# Free Mensuration 02 Practice Test - 8th Grade

### Question 1

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

1:3

3:1

5:1

4:1

#### SOLUTION

Solution :D

Let the length of one diagonal be x units.

The length of the other diagonal is 2x units

Applying Pythagoras theorem,

2×(side of square)2=(length of diagonal)2

⇒(side of square)2

= 12×(length of diagonal)2

Sides of the squares will be √4x22=2x√2

and √x22=x√2

Area of square = (side of square)2

Hence the ratio of the area of squares is (2x√2)2 and (x√2)2

= 12×(2x)2:12×x2

= 4: 1

### Question 2

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√2 meters. What is the area calculated by Ajay?

256 m2

16 m2

32 m2

None of the above

#### SOLUTION

Solution :A

Since area of a square has to be calculated, whose diagonal length is 16√2 meters. The relation between diagonal and side of a square is :

a√2=d, where, a = side of the square and d = length of diagonal (from Pythagorean theorem)

⇒a√2=16√2

⇒ a=16m.

Therefore, area is given by:

A = a2

⇒A=162=256m2.

### Question 3

The area of the floor of a rectangular hall of length 60 m is 1200 m^{2}. The dimensions of the carpet available is size 8 m × 6 m .Hence 25 such carpets are required to cover the hall.

True

False

#### SOLUTION

Solution :A

First, we need to calculate the area of one carpet:

Area of one carpet = 8 × 6 = 48 m2.

Since total area of the floor of the hall = 1200 m2.

Therefore, number of carpets required = area of floorarea of one carpet = 120048 = 25.

### Question 4

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

384 cm^{2}

192 cm^{2}

48 cm^{2}

96 cm^{2}

#### SOLUTION

Solution :D

The diagonals of a rhombus bisect each other. Therefore, in the given figure:

⇒BE=DE

⇒BD=2×BE

⇒BD=2×8=16 cm

Area of a rhombus equals half the product of length of the two diagonals.

⇒Area of ABCD = 12×AC×BD

⟹Area of ABCD= 12×12×16=96 cm2.

### Question 5

The area of the hexagon, if AB = BC = CD = DE = EF = FA = 13cm and AO = PD = 5cm, is

#### SOLUTION

Solution :In the given figure, ∆AOB is right-angled triangle right angled at O.

So,⇒ AO2 + OB2 = AB2.

⇒ 52 + OB2 = 132.

⇒ OB2 = 169 - 25 = 144

⇒ OB = 12cm

So now the total area of the hexagon = 4× Area of ∆AOB+ 2× Area of rectangle BCPO

⇒ Area = 4 × (12 × OB × AO) + 2 × (BC × OB)

⇒ Area = 4 × (12 × 12 × 5) + 2 × (13 × 12)

⇒ Area = 4 × (30) + 2 × (156)

⇒ Area = 432cm2.

### Question 6

Abhishek has three containers:

a)Cylindrical container A having radius r and height 2r

b)Cubical container B having its edge 34r

c)Cuboidal container C having dimensions r×54r×73r

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

#### SOLUTION

Solution :B

Volume of Cylinder is given by: Vcylinder=π×radius2×height

Volume of Cube is given by: Vcube=edge3

Volume of Cube is given by Vcuboid=length×breadth×height

Therefore,

⇒Vcylinder=227×r2(2r)=447r3

⇒Vcube=(34r)3=2764r3

⇒Vcuboid=r×54r×73r=3512r3

Comparing the above volumes we get that Vcube < Vcuboid < Vcylinder

### Question 7

The surface areas of six faces of a cuboid are 12, 12, 36, 36, 48, 48, (all in cm^{2}). The volume of the solid in cm^{3}, is ____.

144 cm^{3}

169 cm^{3}

64 cm^{3}

216 cm^{3}

#### SOLUTION

Solution :A

Let the dimension of a cuboid be l, b, and h.

Since the six surface areas are given:

⇒ l × b = 12.......................................(1)

⇒ b × h = 36.......................................(2)

⇒ l × h = 48.......................................(3)

Now multiplying equation (1),(2) and (3), we get

⇒(l×b)×(b×h)×)(l×h)=12×36×48

⇒ (l×b×h)2=20736

⇒(l×b×h)=√20736 =144 cm3

Since volume of a cuboid is calculated as ′l×b×h′, the required volume is 144 cm3.

### Question 8

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

#### SOLUTION

Solution :C

Let us assume the height and radius of the original cylinder is l and r respectively.

Let V

_{1}be the original volume and V_{2}be the volume after the change in height and radius.Therefore volume of the original cylinder is given by: V=πr2l

After, height of a cylinder becomes 18of the original height and the radius is doubled

We have, r2=2r and l2=l8

So the new volume is given by:

V2=πr22l2=π(2r)2(l8)

⇒V2=4πr2l8

⇒V2=πr2l2=V12

⇒V2=V12

Hence the volume of the cylinder will be halved.

### Question 9

How many small cuboids with dimensions 20 cm × 25 cm × 40 cm each can be accommodated in a cubical box of edge 2 m ?

4

4000

400

40

#### SOLUTION

Solution :C

Volume of one small cuboid = l × b × h = 20cm × 25cm × 40cm

Since the edge length of the cubical box = 2 m = 200 cm

Now, Volume of the cubical box = 200cm×200cm×200cm

So the number of cuboids that can be just accommodated in the box = Volume of the cubical boxvolume of the cuboid.

⇒ Number of cuboids = 200×200×20020×25×40=400

### Question 10

A trapezium has an area of 570 cm^{2}. The height of the trapezium is 5 cm and the larger parallel side is twice in length as the smaller parallel side.Choose the correct option.

Length of the larger side is 152 cm

Length of the smaller side is 76 cm

#### SOLUTION

Solution :A and B

Let h be the height of a trapezium, 'a' be the length of a larger parallel side and 'b' be the length of a smaller parallel side.

Area of trapezium is given by:

A=12×(a+b)×h

Since larger parallel side is twice in length as the smaller parallel side

⇒a=2b⇒A=12×(2b+b)×h

⇒A=3b×h2

⇒570=3b×52

⇒b=76cm

⇒a=2b=2×76=152cm