Mensuration Test

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Mensuration Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

Mark the correct alternative of the following.
The height of sand in a cylindrical-shaped can drop by $$3$$ inches when $$1$$ cubic foot of sand is pound out. What is the diameter of the cylinder (in inches)?

A
$$\dfrac{24}{\sqrt{\pi}}$$

B
$$\dfrac{48}{\sqrt{\pi}}$$

C
$$\dfrac{32}{\sqrt{\pi}}$$

D
$$\dfrac{48}{\pi}$$

Solution

Let $$r$$ be the radius of the cylinder.
It is given that, the height drops $$3$$ inches, when $$1$$ cubic foot of sand is poured out and $$1$$ foot $$=12\ \text{in}$$
So,
Volume of reduced cone = Volume of 1 cubic foot sand
$$\pi r^2 h=12\times 12\times 12\times 12$$
$$\pi r^2 \times 3=12\times 12\times 12$$
$$r^2=\dfrac{12\times 12\times 4}{\pi}$$

$$r=\dfrac{12\times 2}{\sqrt{\pi}}$$

$$\therefore$$ $$r=\dfrac{24}{\sqrt{\pi}}$$

The diameter of cylinder $$=2r=\dfrac{48}{\sqrt{\pi}}$$
Question 2
1 / -0

The surface area of a $$10\ cm \times 4\ cm\times 6\ cm$$ brick is

A
$$84\ cm^2$$

B
$$124\ cm^2$$

C
$$164\ cm^2$$

D
$$180\ cm^2$$

Solution

We know that the total surface area of cuboid $$=2(lb+bh+hl)$$

So, the surface area of the brick $$=2(10\times 6+4\times 6+ 4\times 10)$$

$$=2(40+30+12)$$

$$=164$$
Question 3
1 / -0

The area of the cardboard needed to make a box of size $$25\ cm\times 15\ cm\times 8\ cm$$ will be

A
$$390\ cm^2$$

B
$$1390\ cm^2$$

C
$$2780\ cm^2$$

D
$$1000\ cm^2$$

Solution

We know that total surface area of cuboid $$=2(lb+bh+hl)$$
$$=2(25\times15+15\times8+8\times25)$$
$$=2(375+120+200)$$
$$=1390\ cm^2$$
Question 4
1 / -0

The parallel sides of a trapezium measure $$14cm$$ and $$18cm$$ and the distance between them is $$9\ cm$$. The area of the trapezium is

A
$$96\ cm^2$$

B
$$144\ cm^2$$

C
$$189\ cm^2$$

D
$$207\ cm^2$$

Solution

Since, length of the parallel sides are $$14cm$$ and $$18cm$$ and height $$(h)=9cm$$
By using the formula
Area of trapezium $$=1/2 \times (sum\ of\ parallel\ sides )\times height$$
$$=1/2 \times (14+18) \times 9$$
$$=144cm^2$$
Question 5
1 / -0

Mark correct against the correct answer in the following:
A cuboid has

A
length only

B
length and breadth only

C
length, breadth and height

D
thickness only

Solution

A cuboid has length, breadth and height since, the cuboid is a three dimensional figure.
Option (c) is the correct answer
Question 6
1 / -0

A covered wooden box has the inner measures as 115 cm, 75 cm and 35 cm and thickness of wood as 2.5 cm. The volume of the wood is

A
85,000 $$ cm^{3} $$

B
80,000 $$ cm^{3} $$

C
82,125 $$ cm^{3} $$

D
84,000 $$ cm^{3} $$

Solution

(c) 82,125 $$ cm^{3} $$
Given, inner measure of a wooden box as 115 cm, 75 cm and 35 cm
Since, thickness of the box is 2.5 cm, then outer measures will be 115+5.75 and 35+5 i.e, 120 cm, 80 cm and 40 cm
Volume of cube = $$ l\times b\times h $$
Therefore, the outer volume = $$ 120 l\times 80\times 40 $$ = 38400 $$ cm^{3} $$
And the inner volume = $$ 115\times 75 \times 35 $$ = 301875 $$ cm $$ ^{3} $$
Therefore, volume of the wood = outer volume - inner volume = 34800-301875=82125 $$ cm ^{3} $$
Question 7
1 / -0

If R is the radius of the base of the hat, then the total outer surface area of the hat is

A
$$ \pi r(2h+R) $$

B
$$ 2\pi r(h+R) $$

C
$$ 2\pi rh+\pi R^2 $$

D
None of these

Solution

(c) $$ 2\pi rh+\pi R2 $$
A cylinderical hat with base radius R and r is radius of the top surface
Now , total surface area of the hat= curved surface area + top surface area+ base surface area
= $$ 2\pi rh+\pi r^{2}+\pi (R^{2}-r^{2}) $$
= $$ 2\pi rh+\pi r^{2}+\pi (R^{2}-\pi r^{2})=2\pi rh+\pi R^{2} $$
Question 8
1 / -0

The surface area of the three coterminous faces of a cuboid are $$6 cm^2$$,$$15 cm^2$$ and $$10 cm^2$$ respectively. The volume of the cuboid is

A
30 $$ cm^{3} $$

B
40 $$ cm^{3} $$

C
20$$ cm^{3} $$

D
35 $$ cm^{3} $$

Solution

(a) 30 $$ cm^{3} $$
If l,b and h are the dimensions of the cuboid, Then,
Volume of the cuboid = $$ l\times b\times h $$
Here, 6 = $$ l\times b $$
15 = $$ l\times h $$
10 = $$ b\times h $$
Therefore, $$ 6\times 15\times 10 $$ = $$ l^{2}\times b^{2}\times h^{2} $$
Therefore, volume = $$ l\times b\times h $$ = $$\sqrt{ 6\times 15\times 10 } $$ = $$ 30 cm^{3} $$
Question 9
1 / -0

The perimeter of a trapezium is 52 cm and its each non-parallel side is equal to 10 cm with its height 8 cm. Its area is

A
124 cm$$ ^{2} $$

B
118 cm$$ ^{2} $$

C
128 cm$$ ^{2} $$

D
112 cm$$ ^{2} $$

Solution

(c) 128 cm$$ ^{2} $$
Given, perimeter of a trapezium is 52 cm and each non-parallel side is of 10 cm.
Then, sum of its parallel sides= 52-(10+10)
=52-20 = 32 cm
Therefore, area of the trapezium = $$ \dfrac{1}{2}(a+b)\times h $$
= $$ \dfrac{1}{2}(a+b)\times 32\times 8 $$ = $$ 128 cm^{2} $$
Question 10
1 / -0

A hollow cylinder has only

A
curved surface area

B
total surface area

C
volume

D
Perimeter

Solution

A hollow cylinder has only curved surface area

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Mensuration Test - 35

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Mensuration Test - 35

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Question 1

$$\dfrac{24}{\sqrt{\pi}}$$

$$\dfrac{48}{\sqrt{\pi}}$$

$$\dfrac{32}{\sqrt{\pi}}$$

$$\dfrac{48}{\pi}$$

Solution

Question 2

$$84\ cm^2$$

$$124\ cm^2$$

$$164\ cm^2$$

$$180\ cm^2$$

Solution

Question 3

$$390\ cm^2$$

$$1390\ cm^2$$

$$2780\ cm^2$$

$$1000\ cm^2$$

Solution

Question 4

$$96\ cm^2$$

$$144\ cm^2$$

$$189\ cm^2$$

$$207\ cm^2$$

Solution

Question 5

length only

length and breadth only

length, breadth and height

thickness only

Solution

Question 6

85,000 $$ cm^{3} $$

80,000 $$ cm^{3} $$

82,125 $$ cm^{3} $$

84,000 $$ cm^{3} $$

Solution

Question 7

$$ \pi r(2h+R) $$

$$ 2\pi r(h+R) $$

$$ 2\pi rh+\pi R^2 $$

None of these

Solution

Question 8

30 $$ cm^{3} $$

40 $$ cm^{3} $$

20$$ cm^{3} $$

35 $$ cm^{3} $$

Solution

Question 9

124 cm$$ ^{2} $$

118 cm$$ ^{2} $$

128 cm$$ ^{2} $$

112 cm$$ ^{2} $$

Solution

Question 10

curved surface area

total surface area

volume

Perimeter

Solution

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