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: C

The radius is \(r_2 =\) \(7 - 4\sqrt{2}\) cm, about 1.34 cm.

Put W at the origin with the square of side 4 cm. \(C_1\) touches the two sides at W, so \(O_1 = (1, 1)\). \(C_2\) touches the two sides at Y, so \(O_2 = (4 - r_2, 4 - r_2)\). Both centres lie on the diagonal WY, and touching circles have centres \(r_1 + r_2\) apart:

\[ \begin{aligned} \sqrt{2}\,(3 - r_2) &= 1 + r_2 \\ r_2 &= \frac{3\sqrt{2} - 1}{\sqrt{2} + 1} \\ &= (3\sqrt{2} - 1)(\sqrt{2} - 1) \\ &= 7 - 4\sqrt{2} \end{aligned} \]

The other values fail a size check: \(4 - 3\sqrt{2}\) is negative, and \(1 + 2\sqrt{2} \approx 3.83\) and \(5 + 3\sqrt{2} \approx 9.24\) would not fit in a 4 cm square beside a 1 cm circle.

Answer: D

The area of the circle is \(\frac{1}{2}\pi d\).

Let the side of the square be \(s\), so \(s^{2} = d\). The diagonal is \(s\sqrt{2}\), and it is the diameter, so the radius is \(\frac{s\sqrt{2}}{2}\).

\[ \begin{aligned} \text{area} &= \pi \left(\frac{s\sqrt{2}}{2}\right)^{2} \\ &= \pi \cdot \frac{2s^{2}}{4} = \frac{1}{2}\pi d \end{aligned} \]

\(\pi d\) uses the diagonal as the radius, and \(\pi d^{2}\) and \(\frac{1}{4}\pi d^{2}\) treat \(d\) as a length rather than an area, so they have the wrong dimensions.

Answer: B

The original area is 2250 m².

Let the length be l and the breadth b. The reduced sides are equal, so l - 10 = b - 5, that is, l = b + 5. The lost area is 650:

\[ \begin{aligned} lb - (l - 10)(b - 5) &= 650 \\ 5l + 10b - 50 &= 650 \\ 5(b + 5) + 10b &= 700 \\ b = 45,\ l &= 50 \end{aligned} \]

So the rectangle is 50 m by 45 m, area 2250 m², and the square left is 40 m by 40 m (1600 m², which is 650 less). 1125 and 4500 are half and double the answer, and 2924 does not fit the conditions.

Answer: B

The ratio is 1 : 1.

In case M, four circles fill the 1 cm square in a 2 by 2 grid, so each has diameter 1/2 cm and radius 1/4 cm. In case N, nine circles fill it in a 3 by 3 grid, so each has radius 1/6 cm. The shaded areas are:

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

Both unshaded areas are therefore \(1 - \pi/4\) square cm, and the ratio is 1 : 1.

In general, \(n^{2}\) touching circles in an \(n\) by \(n\) grid always cover \(\pi/4\) of the square, whatever \(n\) is. The ratios 2 : 3 and 3 : 2 come from comparing the numbers of circles, 4 and 9, or their rows, instead of their areas.