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

10th Grade > Mathematics > Area

AREAS CLUBBED MCQs

Total Questions : 49 | Page 1 of 5 pages

Question 1. Ram has a semicircular disc. He rotates it about its diameter by 360 degrees. When he rotates the disc, a volume of air in his room gets swept.The name of the object/shape that exactly occupies this volume is
___.

Discuss Question

The shape obtained when a semi circular disc is rotated by 360 degrees is a sphere. If it was rotated by 180 degrees instead of 360 degrees, then we obtaina hemisphere. The line segment AB willbe diameter of the sphere formed.

Question 2. There is a coin. Ram attaches a conical attachment to one flat end of coin. The conical attachment has same radius as coin. What is the surface area of the combined solid?

Coin base area + Coin C.S.A + Hemisphere C.S.A
Coin base area + Coin C.S.A + cone C.S.A
Total surface area of coin + total surface area of cone
Total surface area of cone

Discuss Question

Answer: Option B. -> Coin base area + Coin C.S.A + cone C.S.A
:
B
This leaves only the following surfaces for the combined solid-
1. One flat surface of coin
2. Curved surface of coin
3. Curved surface area of cone
These 3 areas add up together to give total surface area of combined solid.

Question 3. In the figure, two small circles touch each other externally at the centre of a bigger circle and these small circles are touched internally by a bigger circle with radius 4

c m

. Find the area of the shaded region in

c m^{2}

In The Figure, Two Small Circles Touch Each Other Externally...

Discuss Question

Answer: Option A. -> 27
:
A
From the figure, two circles of radius 4 cm exist in the rectangle.
Their total area = 2

π r^{2}

= 2

\times

3.14x16 = 100.48

c m^{2}

From figure
Length of rectangle is 4r = 16 cm
Breadth of rectangle is 2r = 8cm
Area of rectangle = 16 x8 = 128

c m^{2}

Required area = [Area of rectangle - Total area of circles]
=128 - 100.48= 27.52

c m^{2}

∴

The area of the shaded region
=27.52

c m^{2}

Question 4. Metallic spheres of radii 6cm, 8cm and 10cm, respectively are melted to form a single solid sphere. The radius of the resulting sphere(in cm) is
___

Discuss Question

:
Radius of first Sphere =

r_{1} = 6 c m

Radius of second Sphere =

r_{2} = 8 c m

Radius of third Sphere =

r_{3} = 10 c m

Let radius of resulting sphere = r cm
According to given condition, we have
Volume of 1stsphere + Volume of 2nd sphere + Volume of 3rd sphere = Volume of resulting sphere

\Rightarrow \frac{4}{3} . π . (r_{1})^{3} + \frac{4}{3} . π . (r_{2})^{3} + \frac{4}{3} . π . (r_{3})^{3} = \frac{4}{3} . π . r^{3}

\Rightarrow \frac{4}{3} ((r_{1})^{3} + (r_{2})^{3} + (r_{3})^{3}) = \frac{4}{3} . r^{3}

\Rightarrow

(6^{3} + 8^{3} + (10)^{3}) = r^{3}

\Rightarrow

( 216 + 512 + 1000 ) =

r^{3}

\Rightarrow

r^{3}

= 1728

\Rightarrow

r = 12 cm

Question 5. The height of a cone is 40 cm. A small cone is cut off at the top by a plane parallel to the base. If the volume of the small cone be

\frac{1}{64}

of the volume of the given cone, at what height ( in cm) above the base is the section made ?

Discuss Question

Answer: Option B. -> 30
:
B

The Height Of A Cone Is 40 Cm. A Small Cone Is Cut Off At Th...

Let

R

be the radius of the given cone,

r

the radius of the small cone,

h

be the height of the frustum and

h_{1}

be the height of the small cone.
In the figure,

△ O N C \sim △ O M A

∴ \frac{O N}{O M} = \frac{N C}{M A}

[Sides of similar triangles are proportional]

\Rightarrow \frac{h_{1}}{40} = \frac{r}{R}

\Rightarrow h_{1} = (\frac{r}{R}) \times 40

.......(i)
We are given that

\frac{V o l u m e o f s m a l l c o n e}{v o l u m e o f g i v e n c o n e} = \frac{1}{64}

\Rightarrow \frac{\frac{1}{3} π r^{2} \times h_{1}}{\frac{1}{3} π R^{2} \times 40} = \frac{1}{64}

\Rightarrow \frac{r^{2}}{R^{2}} \times \frac{1}{40} \times [(\frac{r}{R}) 40] = \frac{1}{64}

[ By (i)]

\Rightarrow {(\frac{r}{R})}^{3} = \frac{1}{64} = {(\frac{1}{4})}^{3}

\Rightarrow

\frac{r}{R} = \frac{1}{4}

...... (ii)
From (i) and (ii)

h_{1} = \frac{1}{4} \times 40 = 10 c m

∴, h = 40 - h_{1} = (40 - 10) c m

\Rightarrow h = 30 c m

Hence, the section is made at a height of 30 cm above the base of the cone .

Question 6. Tick the correct answer in the following:
Area of a sector of angle p (in degrees) of a circle with radius R is

p180×2πR
p180×πR2
p360×2πR
p720×2πR2

Discuss Question

Answer: Option D. -> p720×2πR2
:
D
Area of a sector of angle

p = \frac{p}{360} \times π R^{2} = \frac{p}{360} \times \frac{2}{2} \times π R^{2} = \frac{p}{720} \times 2 π R^{2}

Question 7. An umbrella has 8 ribs which are equally spaced. Assuming the umbrella to be a flat circle of radius 42 cm, find the area in

c m^{2}

between the two consecutive ribs of the umbrella.

An Umbrella Has 8 Ribs Which Are Equally Spaced. Assuming Th...

___

Discuss Question

:
Radius= 42cm
8 ribs implies angle subtend between consecutive ribs =

\frac{360}{8}

45^{\circ}

.
Area between consecutive ribs =

\frac{θ}{360} \times π \times (r a d i u s)^{2}

\frac{45}{360}

\frac{22}{7}

42^{2}

= 693

c m^{2}

Question 8. A bucket is in the form of a frustum of a cone, its depth is 15 cm and the diameters of the top and the bottom are 56 cm and 42 cm respectively. How many litres of water can the bucket hold ? (Take

π

\frac{22}{7}

)
__

Discuss Question

A Bucket Is In The Form Of A Frustum Of A Cone, Its Depth Is...

R = 28 cm
r = 21 cm
h = 15 cm
Capacity of the bucket =

\frac{1}{3} π h (R^{2} + r^{2} + R r)

\frac{1}{3} \times \frac{22}{7} \times 15 \times [(28)^{2} + (21)^{2} + (28) (21)] c m^{3}

\frac{22}{7} \times 5 \times [784 + 441 + 588] c m^{3}

\frac{22}{7} \times 5 \times 1813 c m^{3}

= 22× 5× 259

c m^{3}

= 28490

c m^{3}

\frac{28490}{1000}

liters
= 28.49 liters

Question 9. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. The number of lead shots dropped in the vessel is ______.

Discuss Question

Answer: Option A. -> 100
:
A
Let number of lead shots = n
Height of cone

= h = 8 c m

Radius of cone

= r_{1} = 5 c m

Volume of water present in the cone = Volume of cone

= \frac{1}{3} . π . (r_{1})^{2} . h

= (\frac{1}{3} \times \frac{22}{7} \times 5^{2} \times 8 c m^{2})

Volume of water flown out

= (\frac{1}{4} \times \frac{1}{3} \times \frac{22}{7} \times 5^{2} \times 8) c m^{3}

Radius of lead shot

= r = 0.5 c m

Volume ofeach lead shot ...

= \frac{4}{3} . π . r^{3} = (\frac{4}{3} \times \frac{22}{7} \times (0.5)^{3}) c m^{3}

According to given situation, n lead shots are thrown into cone such that 1/4thof water presentin cone flows out.
It means volume of n lead shots = volume of water flown out

\Rightarrow

(n \times \frac{4}{3} \times \frac{22}{7} \times (0.5)^{3}) = (\frac{1}{4} \times \frac{1}{3} \times \frac{22}{7} \times 5^{2} \times 8)

\Rightarrow

n = \frac{1}{4} \times \frac{1}{3} \times 5^{2} \times 8 \times \frac{1}{({0.5}^{3})} \times \frac{3}{4}

\Rightarrow

n = 100

Question 10. A rectangular paper is folded into a cylinder. The length and breadth of the paper are L and B respectively. The surface area of the cylinder is
___

Discuss Question

:
The cylinder is a 3-D object. Here the 3-D surface of the cylinder can be unrolled into a 2-D rectangular sheet of paper. The area of the 3-D surface and that of the 2-D plane are the same. It is clear that the area of rectangle = Length