Mensuration(8th Grade > Mathematics ) Questions and answers for exam Preparation

8th Grade > Mathematics

MENSURATION MCQs

Total Questions : 57 | Page 3 of 6 pages

Question 21. What will be the ratio of the areas of squares, when the diagonal of one square is twice of the other.

Discuss Question

Answer: Option D. -> 4:1
:
D
Let the length of one diagonal be

x

units.
The length of the other diagonal is

2 x

units
Applying Pythagorastheorem,

2 \times {(s i d e o f s q u a r e)}^{2} = {(l e n g t h o f d i a g o n a l)}^{2}

\Rightarrow {(s i d e o f s q u a r e)}^{2}

\frac{1}{2} \times {(l e n g t h o f d i a g o n a l)}^{2}

Sides of the squares will be

\sqrt{\frac{{4 x}^{2}}{2}} = \frac{2 x}{\sqrt{2}}

and

\sqrt{\frac{x^{2}}{2}} = \frac{x}{\sqrt{2}}

Area of square =

{(s i d e o f s q u a r e)}^{2}

Hence the ratio of the area of squares is

(\frac{2 x}{\sqrt{2}})^{2}

and

(\frac{x}{\sqrt{2}})^{2}

\frac{1}{2} \times (2 x)^{2} : \frac{1}{2} \times x^{2}

= 4: 1

Question 22. If the height of a cylinder becomes one-eighth of the original height and the radius is doubled, then which of the following will be true?

Volume of the cylinder will be doubled.
Volume of the cylinder will remain unchanged.
Volume of the cylinder will be halved.
Volume of the cylinder will be thrice that of the original cylinder

Discuss Question

Answer: Option C. -> Volume of the cylinder will be halved.
:
C
Let us assume the height and radius of the original cylinder is l and r respectively.
Let V₁ be the original volume and V₂ be the volume after the change in height and radius.
Therefore volume of the original cylinder is given by:

V = π r^{2} l

After, height of a cylinder becomes

\frac{1}{8}

of the original height and the radius is doubled
We have,

r_{2} = 2 r a n d l_{2} = \frac{l}{8}

So the new volume is given by:

V_{2} = π r_{2}^{2} l_{2} = π (2 r)^{2} (\frac{l}{8})

\Rightarrow V_{2} = \frac{4 π r^{2} l}{8}

\Rightarrow V_{2} = \frac{π r^{2} l}{2} = \frac{V_{1}}{2}

\Rightarrow V_{2} = \frac{V_{1}}{2}

Hence the volume of the cylinder will be halved.

Question 23. What is the area of the rhombus ABCD below if AC=12 cm and BE= 8 cm?

384 cm2
192 cm2
48 cm2
96 cm2

Discuss Question

Answer: Option D. -> 96 cm2
:
D
The diagonals of a rhombus bisect each other. Therefore, in the givenfigure:

\Rightarrow B E = D E

\Rightarrow B D = 2 \times B E

\Rightarrow B D = 2 \times 8 = 16 c m

Area of a rhombus equalshalf the product of lengthof the two diagonals.

\Rightarrow

Area of ABCD=

\frac{1}{2} \times A C \times B D

⟹

Area of ABCD=

\frac{1}{2} \times 12 \times 16 = 96 c m^{2}

Question 24. The area of the floor of a rectangular hall of length 60 m is 1200 m². The dimensions of the carpet available is size 8 m

\times

6 m .Hence 25 such carpets are required to cover the hall.

True
False
32 m2
None of the above

Discuss Question

Answer: Option A. -> True
:
A
First, we need to calculate the area of one carpet:
Area of one carpet = 8

\times

6 = 48

m^{2}

.
Sincetotal area of the floor of the hall = 1200

m^{2}

.
Therefore, number of carpets required =

\frac{a r e a o f f l o o r}{a r e a o f o n e c a r p e t}

\frac{1200}{48}

= 25.

Question 25. Abhishek has three containers:
a)Cylindrical container A having radius r and height 2r
b)Cubical container B having its edge

\frac{3}{4} r

c)Cuboidal container C having dimensions

r \times \frac{5}{4} r \times \frac{7}{3} r

The arrangement of the containers in the increasing order of their volumes is

A, B, C
B, C, A
C, A, B
cannot be arranged

Discuss Question

Answer: Option B. -> B, C, A
:
B
Volume of Cylinder is given by:

V_{cylinder} = π \times {radius}^{2} \times height

Volume of Cubeis given by:

V_{cube} = {edge}^{3}

Volume of Cube is given by

V_{cuboid} = length \times breadth \times height

Therefore,

\Rightarrow V_{cylinder} = \frac{22}{7} \times r^{2} (2 r) = \frac{44}{7} r^{3}

\Rightarrow V_{cube} = (\frac{3}{4} r)^{3} = \frac{27}{64} r^{3}

\Rightarrow V_{cuboid} = r \times \frac{5}{4} r \times \frac{7}{3} r = \frac{35}{12} r^{3}

Comparing the above volumes we get that

V_{cube}

V_{cuboid}

V_{cylinder}

Question 26. How many small cuboids with dimensions 20 cm

\times

25 cm

\times

40 cm each can be accommodated in a cubical box of edge 2 m ?

4
4000
400
40

Discuss Question

Answer: Option C. -> 400
:
C
Volume of one small cuboid= l

\times

\times

h = 20cm

\times

25cm

\times

40cm
Since the edge length of the cubical box = 2 m = 200 cm
Now, Volume of the cubical box =

200 c m \times 200 c m \times 200 c m

So the number of cuboids that can be justaccommodated in the box =

\frac{V o l u m e o f t h e c u b i c a l b o x}{v o l u m e o f t h e c u b o i d}

\Rightarrow

Number of cuboids =

\frac{200 \times 200 \times 200}{20 \times 25 \times 40} = 400

Question 27. Ajay's father asks him to find the area covered by the grass in his garden. Ajay is informed that the shape of the area covered by the grass is square which has a diagonal length of

16 \sqrt{2}

meters. What is the area calculated by Ajay?

256 m2
16 m2
32 m2
None of the above

Discuss Question

Answer: Option A. -> 256 m2
:
A
Since area of a square has to be calculated, whose diagonal length is

16 \sqrt{2}

meters. The relation between diagonal and side of a square is :

a \sqrt{2} = d

,where, a = side of the square and d = length of diagonal (from Pythagorean theorem)

\Rightarrow a \sqrt{2} = 16 \sqrt{2}

\Rightarrow

a=16m.
Therefore, area is given by:
A =

a^{2}

\Rightarrow A = 16^{2} = 256 m^{2}

Question 28.

A flooring tile has the shape of a rectangle whose dimensions are 10cm × 6cm. How many such tiles are required to cover a floor of area 30 × 50 cm²? (If required you can split the tiles in whatever way you want to fill up the corners).

20 tiles
25 tiles
35 tiles
50 tiles

Discuss Question

Answer: Option B. -> 25 tiles
:
B

Number of tiles required
= $\frac{a r e a o f t h e f l o o r}{a r e a o f o n e t i l e}$
= $\frac{(30 \times 50)}{(10 \times 6)}$
= 25 tiles

Question 29.

If a square of side 8m and a rectangle of length 12m have the same perimeter, then find the area of the rectangle(in m²)?

36 m²
72 m²
96 m²
48 m²

Discuss Question

Answer: Option D. -> 48 m²
:
D

Let the side of a square be 'a' and length and breadth of a rectangle be 'l' and 'b' respectively.
Given,
Perimeter of square = Perimeter of rectangle

4a = 2(l +b)

4 $\times$ 8 = 2(12 + b)

32 = 24 + 2b

8 = 2b

Hence b = 4 m

Area of rectangle = l $\times$ b = 12 $\times$ 4 = 48 $m^{2}$

Question 30.

The perimeter of above shown figure is 9.5 cm. State whether the given answer is true or false.

True
False
96 m²
48 m²

Discuss Question

Answer: Option A. -> True
:
A

The perimeter of the figure given above is the sum of arc length BC + length AB + length AC.
Perimeter of the given shape = πr + 2 + 2 = [ $\frac{22}{7}$ × $\frac{3.5}{2}$ ] + 4