Area Questions and Answers

Q: The ratio between the perimeter and breadth of a rectangle is 5:1. If the area of the rectangle is 216 sq. cm, then the length of the rectangle is

The length is 18 cm.Let the length be \(l\) and the breadth \(b\). The perimeter is five times the breadth:\[ \begin{aligned} 2(l + b) &= 5b \Rightarrow l = 1.5b \\ l \times b &= 1.5b^{2} = 216 \\ b^{2} &= 144,\ b = 12 \end{aligned} \]So \(l = 18\) cm. Check: the perimeter is \(2(18 + 12) = 60\) cm, and 60 : 12 = 5 : 1. 24 cm would need a breadth of 9 cm, which gives a perimeter ratio of 66 : 9, not 5 : 1.

Q: If the area of a circular field is 616 m2 then the diameter of the field is

The diameter is 28m.Use \(\text{area} = \pi r^{2}\) with \(\pi = \frac{22}{7}\):\[ \begin{aligned} \frac{22}{7} r^{2} &= 616 \\ r^{2} &= 196,\ r = 14\ \text{m} \\ d &= 2r = 28\ \text{m} \end{aligned} \]14 m is the radius, which is the usual slip; 16, 18 and 20 m do not give an area of 616 m2.Taking \(\pi = \frac{22}{7}\) is the cue: 616 is \(22 \times 28\), so \(r^{2} = 28 \times 7 = 196\), a perfect square.

Q: If the area of a rhombus is 96 cm2 and one of its diagonals is 16cm, find the other diagonal.

The other diagonal is 12 cm.The area of a rhombus is half the product of its diagonals.\[ \begin{aligned} \tfrac{1}{2} \times 16 \times d &= 96 \\ 8d &= 96 \\ d &= 12\ \text{cm} \end{aligned} \]Forgetting the half gives 6 cm, which is not offered; 14, 18 and 20 cm would give areas of 112, 144 and 160 square centimetres.

Q: If the circumference of a circle is 88 cm, the area of the circle is

The area is 616 square centimetres.First find the radius from the circumference \(2\pi r\), taking \(\pi = \frac{22}{7}\).\[ \begin{aligned} 2 \times \tfrac{22}{7} \times r &= 88 \\ r &= 14\ \text{cm} \\ \text{area} &= \tfrac{22}{7} \times 14^{2} \\ &= 616\ \text{cm}^{2} \end{aligned} \]352 is \(\frac{22}{7} \times 112\), a slip in squaring 14, and 600 and 432 do not come from any correct radius.

Q: If the radius of a wheel is 14 cm, the number of revolutions it makes in covering a distance of 1056 cm is?

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.

Q: The length of a rectangle is increased by 25% and its breadth is decreased by 20%. What is the effect on its area?

Multiply the two factors and compare with 1.\[ \frac{A'}{A} = (1 + 0.25)(1 - 0.20) = 1.25 \times 0.80 = 1.00 \]The area is therefore unchanged.The percentages look unequal but exactly offset each other, because \( \tfrac{5}{4} \times \tfrac{4}{5} = 1 \). Adding them (\( +25 - 20 = +5 \)) would suggest a 5% rise, which is the distractor.

Important Instructions

31.

The ratio between the perimeter and breadth of a rectangle is 5:1. If the area of the rectangle is 216 sq. cm, then the length of the rectangle is

A.18 cm
B.24 cm
C.16 cm
D.20 cm

Exam Year(s)

Tripura Public Service Commission - 2019,

Answer: A

The length is 18 cm.

Let the length be \(l\) and the breadth \(b\). The perimeter is five times the breadth:

\[ \begin{aligned} 2(l + b) &= 5b \Rightarrow l = 1.5b \\ l \times b &= 1.5b^{2} = 216 \\ b^{2} &= 144,\ b = 12 \end{aligned} \]

So \(l = 18\) cm. Check: the perimeter is \(2(18 + 12) = 60\) cm, and 60 : 12 = 5 : 1. 24 cm would need a breadth of 9 cm, which gives a perimeter ratio of 66 : 9, not 5 : 1.

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32.

If the area of a circular field is 616 m² then the diameter of the field is

A.16m
B.28m
C.20m
D.18m

Exam Year(s)

Tripura Public Service Commission - 2019,

Answer: B

The diameter is 28m.

Use \(\text{area} = \pi r^{2}\) with \(\pi = \frac{22}{7}\):

\[ \begin{aligned} \frac{22}{7} r^{2} &= 616 \\ r^{2} &= 196,\ r = 14\ \text{m} \\ d &= 2r = 28\ \text{m} \end{aligned} \]

14 m is the radius, which is the usual slip; 16, 18 and 20 m do not give an area of 616 m².

Taking \(\pi = \frac{22}{7}\) is the cue: 616 is \(22 \times 28\), so \(r^{2} = 28 \times 7 = 196\), a perfect square.

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33.

If the area of a rhombus is 96 cm² and one of its diagonals is 16cm, find the other diagonal.

A.14 cm
B.18 cm
C.12 cm
D.20 cm

Exam Year(s)

Tripura Public Service Commission - 2019,

Answer: C

The other diagonal is 12 cm.

The area of a rhombus is half the product of its diagonals.

\[ \begin{aligned} \tfrac{1}{2} \times 16 \times d &= 96 \\ 8d &= 96 \\ d &= 12\ \text{cm} \end{aligned} \]

Forgetting the half gives 6 cm, which is not offered; 14, 18 and 20 cm would give areas of 112, 144 and 160 square centimetres.

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34.

If the circumference of a circle is 88 cm, the area of the circle is

A.\(352\ \text{cm}^{2}\)
B.\(600\ \text{cm}^{2}\)
C.\(616\ \text{cm}^{2}\)
D.\(432\ \text{cm}^{2}\)

Exam Year(s)

Tripura Public Service Commission - 2019,

Answer: C

The area is 616 square centimetres.

First find the radius from the circumference \(2\pi r\), taking \(\pi = \frac{22}{7}\).

\[ \begin{aligned} 2 \times \tfrac{22}{7} \times r &= 88 \\ r &= 14\ \text{cm} \\ \text{area} &= \tfrac{22}{7} \times 14^{2} \\ &= 616\ \text{cm}^{2} \end{aligned} \]

352 is \(\frac{22}{7} \times 112\), a slip in squaring 14, and 600 and 432 do not come from any correct radius.

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35.

If the radius of a wheel is 14 cm, the number of revolutions it makes in covering a distance of 1056 cm is?

A.15
B.10
C.14
D.12

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.

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36.

The Diagonals of two squares are in the ratio of 2:5. find the ratio of their areas.

A.4:15
B.4:25
C.3:15
D.3:25

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.

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37.

The side and one of the diagonals of a rhombus are 13cm and 24cm respectively. The area of the rhombus (in sq.cm.) is

A.156
B.240
C.120
D.130

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.

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38.

The length of a rectangular plot is 20 metres more than its breadth. If fencing the plot at Rs. 26.50 per metre costs Rs. 5300, what is the length of the plot in metres?

A.40 m
B.50 m
C.120 m
D.None of these

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.

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39.

The ratio of the area of a square to that of the square drawn on its diagonal is?

A.2:5
B.3:4
C.3:5
D.1:2

Answer: D

Let the original square have side \( a \), so its diagonal is \( a\sqrt{2} \).

The second square is built on that diagonal as its side:

\[ A_{1} = a^{2}, \qquad A_{2} = \left(a\sqrt{2}\right)^{2} = 2a^{2} \]

\[ \frac{A_{1}}{A_{2}} = \frac{a^{2}}{2a^{2}} = \frac{1}{2} \quad\Longrightarrow\quad 1 : 2 \]

The square on the diagonal is always exactly double - the diagonal is longer by \( \sqrt{2} \), and areas scale by the square of that.

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40.

The length of a rectangle is increased by 25% and its breadth is decreased by 20%. What is the effect on its area?

A.Remains same
B.5% decrease
C.5% increase
D.10% increase

Answer: A

Multiply the two factors and compare with 1.

\[ \frac{A'}{A} = (1 + 0.25)(1 - 0.20) = 1.25 \times 0.80 = 1.00 \]

The area is therefore unchanged.

The percentages look unequal but exactly offset each other, because \( \tfrac{5}{4} \times \tfrac{4}{5} = 1 \). Adding them (\( +25 - 20 = +5 \)) would suggest a 5% rise, which is the distractor.

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