sides, so its perimeter is six times the side.

\[ 6\times8=48\text{ cm} \]

47, 49 and 50 cm are not six times 8.

A regular hexagon has six equal sides and six equal angles of 120 degrees each. Its perimeter is the sum of all six sides, 8 + 8 + 8 + 8 + 8 + 8 = 48 cm. The area would use a different formula, which is not needed here.

Answer: B

The diagonal is 10 cm.

A square with diagonal d has area d²/2. The first square has d = 5√2, so its area is 25. The second has area 50, so its diagonal satisfies d²/2 = 50.

\[ d^{2}=100 \Rightarrow d=10\text{ cm} \]

12, 8 and 7 cm would give areas of 72, 32 and 24.5, none of which is 50.

When the area doubles, the side grows by a factor of root 2, and so does the diagonal. The first diagonal is 5√2, so the new one is 5√2 × √2 = 10 cm. This is a quick way to check the answer.

Answer: C

The cost of fencing is ₹ 3,792. Let the longer side be \(l\), the shorter side \(b\) and the diagonal \(d\). The condition says \(b = \frac{l + d}{8}\), so \(d = 8b - l\). Squaring and using \(d^{2} = l^{2} + b^{2}\): \[ 64b^{2} - 16bl + l^{2} = l^{2} + b^{2} \] \[ 63b^{2} = 16bl \Rightarrow l = \frac{63b}{16} \] With area \(lb = 1008\): \[ \begin{aligned} \frac{63b^{2}}{16} &= 1008 \Rightarrow b^{2} \\ &= 256,\ b \\ &= 16,\ l \\ &= 63 \end{aligned} \] Perimeter \(= 2(63 + 16) = 158\) m, and \(158\times 24 = 3792\).

Answer: B

RS is 28/5 cm.

The area of a parallelogram is any side times the perpendicular distance to it from the opposite side, and it is the same whichever pair of sides you use. PT is the height onto base QR, and QR = PS = 7 cm. PV is the height onto side RS.

\[ \begin{aligned} \text{Area} &= QR \times PT = 7 \times 4 = 28 \\ \text{Area} &= RS \times PV = 5\,RS \\ RS &= \frac{28}{5}\ \text{cm} \end{aligned} \]

The distractor 20/7 comes from 4 × 5 / 7, mixing the two heights; 35/4 comes from 7 × 5 / 4, putting the heights the wrong way round; 9/2 is simply (4 + 5)/2.

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.

Answer: B

A square's diagonal and side are related by \( d = a\sqrt{2} \), so the area can be written directly in terms of the diagonal:

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

The factor \( \tfrac{1}{2} \) is common to both squares and cancels:

\[ \frac{A_{1}}{A_{2}} = \frac{d_{1}^{2}}{d_{2}^{2}} = \frac{2^{2}}{5^{2}} = \frac{4}{25} \]

So the ratio is \( 4 : 25 \). Any two similar figures have areas in the square of their linear ratio, whichever length is used to compare them.

Answer: C

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

\[ \text{half of the known diagonal} = \frac{24}{2} = 12 \text{ cm} \]

\[ \text{half of the other} = \sqrt{13^{2} - 12^{2}} = \sqrt{169 - 144} = 5 \text{ cm} \]

\[ \text{other diagonal} = 10 \text{ cm} \]

\[ \text{area} = \frac{1}{2} d_{1} d_{2} = \frac{1}{2} \times 24 \times 10 = 120 \text{ cm}^{2} \]

5–12–13 is a Pythagorean triple, which keeps the arithmetic exact. Forgetting the \( \tfrac{1}{2} \) gives 240, the second option.

Answer: D

The fencing runs round the perimeter, so divide the cost by the rate.

\[ \text{perimeter} = \frac{5300}{26.50} = 200 \text{ m} \]

\[ 2\big((b + 20) + b\big) = 200 \quad\Longrightarrow\quad 2b + 20 = 100 \]

\[ b = 40 \text{ m}, \qquad l = 60 \text{ m} \]

The length is 60 m, which appears nowhere among the choices, so the answer is none of these. The 40 m on offer is the breadth - a genuine intermediate value, but not what was asked.