Areas Clubbed(10th Grade > Mathematics > Area ) Questions and answers for exam Preparation

10th Grade > Mathematics > Area

AREAS CLUBBED MCQs

Total Questions : 49 | Page 3 of 5 pages

Question 21. If the curved surface area of a cylinder of height 14 cm is 88 sq. cm, then find the diameter of the cylinder.

2 cm
1 cm
4 cm
3 cm

Discuss Question

Answer: Option A. -> 2 cm
:
A
CSA of a cylinder

= 2 π r h

Given,
CSA

= 88 {cm}^{2}

and

h = 14 cm

2 π r h = 88 2 r = \frac{88}{π h} = 88 \times \frac{7}{22} \times \frac{1}{14} = 4 \times 7 \times \frac{1}{14} 2 r = 2 cm

Therefore, diameter = 2r = 2 cm.

Question 22. What is the diameter (in cm) of a sphere whose surface area is 616

{cm}^{2}

Discuss Question

Answer: Option D. -> 14
:
D
Surface area of sphere =

4 π r^{2} = 616 \Rightarrow r^{2} = 616 \times \frac{7}{22} \times \frac{1}{4} \Rightarrow r^{2} = 49 \Rightarrow r = \sqrt{49} \Rightarrow r = 7 cm

∴

The diameter,

D = 2 r = 2 \times 7 = 14 cm

Question 23. A conical vessel of radius 6 cm and height 8 cm is completely filled with water. A sphere is lowered into the water and its size is such that when it touches the sides, it is just immersed. What fraction of water overflows?

0.375
0.625
0.125
0.875

Discuss Question

Answer: Option A. -> 0.375
:
A

A Conical Vessel Of Radius 6 Cm And Height 8 Cm Is Completel...

Radius of the conical vessel,

R = A C = 6 c m

Height of the conical vessel,

h = O C = 8 c m

Radius of the sphere,

P D = P C = r

∴ P C = P D = r A C = A D = 6 cm

[Since, lengths of two tangents from an external point to a circle are equal]

△ O C A & △ O P D

areright triangle.
[

∵

Tangent andradius are perpendicular to each other]

O A = \sqrt{O C^{2} + A C^{2}} = \sqrt{8^{2} + 6^{2}} = \sqrt{100} = 10 cm O P^{2} = O D^{2} + P D^{2}

O D = O A - A D = 10 - 6 = 4 cm O P = O C - P C = 8 - r (8 - r)^{2} = 4^{2} + r^{2} 64 - 16 r + r^{2} = 16 + r^{2} 16 r = 48 \Rightarrow r = 3 cm

.
Volume of water overflown = Volume
of sphere

= \frac{4}{3} π r^{3} = \frac{4}{3} π \times (3)^{3} = 36 π {cm}^{3}

Original volume of water=volume of
cone

= \frac{1}{3} π r^{2} h = \frac{1}{3} π \times 6^{2} \times 8 = 96 π {cm}^{3}

∴

Fraction of water overflown

= \frac{Volumeofwater overflown}{Original volume of water} = \frac{36 π}{96 π} = \frac{3}{8} = 0.375

∴

Fraction of water overflown is

0.375

Question 24. Anita buys a new salt cellar in the shape of a cylinder topped by a hemisphere as shown below. The cylinder has a diameter of 6 cm and a height of 10 cm. She pours the salt into the salt cellar, so that it takes up half the total volume of the cellar. Find the depth of the salt, marked with x in the diagram

3 cm
9 cm
6 cm
12 cm

Discuss Question

Answer: Option C. -> 6 cm
:
C
Let the depth of the salt in the cellarbe

^{'} x^{'}

Volume of = volume ofcylinder +
salt cellar volume of hemisphere

= π r^{2} h + \frac{2}{3} π r^{3} = π \times 3^{2} \times 10 + \frac{2}{3} π \times 3^{3} = π [90 + 18] = 108 π {cm}^{3}

So height

x

will come on the cylinder.
Half oftotal volume

= 54 π {cm}^{3}

π r^{2} x = 54 π π \times 3^{2} \times x = 54 π 9 x = 54 \Rightarrow x = 6

∴

The salt will be 6 cm deep.

Question 25. ABCD is a flower bed. If OA = 21m and OC = 14m. Find the area of the bed in

m^{2}

180.30
921
190
192.50

Discuss Question

Answer: Option D. -> 192.50
:
D
Area of a sector with angle

θ

is given by =

\frac{θ}{360 °}

\times

π

\times

r^{2}

( where

θ

is the angle made by the sector )
Required area = Area of sector OABO – Area of sector OCDO
The angle formed by the sector OABO and sector OCDO = 90°

= \frac{90}{360} \times \frac{22}{7} \times (21)^{2} - \frac{90}{360} \times \frac{22}{7} \times (14)^{2}

= \frac{1}{4} \times \frac{22}{7} \times {(21)^{2} - (14)^{2}}

= \frac{1}{4} \times \frac{22}{7} \times {(35) (7)}

[a^{2} - b^{2} = (a + b) (a - b)]

=192.50

m^{2}

Question 26. The given figure is a sector of a circle of radius 20 cm. Find the perimeter of the sector.
(Take

π

= 3.14)

The Given Figure Is A Sector Of A Circle Of Radius 20 Cm. Fi...

55.25 cm
60.93 cm
65.48 cm
70.17

Discuss Question

Answer: Option B. -> 60.93 cm
:
B
The circumference i.e , perimeter of a sector of angle

P^{\circ}

of a circle with radius R is given by

\frac{P^{\circ}}{360} \times 2 π r + 2 R

= \frac{P^{\circ}}{360} \times 2 π R + 2 R

= \frac{60^{\circ}}{360} \times 2 π (20) + 2 (20)

= 20.93 + 40
= 60.93 cm

Question 27. State true or false.
Area of a circle is inversely proportional to the radius of circle.

True
False
190
192.50

Discuss Question

Answer: Option B. -> False
:
B
Area of a circle is directly proportionaltothe square of the radius of circle.

Question 28. State true or false.
If the radius of a circle is

\frac{7}{\sqrt{π}}

cm, then the area of the circle in

c m^{2}

\frac{49}{π^{2}}

True
False
190
192.50

Discuss Question

Answer: Option B. -> False
:
B
A =

π r^{2}

Given r = \frac{7}{\sqrt{π}} c m

A = π \times \frac{7}{\sqrt{π}} \times \frac{7}{\sqrt{π}}

= 49 c m^{2}

Question 29. A road roller was used for levelling a road of width 2 m. It was observed that, the road roller required 25 complete revolutions to level the entire road. If the radius and the length of the roller is 7 m and 2 m respectively, then length of the road is ____. [Take

π

= 3.14].

1200 m
2100 m
1100 m
1234 m

Discuss Question

Answer: Option C. -> 1100 m
:
C
Given, Radius of Cylinder,

r = 7 m

length of roller,

h = 2 m

Width of road,

B = 2 m

Length of road be

^{'} L^{'}

Number of revolution taken = 25
Roller rolls on its curved surface area.

∴

Surface areaof = CSA of cylinder
one revolution

= 2 π r h = 2 \times \frac{22}{7} \times 7 \times 2 = 88 m^{2}

Total area covered in 25 revolution

= 25 \times 88 = 2200 m^{2}

This area willbe equal to the area of the road levelled

Area of road = L \times B = 2200 \Rightarrow L \times 2 = 2200 \Rightarrow L = 1100 m

∴

Length of the level road is 1100 m.

Question 30. A cylindrical pipe has inner diameter of 7 cm and water flows through it at the rate of 192.5 litres per minute. Find the rate of flow in kilometres per hour.

Discuss Question

Answer: Option B. -> 3
:
B
In one minute 192.5 litres of water flows.
So, The Volume of water that flows in onehour = (192.50