Mensuration Questions and Answers

Q: If the area of a circle, inscribed in an equilateral triangle is 4π cm², then what is the area of the triangle?

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.

Q: Find the length of the longest pole that can be placed in a room 12 m long, 8m broad and 9 m high.

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.

Q: Radius of a cylinder is equal to the side of an equilateral triangle having area 16√3 cm² and height of the cylinder is equal to the perimeter of the triangle. Then find the volume of cylinder.

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.

Important Instructions

141.

If the dimensions of a room are \( l \), \( b \) and \( h \) (length, breadth and height respectively), then which of the following is the area of its four walls?

A.\( 2h(l+b) \)
B.\( 2h(l+h) \)
C.\( 2l(h+b) \)
D.\( 2h + l + b \)

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.

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142.

If the area of a circle, inscribed in an equilateral triangle is 4π cm², then what is the area of the triangle?

A.2√3 cm²
B.9√3 cm²
C.8√3 cm²
D.18 cm²

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.

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143.

A hemisphere is made of a sheet of a metal 1 cm thick. If the outer radius is 5 cm, what is the weight of the hemisphere (1 cm³ of the metal weigh 9 g)?

A.54π g
B.366π g
C.122π g
D.108π g

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.

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144.

Find the area of a parallelogram with base 24 cm and height 16 cm.

A.262 cm²
B.384 cm²
C.192 cm²
D.131 cm²

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.

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145.

Given that one cubic cm of marble weighs 25 gms, the weight of a marble block 28 cm in width and 5 cm thick is 112 kg. The length of the block is:

A.26.5 cm
B.32 cm
C.36 cm
D.37.5 cm

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.

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146.

If the angle of a sector is \( z^{\circ} \), then the formula used to calculate the length of its arc is

A.z°⁄360° x area of major arc
B.z°⁄360° x area of semicircle
C.z°⁄360° x area of circle
D.z°⁄360° x circumference

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.

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147.

Find the length of the longest pole that can be placed in a room 12 m long, 8m broad and 9 m high.

A.16 m
B.17 m
C.18 m
D.19 m

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.

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148.

Radius of a cylinder is equal to the side of an equilateral triangle having area 16√3 cm² and height of the cylinder is equal to the perimeter of the triangle. Then find the volume of cylinder.

A.1536π cu.cm
B.1518π cu.cm
C.1620π cu.cm
D.1460π cu.cm

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.

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149.

Each side of an equilateral triangle is 6 cm. Find its area.

A.9√3 sq.cm.
B.6√3 sq.cm.
C.4√3 sq.cm.
D.8√3 sq.cm.

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 \( 1.5\sqrt{3} \).

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150.

There were two solid spherical balls. Ratio between radius of first ball to that of second ball is 4 : 3. Second ball was cut into two equal halves and the difference between total surface area of first ball and total surface area of a part of second ball is 1424.5 cm². Find value of radius of bigger ball ?

A.14 cm
B.16 cm
C.10.5 cm
D.20 cm

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.

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