In this page we have Practice questions for Mensurations Class 8 maths Chapter 11 . Hope you like them and do not forget to like , social share
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Question 1
There are two cuboidal whose dimensions are given below. Which box requires the higher amount of material to make?
Cuboid A: L=23, B=30, H=40
Cuboid B: L=30, B=12, H=44
Solution
Material will be measures by the surface area
Surface Area of Cuboid A= 2(LB + BH+ LH) = 2(23 *30 + 30*40 + 23 *40) =2*2810 cm^{2}
Surface Area of Cuboid B= 2(LB + BH+ LH) = 2(30*12 + 12*44 + 30 *44) =2*2208 cm^{2}
So Cuboid A will require the higher amount of material to make
Question 2
Three cubes, each of edge 2 cm. long are placed together. Find the total surface area of the cuboid so formed?
Solution
L = 2+ 2 + 2= 6 cm
B=2 cm
H=2 cm
Surface Area of Cuboid = 2(LB + BH+ LH) = 2(6 *2 + 2*2 + 6 *2) =56 cm^{2}
Question 3
Find the side of a cube whose surface area is 2400 cm
^{2}.
Solution
Surface of cube= 6a^{2}
So , 6a^{2}=2400
a=20 cm
Question 4
Meghna painted the outside of the cabinet of measure 2 m × 3 m × 2.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet and back side?
Solution
Here L=2 m, B=3 m and H=2.5 m
Total surface area of a cuboid =2LB+2BH+2LH
Now is she left bottom and back side of box area covered is
=2LB+2BH+2LH-LB-LH
= LB+2BH+LH
=(2*3)+2(3*2.5)+(2.5*2)
= 26m^{2}
Question 5
Ahmed is painting the walls and ceiling of a cuboidal hall with length, breadth and height of 25 m, 12 m and 8 m respectively. From each can of paint 200 m² of area is painted. How many cans of paint will she need to paint the room?
Solution
Here L=25 m, B=12 m and H=8 m
Area to be painted= Area of lateral surface + Area of Ceiling
=2(LH) + 2(BH) + LB
=892 m^{2}
Theerefore, no of cans required= 892/200
=4.5 cans
Question 6
A open cylindrical tank of radius 14 m and height 3 m is made from a sheet of metal. How much sheet of metal is required?
Solution
here r=14 m and h=3 m
Sheet of metal require = total surface area of Cylinder = 2 π rh + π r^{2}
=880 m^{2}
Question 7
The lateral surface area of a hollow cylinder is 4224 cm
^{2}. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet?
Solution
Here B= 33 cm
Area of Rectangular Sheet formed = Surface Area of the Cylinder
Area of Rectangle = L × B
Therefore
L × B= 4224
Putting the value of B and solving
L=4224/33=128 cm
Now
Perimeter Of Rectangle=2(L+B) = 2(128+33)
=322cm
Question 8
A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm and length is 1 m.
Solution
Length of roller = 1m = 100cm , Diameter of road roller = 84 cm
Circumference of roller = 2 π r= π D = 22/7 * 84 = 264 cm
Now Length travelled in 1 rev= Circumference of road roller
Therefore
Length of road = Length travelled in 750 = 750 * Circumference of road roller = 198000 cm
Width of road = length of roller = 100 cm
Area of road = Area of rectangle = L * B = 198000 * 100 = 198 * 10^{5} cm ^{2}
Question 9
A rectangular sheet of metal foil is 88 cm. long and 20 cm. wide. A cylinder is made out of it, by rolling the foil along width. Find the volume of the cylinder.
Solution
L = 88 cm
B= 20 cm
As per the question,Sinc ecylinder is made out of it, by rolling the foil along width ,the circumference will be equal to the breadth of the sheet and the length will be the height of the cylinder.
Circumference=2 π r= 20 cm
or r= 10/π
Height of the cylinder = 88 cm
Now volume = π r^2 h = 88 *100/π = 2800 cm^{2}
Question 10
The perimeter of the floor of a hall is 250 m. If the height is 4 m, find the cost of painting the four walls at the rate of Rs. 12 per square meter.
Solution
Surface area to be painted = 2(LH+BH) = 2H(L+B)
given perimeter of floor = 250m = 2(L+B)
and height is 4m
Substituting these values
Surface area=1000 m^{2}
cost of painting 1 square metre=Rs 12
cost of painting 1000 square metre=1000*12= Rs 12000
Question 12
How many times do the volume and surface area of a cube increase if its edges get tripled.
Solution
$V_1=L^3$ and $SA_1= 6L^2$
$V_2= (3L)^3 = 27L^3$ and $SA_2=6(3L)^2 = 9 \times 6L^2$
So, Volume becomes 27 times and Surface Area become 9 times
Question 13
How many times do the volume and surface area of a cylinder increase if its radius doubled and height remains same
Solution
$V= \pi r^2 h$ and $SA= 2 \pi rh$
If radius is doubled and height remains same ,then
$V_2= \pi (2r)^2 h = 4 \pi r^2 h$
$SA_2 = 2 \pi (2r) h =2 \times 2 \pi rh$
So volume becomes four times and Surface area becomes doubled
Question 14
How many times do the volume and surface area of a cylinder increase if its radius remains same and height is doubled
Solution
$V= \pi r^2 h$ and $SA= 2 \pi rh$
if its radius remains same and height is doubled ,then
$V_2= \pi r^2 (2h) = 2 \pi r^2 h$
$SA_2 = 2 \pi r (2h) =2 \times 2 \pi rh$
So volume becomes doubled and Surface area becomes doubled
Question 15
The height of a cylinder is 15 cm. and curved surface area is 660 cm
^{2}. Find the radius of the cylinder
Solution
$2 \pi rh= 660$
$ 2\times (22/7) \times r \times 30 = 660$
r=3.5 cm
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