Surface Area and Volume Solved Problems

The world we live in is three dimensional in nature. Every object can be measured in terms of length, breadth and height. For example, the house that we live in has dimensions. The surface area is the outer part of the figure, whereas volume is the inner part. This article contains solved examples on surface area and volume.

Surface Area and Volume Definitions

The complete area of the surface of a three dimensional solid can be termed as surface area. It is denoted by “S” and measured in square units.

The quantity of a three dimensional solid or the space taken by it is called volume. Volume is denoted by “V” and measured in cubic units.

Surface Area and Volume Formulae

To find the surface area and volume of the solids individually or combined together, the following table can be used for reference.

Name	Diagram of Solid	Lateral or Curved Surface Area	Total Surface Area	Volume	Length of the Diagonal and Explanation of Terms
Cube		4l²	6l²	l³	√3 l = edge of the cube
Cuboid		2h(l +b)	2(lb + bh + hl)	lbh	$l^{2}+b^{2}+h^{2}$ l = length b = breadth h = height
Cylinder		2πrh	2πr²+ 2πh = 2πr(r + h)	πr²h	r = radius h = height
Hollow cylinder		2πh (R + r)	2πh (R + r) + 2πh (R² – r²)	–	R = outer radius r = inner radius
Cone		$\pi rl = \pi \sqrt{h^{2}+r^{2}$	πr² + πrl = πr(r + l)	1/3 πr²h	r = radius h = height l = slant height
Frustum of a cone		πl(R + r)	πR²+πr²+πl(R+r)	⅓ πh[R²+r²+Rr]	–
Sphere		4πr²	4πr²	4/3 πr³	r = radius
Prism		Perimeter of base × Height	Lateral surface area + 2(Area of the end surface)	Area of base × Height	–
Pyramid		1/2 (Perimeter of base) × Slant height	Lateral surface area + Area of the base	1/3 Area of base × Height	–

Surface Area and Volume Examples

Example 1: The radius and height of a cone are in the ratio 4 : 3. The area of the base is 154cm². What is its curved surface area?

Solution:

$\begin{array}{l}\pi[r^2]=154\\ r=7cm\\ \frac{r}{h}=\frac{7}{3}\\ \frac{7}{h}=\frac{4}{3}\\ h=\frac{21}{4}\\ Curved\ surface \ area=\pi[rl]\\ =\pi[r]\sqrt{r^2+h^2}\\ =[\frac{22}{7}]\times 7\times \sqrt{7^2+[\frac{21}{4}]^2}\\ =192.25 cm^2\end{array} $

Example 2: The percentage decrease in the surface area of the cube, when each side decreased to 0.5 times the original length is ____________.

Solution:

The surface area of the cube = 6a²

The new surface area of the cube

$\begin{array}{l} = 6\times [\frac{a}{2}]^2=\frac{6a^2}{4}\end{array} $

Percent change

$\begin{array}{l} = [\frac{6a^2-\frac{6a^2}{4}}{6a^2}]\times 100\end{array} $

$\begin{array}{l}=\frac{18a^2\times 100}{4\times 6a^2}\end{array} $

= 75%

Example 3: A circus tent is bent in the form of a cone over a cylinder. The diameter of the base is 9 m, the height of the cylindrical part is 4.8 m, and the total height of the tent is 10.8 m. What is the amount of canvas required for the tent?

Solution:

The surface area of the conical part

$\begin{array}{l} = \pi r \times \sqrt{r^2+(h_1)^2}\end{array} $

$\begin{array}{l} =\pi \times 4.5 \times \sqrt{[4.5]^2+6^2}\end{array} $

$\begin{array}{l} = \frac{22}{7} \times 4.5 \times 7.5\end{array} $

The surface area of the cylindrical part

$\begin{array}{l}=2\pi*rh{_2}\end{array} $

Total surface area

$\begin{array}{l}=[\pi]r(\sqrt{r^2+[h_1]^2+2h_2})=241.84sq\ m\end{array} $

Example 4: A toy is in the shape of a cone over a hemisphere. The radius of the hemisphere is 3.5 cm, and the total height of the toy is 15.5 cm. Find the total area of the toy.

Solution:

The top is made up of two shapes, one is a hemisphere of radius r = 3.5 cm and the other cone of height h = 15.5 – 3.5 = 12 cm.

The total area of the top

$\begin{array}{l} = 2\pi\times (3.5)^2+[\pi]3.5\times \sqrt{3.5^2+12^2}=214.5 \ sq\ cm\end{array} $

Example 5: The dimensions of the cuboid are in the ratio of 1 : 2 : 3, and its total surface area is 88 cm². What is the lateral surface area?

Solution:

Let the length, breadth and height of the cuboid be l, 2l and 3l, respectively.

Total surface area = 2{(l * 2l) + (2l * 3l) + (3l * l)} = 88 cm²

22l² = 88

l = 2 cm

Hence l = 2 cm, 2l = 4 cm, 3l = 6 cm

Lateral surface area = 2(l * 3l) + (2l * 3l) = 18l² = 72 cm²

Example 6: A sphere, a cylinder and a cone are of the same height and same radius, then the ratio of their curved surfaces is ______.

Solution:

Let the

Let the radius of the sphere be r

Height of the sphere = 2r

Height of the cylinder = 2r

Height of the cone = 2r

Hence, the ratio of the curved surface area is

$\begin{array}{l}=4 \pi r^{2}:2 \pi r (2r):\pi r\sqrt{r^{2}+(2r)^{2}}\\ =4 \pi r^{2}:4 \pi r^{2}:\pi r^{2}\sqrt{5}\\ =4:4:\sqrt{5}\end{array} $

Example 7: Each side of a cube increases by 50%. Then, the surface area of the cube increases by ______.

Solution:

Let each side of the cube be a.

After increment, each sides of the cube

$\begin{array}{l} = a+\frac{50a}{500}\end{array} $

The surface area of the cube = a²

Surface area, after increment

$\begin{array}{l}=6(\frac{a\times 3}{2})^{2}\\ =6a^{2}\times \frac{9}{4}\\\end{array} $

$\begin{array}{l}\text{Increment in surface area} =[\frac{\text{Increment} -\text{Original}}{\text{Original}}\times 100] \%\\ =[\frac{6a^{2}\times \frac{9}{4}-6a^{2}}{6a^{2}}\times 100] \% \\ =[\frac{5}{4}\times 100] \%\\ =125 \%\\\end{array} $

Example 8: The level of water of a swimming pool is rectangular, of which length is 32 metres and width is 9.5 metres, and the depth of water at one end is 1 ½ metres which increases to 4 ½ metres at the other end. Find the volume of water in the swimming pool.

Solution:

The cross-section of the swimming pool is like a trapezium, of which parallel sides are 4 ½ metres and 1 ½ metres, and the vertical perpendicular is 32 metres.

Therefore, the area of cross-section

$\begin{array}{l}=\frac{1}{2}\times 16\times \left( 4\frac{1}{2}+1\frac{1}{2} \right)\end{array} $

= 48 sq.m.

Hence, the volume of water in the swimming pool = area of the cross-section × length

= 48 × 9.5

= 456 cubic metres.

Example 9: The length of a cold storage is double its breadth. The height is 3 metres, and the area of its four walls (including doors) is 108 metre². Find its volume.

Solution:

Let the length, breadth and height of the cold storage be l metres, b metres and h metres, respectively.

Then, l = 2b (given) and h = 3 metres.

Now, the area of the four walls = 108

⇒ 2 (l + b) h = 108

⇒ 2 (2b + b) × 3 = 108

⇒ 18b = 108

⇒ b = 6 metres

∴ l = 2b

⇒ l = 12 metres

Hence, the volume of the cold storage = (lbh) m³ = (12 × 6 × 3) m³ = 216 m³.

Example 10: A cone and a cylinder have the same curved surface area and base area. If the height of the cylinder is 3 m, find the slant height of the cone.

Solution:

The cone and cylinder have the same base area. It implies that they have the same radius.

The curved surface area of cone = Curved surface area of the cylinder

πrl = 2πrh

l = 2h

l = 2 * 3

l = 6

Frequently Asked Questions

What do you mean by the surface area of a solid?

Surface area is the complete area of the surface of a three dimensional solid.

What is the total surface area of a sphere?

The total surface area of a sphere = 4πr², where r is the radius of the sphere.

What is the volume of a cylinder?

The volume of a cylinder = πr²h, where r is the radius, and h is the height of the cylinder.

What is the volume of a sphere?

The volume of a sphere = (4/3)πr³, where r is the radius of the sphere.

What is the volume of a cone?

The volume of a cone = (⅓)πr²h.

Surface Area and Volume Examples

Surface Area and Volume Definitions

Surface Area and Volume Formulae

Surface Area and Volume Examples

Frequently Asked Questions

What do you mean by the surface area of a solid?

What is the total surface area of a sphere?

What is the volume of a cylinder?

What is the volume of a sphere?

What is the volume of a cone?

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