Mensuration Questions and Answers

Answer: C

Pouring conserves volume, so equate the cone's with the cylinder's.

\[ V_{\text{cone}} = \frac{1}{3}\pi (12)^{2}(50) = \frac{1}{3}\pi (7200) = 2400\pi \text{ cm}^{3} \]

\[ V_{\text{cyl}} = \pi (10)^{2} h = 100\pi h \]

\[ 100\pi h = 2400\pi \quad\Longrightarrow\quad h = 24 \text{ cm} \]

The \( \pi \) cancels, so no approximation is needed. The one-third factor is what keeps the liquid from filling the wider cylinder to 50 cm.

Answer: D

W.K.T, the shape of Kite resembles Quadrilateral.
The adjacent sides of a kite are of equal length.

Given, perimeter of Quadrilateral = 100 m
Length of one of the side = x = 20 m
Let the Length of other side = y

W.K.T, perimeter of Quadrilateral = sum of the length of all the four sides
---> 100 = 20 + 20 + y + y
---> 100 = 40 + 2y
---> 60 = 2y
---> y = 30 m

Hence, the length of sides of the kite (ie., Quadrilateral) = 20 m, 20 m, 30 m, 30 m

Answer: A

\[ \pi \text{ radians} = 180^{\circ} \]

\[ 3.5 \text{ rad} = 3.5 \times \frac{180}{\pi} = 3.5 \times 57.2958 \]

\[ = 200.54^{\circ} \]

Since \( \pi \approx 3.14 \) radians is \( 180^{\circ} \), 3.5 radians must be somewhat more than a straight angle - which rules out every option below \( 180^{\circ} \).

Answer: D

Melting conserves volume, so divide the cylinder's by one bullet's. The bullet's radius is \( 1.5 \div 2 = 0.75 \) cm.

\[ V_{\text{cyl}} = \pi (6)^{2}(28) = 1008\pi \text{ cm}^{3} \]

\[ V_{\text{bullet}} = \frac{4}{3}\pi (0.75)^{3} = \frac{4}{3}\pi (0.421875) = 0.5625\pi \]

\[ n = \frac{1008}{0.5625} = 1792 \]

The diameter must be halved before cubing - using 1.5 as the radius would cut the count eightfold.

Answer: A

Given, ratio of the parallel sides of a trapezium = 5 : 3
---> Let the parallel sides (ie., 'a' and 'b') be 5x and 3x respectively.
Given, perpendicular distance = height = 24 m
and Area = 1440 m2

Formula: Area of Trapezium = (1/2) * h * (a + b)
---> 1440 = (1/2) * 24 * (5x + 3x)
---> 1440 = 12 * (8x)
---> 120 = 8x
---> 15 = x

Thus, parallel sides, a = 5x = 5 * 15 = 75 m
and b = 3x = 3 * 15 = 45 m
Hence, Longer parallel side is 'a' = 75 m
Smaller parallel side is 'b' = 45 m.

Answer: C

\[ A = \frac{1}{2}(a + b)h = \frac{1}{2}(20 + 18) \times 15 \]

\[ = \frac{1}{2} \times 38 \times 15 = 285 \text{ cm}^{2} \]

The two parallel sides are averaged, then multiplied by the perpendicular distance between them.

Answer: A

\[ A = \frac{\theta}{360} \times \pi r^{2} \]

\[ 140 = \frac{\theta}{360} \times \pi \times 256 \]

\[ \theta = \frac{140 \times 360}{256\pi} = \frac{50400}{804.25} \]

\[ \theta = 62.66^{\circ} \]

The whole circle covers 804.2 cm², and 140 is a little over a sixth of it - consistent with an angle a little over \( 60^{\circ} \).

Answer: D

Let the length of the brick be 3X, breadth of the brick be 2X and height of the brick be 4X.
Given: Volume of a brick is 3000 cm³
WKT, Volume of brick = l x b x h
Volume of brick = 3X*2X*4X
3000 = 24X³
X = 5
Length of the brick = 3X = 3(5) = 15 cm
Breadth of the brick = 2X = 2(5) = 10 cm
Height of the brick = 4X = 4(5) = 20 cm

WKT, Total surface area of brick = 2(lb + bh + hl)
Total surface area of brick = 2[(15 x 10) + (10 x 20) + (20 x 15)]
= 2[150 + 200 + 300]
= 1300²
The cost = 1300 x (50/100) = Rs. 650.

Answer: B

Angles in opposite segments of a circle - equivalently, the opposite angles of a cyclic quadrilateral - add to \( 180^{\circ} \), so they are supplementary.

\[ \angle A + \angle C = 180^{\circ} \]

This follows from the inscribed-angle theorem: each angle is half the arc it faces, and the two arcs together make the whole \( 360^{\circ} \).

Complementary would mean adding to \( 90^{\circ} \), a different relation altogether.

Answer: C

Circular measure is the system that measures an angle by the arc it subtends on a circle, divided by the radius - which is exactly the radian.

\[ \theta = \frac{\text{arc}}{\text{radius}} \]

It stands beside sexagesimal measure (degrees, minutes, seconds) as the other standard way of quantifying angles.