# Free Mensuration 03 Practice Test - 8th Grade

### Question 1

The sides of a rectangle are in the ratio of 6 : 5 and its area is 750 sq.m. Find the perimeter of rectangle (in m).

120

110

122

None of these

#### SOLUTION

Solution :B

Let 6x and 5x be the sides of a rectangle.

⇒6x×5x=750

⇒30x2=750

⇒x2=75030=25

⇒x=5

Length = 6x = 30m

Breadth = 5x = 25m

Perimeter = 2(l + b) = 2(30 + 25) = 110m

### Question 2

A mat in the shape of a parallelogram has a height of 4 cm and a base of 3 cm. How much would it cost to cover a parallelogram shaped hall with an area of 180 sq. cm with mats, if each mat costs Rs.7?

Rs. 105

Rs. 115

Rs. 135

Rs. 95

#### SOLUTION

Solution :A

Let n be the number of mats required

base of mat = 3 cm

height of mat = 4 cmArea of the parallelogram = base × height

Area of hall = Area of mat × number of mats

⇒ 180 = 4 x 3 x n⇒n=18012=15

Given that each mat costs Rs. 7

Then the cost of 15 mats costs = 7 x 15 = Rs.105

### Question 3

The perimeter of the given figure is

#### SOLUTION

Solution :

A Perimeter is the boundary of any closed geometrical figure.

The perimeter of the given figure is the sum of arc length BC and the slant lengths AB and AC.

Arc is semicircle so its perimeter is πr

Perimeter=πr+2×Slant height

=227×1.4+2×2

⇒4.4+4=8.4cm

### Question 4

PQRS is a quadrilateral in which PQ=5√2cm, QR=5√2cm, RS=8cm, SP=6cm, ∠PSR=∠PQR=90∘. Then its area is

40.4 cm2

46.4 cm2

32.4 cm2

49 cm2

#### SOLUTION

Solution :D

Area of right angled triangle ΔPQR

=12×Base×Height=12×5√2×5√2=25 cm2Area of right angled triangle ΔPSR

=12×8×6=24 cm2Area of quadrilateral PQRS

= Area of ΔPQR+ΔPSR=49 cm2

### Question 5

The capacity of a soda can be calculated using the formula.

2πr(r+h)

2πr

πr2h

All of these

#### SOLUTION

Solution :C

A soda can is a rough example of a cylinder.

The volume of a cylinder is πr2h.

The capacity of any vessel is its internal volume so, the capacity of a soda can be calculated using the formula πr2h.Where,

r = base radius of the canh = height of the can

### Question 6

The volume of a cube whose edge is 24 m is 1296 m^{3} .

True

False

#### SOLUTION

Solution :B

Volume of cube = (edge)3

Hence the volume of cube with edge 24 m is (24)3

= 13824 m3.

### Question 7

A cuboid whose length, breadth and height are equal is called a ___________.

Hexagon

Pentagon

Cube

Cuboid

#### SOLUTION

Solution :C

A cuboid whose length, breadth and height are equal is called a Cube. Cube is a special case of a cuboid in which all the sides are equal.

### Question 8

If the height of a cylinder is halved and its diameter is doubled then, the new volume is

is equal to

is double of

is three times of

is four times of

#### SOLUTION

Solution :B

Let the radius of a cylinder be 'r' and the height of a cylinder be 'h'.

Volume of cylinder=V=πr2hNew radius=2rNew height=h2New volume=π(2r)2h2=2πr2hNew volume=2V

### Question 9

An isosceles trapezium has an area of 36 cm^{2}, the parallel sides are 12 cm and 6 cm respectively. The perimeter of the trapezium should be

28 cm

32 cm

24 cm

20 cm

#### SOLUTION

Solution :A

The area of a trapezium is given by:

A=12×(Sum of parallel sides)×(height)

⇒36=12×(12+6)×h

⇒h=4 cm

So now in the trapezium,

h=4 cm

a=b=6 cm

AB=CD[S-A-S congruency in△AFB and △DEC]

∴2AB+a=12 cm

⇒2AB=12−6

⇒AB=CD=3 cm

So now in△ABF,AB2+h2=d2

d2=32+42=25

d=5 cm

As the trapezium is isosceles, the slant sides of the trapezium are equal in length

d=c=5 cm

∴ the perimeter of the trapezium

=(2×AB)+a+c+b+d

=(2×3)+6+5+6+5=6+22=28 cm

### Question 10

Every quadrilateral has two pairs of parallel sides.

#### SOLUTION

Solution :B

A quadrilateral is a four- sided rectilinear figure and parallelogram is a four-sided plane rectilinear figure with opposite sides parallel. Square, rectangle are considered as parallelogram because both of them have opposite sides parallel. In trapezium, only one side is parallel hence, it is not considered as a parallelogram. Hence, every quadrilateral doesn't have two pairs of parallel sides.