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Mensuration Questions and Answers

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51.

A boy cut a sector containing an angle of 140° from a circle of radius 15 cm and he folded the sector into a cone. What is the curved surface area of the cone \(\left(\pi = \frac{22}{7}\right)\).

Exam Year(s)

TNPSC Group 1 - 2017, TNPSC - 2017,

Answer: C

The curved surface area is 275 sq. cm.

When a sector is folded into a cone, its area becomes the cone's curved surface, so the CSA equals the sector's area:

\[ \begin{aligned} \text{CSA} &= \frac{140}{360} \times \frac{22}{7} \times 15^{2} \\ &= \frac{7}{18} \times \frac{22}{7} \times 225 = 275 \end{aligned} \]

The radius of the circle becomes the slant height (15 cm) of the cone. 572, 527 and 257 are digit rearrangements of the answers and do not come from the formula.

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52.

The area of circle increased by 66 sq.cm. If its radius is increased by 1 cm. Find the area of new circle.

Exam Year(s)

TNPSC Group 1 - 2024, TNPSC - 2024,

Answer: C

The area increases by 66 when the radius goes up by 1, so \(\pi(2r+1)=66\) with \(\pi=22/7\) gives \(2r+1=21\) and \(r=10\). The new radius is 11, and the new area is \(\tfrac{22}{7}\times121=\tfrac{2}{7}\times11^{3}\), option C.

The options with 10, 9 and 12 would come from other radii.

Because \(\pi\times(r+1)^{2}-\pi r^{2}=66\) simplifies to \(2r+1=21\), the radius is 10 cm and the new radius 11 cm. The new area \(\tfrac{22}{7}\times 11^{2}\) can be written as \(\tfrac{2}{7}\times 11^{3}\), which is why option C is printed in that odd form.

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53.

The radius of a circle is increased so that its circumference is increased by 5%. Find the percentage of increase in its area.

Exam Year(s)

TNPSC Group 1 - 2024, TNPSC - 2024,

Answer: A

If the circumference increases by 5%, the radius also increases by 5%, so the new area is 1.05 squared = 1.1025 times the old area.

The increase is 10.25%, option A. 10% and 5.5% neglect the squaring, and 25% is far too large.

Circumference is proportional to the radius, so a 5 per cent increase in the circumference means a 5 per cent increase in the radius. The area depends on the square of the radius, giving 1.05 squared = 1.1025, an increase of 10.25 per cent. The common mistake is to double the 5 per cent and answer 10.

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54.

Three circles of radius 3.5 cm are placed in such a way that each circle touches the other two. The area of the portion enclosed by the circles is

Exam Year(s)

TNPSC Group 1 - 2025, TNPSC - 2025,

Answer: A

The centres of the three circles form an equilateral triangle of side 7 cm, whose area is \(\tfrac{\sqrt{3}}{4}\times 49\approx 21.217\) cm\(^2\). The three circles cover three 60 degree sectors, together half a circle: \(\tfrac{1}{2}\pi r^{2}\).

With \(\pi=\tfrac{22}{7}\) this is \(\tfrac{1}{2}\times\tfrac{22}{7}\times12.25=19.25\). The enclosed area is \(21.217-19.25\approx1.967\) cm\(^2\), option A.

With the exact value of \(\pi\) the result is 1.975, which is printed as option B, so the key assumes \(\pi=\tfrac{22}{7}\), the usual choice with a radius of 3.5.

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55.

The perimeter at a rectangle is 144 cm. It one of the two adjacent sides is 20 cm longer than other, then the lengths of the sides are

Exam Year(s)

Tripura Public Service Commission - 2021,

Answer: D

The sides are 26 cm, 46 cm.

Half the perimeter, 72 cm, is the sum of two adjacent sides. If the shorter side is x, the other is x + 20:

\[ \begin{aligned} x + (x + 20) &= 72 \\ 2x &= 52,\ x = 26 \end{aligned} \]

So the sides are 26 cm and 46 cm. Check: \(2(26 + 46) = 144\). In each of the other options either the two sides do not differ by 20 cm or they do not add up to 72 cm.

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56.

The circumference and the area of a circle at radius 21 cm are

Exam Year(s)

Tripura Public Service Commission - 2021,

Answer: A

They are 132 cm and 1386 \(\text{cm}^{2}\), in that order.

\[ \begin{aligned} C &= 2\pi r = 2 \times \frac{22}{7} \times 21 = 132\ \text{cm} \\ A &= \pi r^2 = \frac{22}{7} \times 441 = 1386\ \text{cm}^2 \end{aligned} \]

The option 1386 cm2, 132 cm reverses the order the question asks for. The other two give a length and an area the same value or the wrong units: a circumference is a length in cm, and an area is in square centimetres.

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57.

If the edge of a cube is 1 cm then which of the following is its volume?

Answer: C

\[ V = a^{3} = 1^{3} = 1 \text{ cm}^{3} \]

A cube of unit edge has unit volume; its total surface area, a different quantity, would be \( 6a^{2} = 6 \) cm².

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Answer: C

The diagonals of a rhombus bisect each other at right angles, cutting it into four congruent right triangles. That gives

\[ A = \frac{1}{2} d_{1} d_{2} \]

Each of the four triangles has area \( \tfrac{1}{2} \cdot \tfrac{d_{1}}{2} \cdot \tfrac{d_{2}}{2} = \tfrac{d_{1}d_{2}}{8} \), and four of them make \( \tfrac{d_{1}d_{2}}{2} \).

The same half-product formula works for any quadrilateral whose diagonals are perpendicular, including the square and the kite.

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59.

Value of sin 2π⁄3 considering radian is

Answer: D

Convert to degrees first.

\[ \frac{2\pi}{3} = \frac{2 \times 180^{\circ}}{3} = 120^{\circ} \]

\[ \sin 120^{\circ} = \sin(180^{\circ} - 120^{\circ}) = \sin 60^{\circ} = \frac{\sqrt{3}}{2} = 0.866 \]

Sine never exceeds 1, so any option above it can be discarded immediately.

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60.

Considering radian, value of cos 2.3π is

Answer: B

Convert the radian measure to degrees, using \( \pi \text{ radians} = 180^{\circ} \).

\[ 2.3\pi = 2.3 \times 180^{\circ} = 414^{\circ} \]

Subtracting a full turn:

\[ 414^{\circ} - 360^{\circ} = 54^{\circ} \]

\[ \cos 54^{\circ} = 0.587 \]

Cosine never exceeds 1, so the options above 1 can be discarded before any calculation.

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