Mensuration Questions and Answers

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.

Answer: D

\[ \text{side of square} = \sqrt{4096} = 64 \text{ cm} \]

\[ \text{length} = 2 \times 64 = 128 \text{ cm}, \qquad \text{breadth} = 64 - 24 = 40 \text{ cm} \]

\[ \text{breadth} : \text{length} = 40 : 128 = 5 : 16 \]

That ratio is not among the choices, so the answer is none of these. The 7 : 16 on offer is close enough to catch a careless division.

Answer: D

Let the sides of the rectangle be l and b respectively.

From the given data, 

(√l² + b²) = (1 + 108 1/3 %)lb 

=> l² + b² = (1 + 325/3 * 1/100)lb

= (1 + 13/12)lb

= 25/12 lb

=> (l2 + b²)/lb = 25/12 

12(l2 + b²) = 25lb 

Adding 24lb on both sides 

12l2 + 12b² + 24lb = 49lb 

12(l2 + b² + 2lb) = 49lb 

but 2(l + b) = 28 => l + b = 14 

12(l + b)² = 49lb 

=> 12(14)² = 49lb 

=> lb = 48 

Since l + b = 14, l = 8 and b = 6 

l - b = 8 - 6 = 2m.

Answer: B

Correct answer: 100 sq. cm

For a square of side \(a\), the diagonal is \(a\sqrt{2}\).

\[a\sqrt{2} = 10\sqrt{2} \quad\Longrightarrow\quad a = 10 \text{ cm}\]

\[\text{Area} = a^2 = 100 \text{ sq. cm}\]

Equivalently, the area of a square is half the square of its diagonal: \(\tfrac{1}{2}(10\sqrt{2})^2 = 100\).

Answer: C

Compare each shape's area with its perimeter.

• Equilateral triangle, side 1: area \( \tfrac{\sqrt{3}}{4} = 0.43 \), perimeter 3
• Square, side 1: area 1, perimeter 4
• Regular pentagon, side 1: area 1.72, perimeter 5
• Square, side 4: area \( 4^{2} = 16 \), perimeter \( 4 \times 4 = 16 \) ✓

Only the square of side 4 cm has the two numerically equal. For a square, \( a^{2} = 4a \) forces \( a = 4 \) - a single size, and the reason unit-sided shapes cannot work.