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 m2. 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 cm2
192 cm2
48 cm2
96 cm2
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 cm2). The volume of the solid in cm3, is ____.
144 cm3
169 cm3
64 cm3
216 cm3
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 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=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 cm2. 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