Surface Area of a Combination of Solids: In our daily lives, we come across various solids of different shapes and sizes for which we can calculate surface area. We know how to calculate the surface area of the items in our surroundings. But what if these basic forms combine to produce a new shape that isn’t the same as the original? 

The challenge now is figuring out how to determine the surface and area of new objects. We must notice the new form while calculating the surface area of these new forms. In this article, we will discuss the combination of solids and their surface area in detail.

Surface Area of a Combination of Solids: Introduction

Have you ever come across a capsule? It consists of two hemispheres joined by a cylinder.

In everyday life, we’ll see shapes that are made up of different solids. These forms can be hollow or solid, and they can be studied as a combination of solids.

Volume and Surface Area of Solids

The surface area of a three-dimensional figure is the amount of exterior space that it covers. A solid shape’s surface area can be the lateral surface area, curved surface area, or total surface area.

A solid shape’s volume is defined as the amount of space it takes up. It is the space bounded by a border, occupied by an object, or capable of holding something.

Combination of Solids

Solid forms are three-dimensional structures that are otherwise planar shapes. When converted to a three-dimensional structure, a square becomes a cube, a rectangle becomes a cuboid, and a triangle becomes a cone. In the case of planar shapes, we can only measure the area of the shape.

But when we deal with \(3-D\) shapes, we intend to measure their volume, surface area or curved surface area. Solid shapes, as already said, are \(3-D\) counterparts of their planar shape. But what if these solid shapes join together to form a new shape? A different level of measurement is obtained by combining solids.

We observe many shapes that combine different shapes in our daily lives, like huts, tents, capsules, and ice-cream-filled cones. So, what precisely is a solid combination? The figure created by combining two or more different solids is known as a combination of solids.

Examples of Combinations of Solids

A Circus Tent or a Hut

A circus tent is a cylinder and a cone combination. A cuboid and a cone can also be seen in circus tents. A hut is a type of kutcha home that resembles a tent.

An Ice Cream Cone

A right circular cone is combined with a hemisphere with the same circular base as the cone to form an ice cream cone.

A Dome on a Solid Shape

A dome is usually constructed on top of a structure or a tent. The upper part of the hemisphere is known as a dome. Now, if a structure has a dome, we combine the solid shape of the structure with the dome shape. An apex dome, which is a combination of a hemisphere and a cylinder, is seen below.

Mushroom

Mushrooms have a cylinder-shaped body with a cone-shaped top.

Funnel

A funnel is a combination of a frustum of a cone and a cylinder.

Pencil

A sharpened pencil is a combination of a cone and a cylinder.

Surface Areas Formulas

The surface area of various solid shapes are given below:

1. Cuboid

Lateral Surface Area \( = 2\left( {l + b} \right)h\)

Total Surface Area \(=2(l b+b h+h l)\)

Where \(l, b\), and \(h\) are the length, breadth and height of a cuboid.

2. Cube

Lateral Surface Area \(=4 a^{2}\)

Total Surface Area \(=6 a^{2}\)

Where \(a\) is the side of the cube.

3. Cylinder

Curved Surface Area \(=2 \pi r h\)

Total Surface Area \(=2 \pi r(r+h), r\) is the radius of the circular base, and \(h\) is the height of the cylinder.

4. Cone

Curved Surface area \(=\pi r l\)

Total Surface Area \(=\pi r(l+r), r\) is the radius of the circular base, \(l\) is the slant height of the cone.

5. Sphere

Curved Surface Area \(=4 \pi r^{2}\)

Total Surface Area \(=4 \pi r^{2}\)

Where \(r\) is the radius of the sphere

6. Hemisphere

Curved Surface Area \(=2 \pi r^{2}\)

Total Surface Area \(=3 \pi r^{2}\)

Where \(r\) is the radius of the hemisphere.

7. Frustum of a Cone

Lateral/Curved Surface Area \(=\pi l\left(r_{1}+r_{2}\right)\)

Total Surface Area \(=\pi l\left(r_{1}+r_{2}\right)+\pi r_{1}^{2}+\pi r_{2}^{2}, r_{1}\) and \(r_{2}\) are the radius of the circular bases, \(l\) is the slant height.

Surface Area of Combination of Solids

When working with solid structure calculations, we need to be extra cautious with our measurements. Finding the surface area of a solid combination requires logic and expertise. In such computations, the initial step is to figure out what shapes have combined to form the structure.

Finding the surface area for a structure becomes simple and quick once you’ve figured out the basic shapes. You must combine the surface areas of the constituting structures to determine the surface area of a solid structure created by combining two or more solids.

To calculate the surface area of a circus tent, combine the cone and cylinder surface areas. In a tent made up of a cone and a cylinder, we first compute the surface area of each cone and cylinder separately, then combine them.

For example, consider three cubes each of \(5 \,\text {cm}\) edge are joined end to end. Find the surface area of the resulting cuboid.

If three cubes are joined end to end, we get a cuboid such that,

length of the resulting cuboid, \(l=5 \mathrm{~cm}+5 \mathrm{~cm}+5 \mathrm{~cm}=15 \mathrm{~cm}\)

the breadth of the resulting cuboid, \(b=5 \mathrm{~cm}\)

height of the resulting cuboid, \(h=5 \mathrm{~cm}\)

The surface area of the cuboid \(=2(l b+b h+l h)\)

\(=2(15 \times 5+5 \times 5+5 \times 15)\)

\(=2(75+25+75)=2(175)=350 \mathrm{~cm}^{2}\)

Solved Examples – Surface Area of a Combination of Solids

Q.1. Rice is stored in a cylinder-shaped container mounted by a hollow hemisphere. Suppose the diameter of the hemisphere is \(10 \,\text {cm}\). Find the total exterior surface area of the container if the entire height of the vessel is \(12 \,\text {cm}\).
Ans: Given, the diameter of the hemisphere \(=10 \mathrm{~cm}\)
Height of the vessel \(=12 \mathrm{~cm}\)

The radius of the cylindrical vessel \(=\frac{d}{2}=\frac{10}{2}=5 \mathrm{~cm}\)
Height of the cylinder \(=12-5=7 \mathrm{~cm}\)
The surface area of the vessel \(=\) Curved surface area of hemisphere \(+\) Curved surface area of the cylinder \(+\) Area of base
\(=2 \pi r^{2}+2 \pi r(r+h)+\pi r^{2}\)
\(=3 \pi r^{2}+2 \pi r(r+h)\)
\(=3 \times \frac{22}{7} \times 5^{2}+2 \times \frac{22}{7} \times 5(5+7) \quad \mathrm{cm}^{2}\)
\(=235.71+377.14 \mathrm{~cm}^{2}\)
\(=612.85 \mathrm{~cm}^{2}\)

Q.2. A capsule is formed like a cylinder with two hemispheres attached to each end. What is the surface area of the capsule if the length of the capsule is \(10 \,\text {mm}\) and the width is \(6 \,\text {mm}\) ?

Ans: Given, the diameter of hemisphere \(=6 \mathrm{~mm}\)
The radius of hemisphere \(=\frac{6}{2}=3 \mathrm{~mm}\)
Height of entire capsule \(=10 \mathrm{~mm}\)
Height of cylinder \(=10-6=4 \mathrm{~mm}\)
The surface area of capsule \(=2 \times\) Surface area of hemisphere \(+\) Surface area of the cylinder
\(=2 \times 2 \pi r^{2}+2 \pi r(r+h)\)
\(=4 \times \frac{22}{7} \times 3^{2}+2 \times \frac{22}{7} \times 3(3+4)\)
\(=113.14+132\)
\(=245.14 \mathrm{~mm}^{2}\)

Q.3. As shown in the figure, a cubical block of side \(8 \,\text {cm}\) is surmounted by a hemisphere. Find the surface area of the solid.

Ans: Given, side of cubical block \(=8 \mathrm{~cm}\)
The surface area of solid \(=\) Surface area of cube \(+\) Curved surface area of the hemisphere Area of the base of the hemisphere
\(=6 a^{2}+2 \pi r^{2}-\pi r^{2}\)
\(=6 a^{2}+\pi r^{2}\)
\(=6(8)^{2}+\left(\frac{22}{7}\right)\left(\frac{8}{2}\right)^{2}\)
\(=384+50.28\)
\(=434.28 \mathrm{~cm}^{2}\)

Q.4. A shuttlecock used for playing badminton has the shape of a frustum of a cone mounted on a hemisphere. The diameters of the frustum are \(5 \,\text {cm}\) and \(2 \,\text {cm}\). The height of the entire shuttlecock is \(7 \,\text {cm}\). Find its external surface area.

Ans: Surface area of shuttlecock \(=\) curved surface area of frustum cone \(+\) curved surface area of hemisphere
\(=\pi(R+r) l+2 \pi r^{2} \quad \cdots-(1)\)
Height of shuttlecock \(=7 \mathrm{~cm}\)
The radius of hemisphere \(+\) Height of frustum cone \(=7\)
\(1+h=7\)
\(h=6\)
\(l=\sqrt{\left(h^{2}+(R-r)^{2}\right)}\)
\(l=\sqrt{6^{2}+\left(\left(\frac{5}{2}\right)-1\right)^{2}}\)
\(l=\sqrt{36+\left(\frac{9}{4}\right)}\)
\(l=\frac{\sqrt{153}}{2}\)
\(l=\frac{12.36}{2}\)
\(l=6.18\)
By applying the value of \(l\) in \((1)\), we get
\(=\pi\left(\left(\frac{5}{2}\right)+1\right) l+2 \pi r^{2}\)
\(=\pi\left[\left(\frac{7}{2}\right)(6.18)+2(1)^{2}\right]\)
\(=\left(\frac{22}{7}\right)[(21.63+2]\)
\(=74.26 \mathrm{~cm}^{2}\)

Q.5. A jewel box is in the shape of a cuboid of dimensions \(20 \mathrm{~cm} \times 15 \mathrm{~cm} \times 10 \mathrm{~cm}\) surmounted by a half part of a cylinder, as shown in the figure. Find the total surface area (TSA) of the box except the base.

Ans: Let \(l, b\) and \(h_{1}\) be the length, breadth, and height of the cuboid. Also, let us take \(r\) and \(h_{2}\) be the radius and height of the cylinder.
Now, TSA of the box \(=\) CSA of the cuboid \(+\frac{1}{2}\) (CSA of the cylinder)
\(=2(l+b) h_{1}+\frac{1}{2}\left(2 \pi r h_{2}\right)+\frac{1}{2} \pi r^{2}+\frac{1}{2} \pi r^{2}\)
\(=2(20+15) 10+\left(\frac{22}{7} \times \frac{15}{2} \times 20\right)+\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\)
\(=2 \times 350+471.42+38.5\)
\(=700+471.42+38.5\)
\(=1209.92 \mathrm{~cm}^{2}\)
Therefore, the TSA of the box is \(1107.42 \mathrm{~cm}^{2}\).

Summary

This article discussed the combination of solids, surface area formula, the surface area of a combination of solids, the difference between total surface area and volume, examples of the surface area of a combination of solids, and solved examples.

FAQs About Surface Area of a Combination of Solids

Let’s look at some of the commonly asked questions about surface area of a combination of solids:

Q.1. What is the volume and surface area of solids?
Ans: The surface area of a three-dimensional form is the amount of exterior space that it covers. A solid shape’s surface area is divided into the lateral surface area, curved surface area, and total surface area.
A solid shape’s volume is defined as the amount of space it takes up. It is the space bounded by a border, occupied by an object, or capable of holding something.

Q.2. What is the difference between total surface area and volume?
Ans: The surface area measures the area occupied by the upper layer of a surface, or, to put it another way, the area of all the shapes/planes that make up the figures/solids, whereas volume is the measure of a figure’s carrying capacity or the space enclosed within the figure.

Q.3. What is the surface area formula?
Ans: The surface area of various solid shapes are given below:
1. Cuboid:
Lateral Surface Area \( = 2\left( {l + b} \right)h\)
Total Surface Area \(=2(l b+b h+h l)\)
Where \(l, b\), and \(h\) are the length, breadth and height of a cuboid.
2. Cube:
Lateral Surface Area \(=4 a^{2}\)
Total Surface Area \(=6 a^{2}\)
Where \(a\) is the side of the cube.
3. Cylinder:
Lateral/Curved Surface Area \(=2 \pi r h\)
Total Surface Area \(=2 \pi r(r+h), r\) is the radius of circular base and \(h\) is the height of the cylinder.
4. Cone:
Lateral/Curved Surface area \(=\pi r l\)
Total Surface Area \(=\pi r(l+r), r\) is the radius of the circular base, \(l\) is the slant height of the cone.
5. Sphere:
Lateral/Curved Surface Area \(=4 \pi r^{2}\)
Total Surface Area \(=4 \pi r^{2}, r\) is the radius of the sphere.
6. Hemisphere:
Lateral/Curved Surface Area \(=2 \pi r^{2}\)
Total Surface Area \(=3 \pi r^{2}, r\) is the radius of the hemisphere.
7. Frustum of a cone:
Lateral/Curved Surface Area \(=\pi l\left(r_{1}+r_{2}\right)\)
Total Surface Area \(=\pi l\left(r_{1}+r_{2}\right)+\pi r_{1}^{2}+\pi r_{2}^{2}, r_{1}\) and \(r_{2}\) are the radius of the circular bases, \(l\) is the slant height of the cone.

Q.4. How do you find the surface area of two combined solids?
Ans: The total surface area of a combined solid is the sum of the total surface areas of the separate solids that make up the combined solid, eliminating overlapping sections from each solid.

Q.5. How do you find the volume of a combined solid?
Ans: The volume of a combined solid is the sum of the volumes of the separate solids that make it up.

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