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The surface area of an object is defined as the area covered by the object when placed in space. The volume is defined as the capacity of an object to hold any material. This article provides some surface area and volume examples to help you understand the concept.
Solved Surface Area And Volume Examples
Example 1: Suraj wants to buy wrapping paper to cover a cardboard box to pack a present. If the cardboard box has the length, breadth, and height of 20 cm, 30 cm, and 15 cm, how much square wrapping paper of 20 cm should Suraj get?
Solution: We are given the dimensions of the cardboard box and the wrapping paper,
From the question,
length, l = 20 cm,
breadth, b = 30 cm,
height, h = 15 cm,
The total surface area of the cardboard box,
A = 2(l.b + b.h + h.l)
A = 2(20.30 + 30.15 + 15.20)
A = 2(600 + 450 + 300)
A = 2(1350)
A = 2700 cm2
The area of the wrapping paper,
A = 20 x 20
A = 400 cm2
The number of wrapping paper required is given by,
= (surface area of the box)(area of the wrapping paper)
= 2700400
= 6.75 ≈ 7 wrapping papers.
Example 2: Find the curved surface area of a cylindrical roller, with a diameter of 40 cms and height of 30 cms.
Solution: The curved surface area of a cylinder is given by the following formula,
A = 2.π.r.h
In the question, we are given the diameter of the cylinder,
Thus, the radius,
r = 402
r = 20 cm.
The curved surface area of the cylindrical roller,
A = 2πrh,
A = 2 x π x 20 x 30,
A = 1200 x π,
Taking the value of π as 3.14,
A = 1200 x 3.14,
A = 3768 cm2.
Example 3: Find the curved surface area of a cone with a slant height (l) 12 cm and base diameter (d) of 50 cm.
Solution: In the given question,
Slant height of cone (l) = 12 cm,
Diameter of cone (d) = 50 cm,
Therefore, the base radius of the cone (r) = 502
r = 25 cm
The curved surface area of a cone is given by,
A = π.r.l
Implying the formula with the given values,
Curved surface area = π x 25 x 12,
= π x 300,
Taking the value of π as 3.14,
= 3.14 x 300
= 942 cm2
The curved surface area of the given cone is 942 cm2.
Example 4: Find the area of a hollow sphere with a radius of 7 cm.
Solution: The surface area of a sphere is given by,
A = 4.π.r2.
In the question, we are given the radius r = 7 cm.
Then, the surface area of the sphere,
A = 4 x π x (7)2
Taking the value of π as 22/7,
A = 4 x 22/7 x 7 x 7,
A = 4 x 22 x 7
A = 616 cm2
The surface area of the given sphere is 616 cm2.
Example 5: A bowl in the shape of a hemisphere has a diameter of 10 cm; find the bowl’s volume.
Solution: The hemispherical bowl has a diameter d = 10 cm,
The radius of the bowl r = 102
r = 5 cm,
The volume of a hemisphere is given by,
V = 23.π r3,
V= 23. π (5)3
V = 23.π x 125,
Taking the value of π as 3.14
V = 23 x 3.14 x 125,
V = 261.67 cm3.
Therefore, the volume of the hemispherical bowl is 261.67 cm3.
Example 6: Find the volume of a hollow sphere with a radius of 9 cm.
Solution: The volume of a sphere is given by,
V = 43. π.r3
The radius of the given sphere r = 9 cm,
The volume of the sphere,
V = 43. π x (9)3,
V = 43 x π x 9 x 9 x 9,
V = 4 x π x 3 x 81,
Taking the value of π as 3.14,
V = 4 x 3.14 x 3 x 81,
V = 3052.08 cm3
The volume of the given sphere is 3052.08 cm3.
Example 7: A building is standing on a cylindrical pillar of base radius 10 m; if the pillar’s height is 10 m, what amount of building material will be needed to make ten more such pillars?
Solution: The radius of the pillar h = 10 m,
Height of the pillar h = 10 m,
the amount of building material used in the pillar,
Volume of the pillar = π.r2.h
= π x 102 x 10,
Taking the value of π as 3.14,
= 3.14 x 1000
The total volume of the pillar,
V = 3140 m3,
The total amount of building material required to make ten such pillars,
Volume of one pillar x 10,
= 3140 x 10,
= 31400 m3
Ten pillars would require 31400 m3 building material.
Conclusion
The area is measured in unit squares, and volume is measured in the square cube. The volume calculation measures an object in three dimensions- length, breadth, and height, while surface area is measured in two dimensions- length and breadth. The formulas of the area and volume of different shapes are determined by evaluating the measures of the object.
