Mensuration Questions and Answers

Answer: A

The four walls form a band running round the floor at height \( h \). Two of them measure \( l \times h \) and two \( b \times h \):

\[ A = 2lh + 2bh = 2h(l + b) \]

Equivalently it is the floor's perimeter \( 2(l+b) \) multiplied by the height. The floor and ceiling are excluded, so \( lb \) never appears.

Answer: A

Area of a circle = 4π cm²
πr² = 4π
r² = 4
r = 2 cm.
WKT, Radius of incircle = a/2√3
2 = a/2√3
a = 4√3.

WKT, Area of equilateral triangle = (√3/4)a²
Area of equilateral triangle = (√3/4)(4√3)²
= 12√3 cm².

Answer: B

The sheet is 1 cm thick, so the outer radius is 5 cm and the inner 4 cm. The metal is the shell between them.

\[ V = \frac{2}{3}\pi\left(R^{3} - r^{3}\right) = \frac{2}{3}\pi(125 - 64) \]

\[ = \frac{2}{3}\pi(61) = \frac{122\pi}{3} \text{ cm}^{3} \]

\[ \text{weight} = \frac{122\pi}{3} \times 9 = 366\pi \text{ g} \]

Treating the hemisphere as solid would give \( \tfrac{2}{3}\pi(125) \times 9 = 750\pi \) g - the cavity must be subtracted.

Answer: B

\[ A = \text{base} \times \text{height} = 24 \times 16 = 384 \text{ cm}^{2} \]

There is no factor of one half - that belongs to the triangle, and using it would give the 192 cm² on offer.

Answer: B

Turn the weight into a volume, then divide out the two known dimensions.

\[ 112 \text{ kg} = 112{,}000 \text{ g} \quad\Longrightarrow\quad V = \frac{112000}{25} = 4480 \text{ cm}^{3} \]

\[ \text{length} = \frac{4480}{28 \times 5} = \frac{4480}{140} = 32 \text{ cm} \]

The density is per gram, so the kilograms must be converted first.

Answer: D

An arc is the fraction of the circumference that its central angle is of a full turn.

\[ \text{arc length} = \frac{z^{\circ}}{360^{\circ}} \times \text{circumference} = \frac{z}{360} \times 2\pi r \]

The same fraction applied to the area of the circle gives the sector's area instead:

\[ \text{sector area} = \frac{z}{360} \times \pi r^{2} \]

Length comes from circumference and area from area - matching the two is what the options are testing.

Answer: B

The longest pole runs along the space diagonal of the room.

\[ d = \sqrt{l^{2} + b^{2} + h^{2}} \]

\[ = \sqrt{12^{2} + 8^{2} + 9^{2}} = \sqrt{144 + 64 + 81} = \sqrt{289} \]

\[ = 17 \text{ m} \]

The floor diagonal alone would be \( \sqrt{144 + 64} = 14.4 \) m; the height adds the third term.

Answer: A

Area of equilateral triangle = 16√3 cm²
WKT, Area of equilateral triangle = √3/4 a2
√3/4 a² = 16√3
a² = 64
a = 8 cm

Side of an equilateral triangle = Radius of a cylinder
a = r
r = 8cm
Perimeter of equilateral triangle = 3a = 3(8) = 24 cm

Perimeter of equilateral triangle = Height of cylinder
So, h = 24 cm
Volume of cylinder = πr²h
= π x 8 x 8 x 24
= 1536π cm³.

Answer: A

\[ A = \frac{\sqrt{3}}{4}a^{2} = \frac{\sqrt{3}}{4} \times 36 \]

\[ = 9\sqrt{3} \approx 15.6 \text{ sq. cm} \]

The side is squared before the constant is applied - multiplying 6 by \( \tfrac{\sqrt{3}}{4} \) directly would give \( 1.5\sqrt{3} \).

Answer: A

Ratio between radius of first ball to that of second ball is 4 : 3.
Let the radius of first ball be '4r' and radius of second ball be '3r'.
When we cut the second ball it become hemisphere.

WKT, Total surface area of sphere = 4π r²
Total surface area of hemisphere = 3π r²

As per the question,
4 x (22/7) x (4r)² - 3 x (22/7) x (3r)² = 1424.5
(22/7)*r²[64 - 27] = 1424.5
r² = [1424.5*7]/[37*22]
r² = 12.25
r = 3.5 cm
Therefore, radius of bigger ball = 4r = 4(3.5) = 14 cm.