Mensuration Questions and Answers

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.

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151.

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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152.

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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153.

The volume of a brick is 3000 cm³ and its dimensions are in the ratio 3 : 2 : 4. If its entire surface is painted at 50 paisa per cm². The cost will be (in Rs.)

A.Rs. 580
B.Rs. 620
C.Rs. 680
D.Rs. 650

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.

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154.

If the radius of sphere is increased by 9.09 %, then surface area of sphere is increased by how much percentage.

A.19%
B.17%
C.21%
D.23%

Answer: A

An increase of \( 9.09\% \) is a factor of \( \tfrac{1}{11} \), since \( \tfrac{1}{11} = 0.0909 \). The radius therefore becomes \( \tfrac{12}{11}r \).

Surface area varies as the square of the radius:

\[ \frac{S'}{S} = \left(\frac{12}{11}\right)^{2} = \frac{144}{121} = 1.1901 \]

\[ \text{increase} = 19.01\% \approx 19\% \]

Recognising 9.09% as \( \tfrac{1}{11} \) keeps the arithmetic exact; the decimal route gives 18.998% and invites a rounding error.

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155.

A special type of cylindrical vessel with radius and height of 24.5 cm and 5 cm respectively is used to hold Cognac. The vessel is filled upto 80% of its capacity and then total Cognac from cylindrical vessel transferred to 9 cuboidal vessels whose length and breadth is 7 cm & 8 cm respectively. Find the height of each cuboidal vessel?

A.18 cm
B.25 cm
C.23 cm
D.14.9 cm

Answer: D

Cylinder: Radius - 24.5 cm; Height - 5 cm
Cuboid: Length - 7 cm; Breadth - 8 cm
WKT, Volume of cylinder = πr²h
Volume of cubiod = l x b x h

Volume of 80% of Cognac in cylindrical vessel = Volume of 9 cuboidal vessels
(22/7) x 24.5 x 24.5 x 5 x 0.8 = 9 x 7 x 8 x h
h = [(22/7) x 24.5 x 24.5 x 5 x 0.8 ]/[9 x 7 x 8]
h = 7546/504
h = 14.9 cm.

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156.

The length and breadth of a rectangle are in the ratio of 9:5. When the sides of the rectangle are extended on each side by 5 m the ratio of length to breadth becomes 5:3. What is the area of the original rectangle?

A.1160 sq.m
B.1565 sq.m
C.1125 sq.m
D.1045 sq.m

Answer: C

Let the length and breath of rectangle be 'X' m and 'Y' m respectively.
Given, length and breadth of a rectangle are in the ratio of 9 : 5.
X/Y = 9/5
5X = 9Y
5X - 9Y = 0 ....(i)

If sides of the rectangle are extended by 5 m, ratio becomes 5 : 3.
(X + 5)/(Y + 5) = 5/3
3(X + 5) = 5(Y + 5)
3X + 15 = 5Y + 25
3X - 5Y = 10 ....(ii)
By solving equation (i) and (ii),
X = 45 m ; Y = 25 m
So, length of rectangle = 45 m
Breadth of rectangle = 25 m

WKT, Area of rectangle = l x b
Area of rectangle = 45 x 25 = 1125 sq.m.

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