Conversion and Combination of Solids: Meaning, Formula, Solved Examples

Conversion and Combination of Solids: Let’s say we have a box of ice cream that we want to deliver to students by scooping it into a cone. We’re going to change the shape of a solid from a cuboid to a cone with a hemisphere. In our daily lives, we come across several examples of transforming shapes from one shape to another.

Different solid shapes are formed by combining different solids such as ice cream formed by cone and a hemisphere, capsule formed by two hemispheres and a cylinder. So there is a need to find the surface area or volume of such a solid figure. This article describes the conversion and combination of solids in daily life with solved examples.

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Solid Shapes

Solid shapes are three-dimensional objects of otherwise planar shapes. A square becomes a cube when changed to a \(3 -{\text{D}}\) figure. Similarly, a rectangle is a cuboid, and a triangle may become a cone. For planar shapes, our measurement is restricted to only the area of that shape. But when we want to measure \(3 -{\text{D}}\) shapes, we can measure their volume, surface area or curved surface area. Solid shapes, as already said, are \(3 -{\text{D}}\) figures equivalent to their planar shape.

We see many objects in our day-to-day life such as compass boxes, books, ice cream cones, footballs and gas cylinders are all different solid shapes. These objects occupy a certain shape in the space and have three dimensions – breadth, length and height or depth.

Formulae for Various Shapes

To study the combination and conversion of the solid shapes, we need the formulae related to various figures. First, let us recall the basic formulae related to various shapes. The formulae are the basics of calculating surface area or volumes of solid structures formed by combining different solids.

Let’s recall the formula for various shapes that will help solve the examples.

Surface Areas Formulas

The formulae for a surface area of different solid shapes are given below:

1. Cuboid: \(2\left({lb + bh + hl} \right),\) where \(l,b,\) and \(h\) are the length, breadth and height of a cuboid.
2. Cube: \(6{a^2}\) where \(a\) is the side of the cube.
3. Cylinder: \(2\pi r\left({r + h} \right)\) where \(r\) is the radius of the circular base and \(h\) is the height of the cylinder.
4. Cone: \(\pi r\left({l + r} \right)\) Where \(r\) is the radius of the circular base, \(l\) is the slant height of the cone.
5. Sphere: \(4\pi {r^2}\) where \(r\) is the radius of the sphere.
6. Hemisphere: \(3\pi {r^2}\) where \(r\) is the radius of the hemisphere.

Volumes Formulas

Volume is the capacity of any solid shape. The formulae for volumes of different shapes are:

1. Cuboid: \(l \times b \times h,\) where \(l,b\) and \(h\) are the length, breadth and height of a cuboid.
2. Cube: \({a^3}\) where \(a\) is the side of the cube.
3. Cylinder: \(\pi {r^2}h\) where \(r\) and \(h\) is the radius of the circular base and the height of the cylinder, respectively.
4. Cone: \(\frac{1}{3}\pi {r^2}h\) where \(r\) and \(h\) is the radius of the circular base and height of the cone, respectively.
5. Sphere: \(\frac{4}{3}\pi {r^3}\) where \(r\) is the radius of the sphere.
6. Hemisphere: \(\frac{2}{3}\pi {r^3}\) where \(r\) is the radius of the hemisphere.

Conversion of Solid from One Shape to Another

Conversion of solids from one shape to another means the volume of the solids remains the same. We will only transform the shape of the solid. Thus, remember that the original and the new solid volume will be the same when any solid is changed to another solid.

There are many circumstances in our daily life where we want to convert the shape of solids.

Examples of Conversion of Solids

1. You have a gold bar, and you want to get jewellery cast out of it. You will reach the jewellery merchant for the design, and he will mould the gold bar to whatsoever jewellery and shape you want.
2. You have a wax that you want in different shapes. You will have to melt the wax to pour it into the shape you want it to be. Thus, the conversion of solids from one shape to another do occur in our daily life.

Combination of Solids

Shapes that are formed by combining different basic shapes are the combined solid. These may be solid or hollow. Combination of solids heads to a different level of measurement. You must have noticed the shape of ice cream or a capsule. Ice cream has a cone, and a hemisphere mounted on it, while a capsule looks like a combination of two hemispheres fixed to a cylinder on either end. Such solids may be studied under the combination of solids whose volume and surface area are known.

Examples of Combinations of Solids

Few examples of combined solids in our daily life that we come across are

A Capsule: A capsule is a combination of a cylinder and two hemispheres on both ends of a cylinder.

A Circus Tent or a Hut: A circus tent is formed by the combination of a cylinder and a cone. Some circus tents also constitute a cuboid and cone. 

An Ice Cream Cone: An ice cream cone is formed by the combination of a cone and a hemispherical shape.

Surface Area and Volume of Combined Solids

To figure out the volume and surface area of solids made up of various solids, we must first figure out what forms make up that solid. Finding volume or surface area will be simple once you have figured out the basic shapes. To apply the formulas directly, we must ensure that the constituent forms are fundamental shapes such as cube, cuboid, cone, cylinder, etc.

We need to add the volume of two or more shapes that are components of the combined solid. Similarly, to find the surface area of combined solids, we need to add the surface areas of elementary shapes.

Solved Examples – Conversion and Combination of Solids

Q.1. A solid ball of radius \(21\,{\text{cm}}\) is melted and recast into \(27\) smaller balls of equal radii. Find the radius of each ball.
Ans: We know that the volume of a sphere \( = \frac{4}{3}\pi {R^3}\)
The volume of the solid ball \( = \frac{4}{3}\pi {\left({21} \right)^3} = 38,808\,{\text{c}}{{\text{m}}^3}\)
Assume the radius of the smaller balls as \(r.\)
The volume of solid ball \( = 27 \times \) volume of smaller balls
\( \Rightarrow 27 \times \frac{4}{3}\pi {r^3} = 38808\)
\( \Rightarrow \frac{{22}}{7}{r^3} = \frac{{38808}}{{36}}\)
\( \Rightarrow {r^3} = 343\,{\text{c}}{{\text{m}}^3}\)
\( \Rightarrow r = 7\,{\text{cm}}\)
Therefore, the radius of the smaller balls is \(7\,{\text{cm}}\) each.

Q.2. A solid vessel is a cylinder covered by a hollow hemisphere used to store rice. The hemisphere has a diameter of \(16\,{\text{cm}}{\text{.}}\) If the total height of the vessel is \(15\,{\text{cm}}{\text{,}}\) calculate the volume of rice stored in it?

Ans: The radius of the cylindrical vessel \( = \frac{d}{2} = \frac{{16}}{2} = 8\,{\text{cm}}.\)
Total height of the vessel \( = 14\,{\text{cm}}\)
Height of the cylinder part \( = 15 – 8 = 7\,{\text{cm}}\)
The volume of rice stored in the container \( = \) volume of cylindrical vessel \( + \) volume of the hemisphere
\( = \pi {r^2}h + \frac{2}{3}\pi {r^3}\)
\( = \left({\frac{{22}}{7} \times 8 \times 8 \times 7}\right) + \left({\frac{2}{3} \times \frac{{22}}{7} \times 8 \times 8 \times 8} \right)\)
\(= 1408 + 1072.76\)
\( = 2480.76\,{\text{c}}{{\text{m}}^3}.\)
The volume of rice stored in the container is \( 2480.76\,{\text{c}}{{\text{m}}^3}.\)

Q.3. How many candles (cylindrical in shape), \(1.75\,{\text{cm}}\) in diameter and of thickness \(2\,{\text{mm}},\) have to be melted to form a cuboid candle of \(5.5\,{\text{cm}} \times {\text{10}}\,{\text{cm}} \times 3.5\,{\text{cm}}\)?
Ans: Radius of candle \( = \frac{{1.75}}{2} = 0.875\,{\text{cm}}\)
We know that the volume of a cylinder \( = \pi {r^2}h\)
The volume of one cylindrical candle \( = \frac{{22}}{7} \times {\left({0.875} \right)^2} \times \left({0.02} \right){\text{c}}{{\text{m}}^3}\)
\( = 0.048125\,{\rm{c}}{{\rm{m}}^3}\)
Volume of cuboid candle \( = 5.5 \times 10 \times 3.5 = 192.5\,{\text{c}}{{\text{m}}^3}\)
Thus, number of cylindrical candles \( = \frac{{{\text{the}}\,{\text{volume}}\,{\text{of}}\,{\text{a}}\,{\text{cuboid}}\,{\text{candle}}}}{{{\text{the}}\,{\text{volume}}\,{\text{of}}\,{\text{one}}\,{\text{cylindrical}}\,{\text{candle}}}}\)
\(= \frac{{192.5}}{{0.048125}}\)
\( = 4000\)

Q.4. Tissue boxes of size \(10\,{\text{cm}} \times 8\,{\text{cm}} \times 9\,{\text{cm}}\) can be adjusted inside a cupboard box of size \(36\,{\text{cm}} \times 40\,{\text{cm}} \times 100\,{\text{cm}}{\text{.}}\) How many can such boxes be adjusted in the cupboard box?
Ans: Volume of the tissue box \( = 10 \times 8 \times 9\,{\text{c}}{{\text{m}}^3} = 720\,{\text{c}}{{\text{m}}^3}\)
Volume of the cupboard \( = 36\, \times 40\, \times 100\,{\text{c}}{{\text{m}}^3} = 144,000\,{\text{c}}{{\text{m}}^3}\)
Number of tissue boxes that can be adjusted \( = \frac{{144000}}{{720}} = 200\) boxes
Thus, \(200\) tissue boxes can be adjusted in the cupboard box.

Q.5. A toy of the shape of a cone mounted over a hemisphere. The total height of the toy is \(15.5\,{\text{cm}}\) and the radius of the hemisphere is \(3.5\,{\text{cm}}.\) Calculate the total surface area of the toy.
Ans: Given, the hemisphere of radius \(r = 3.5\,{\text{cm}},\) cone of height \(h = 15.5 – 3.5 = 12\,{\text{cm}}\)
The total surface area of the toy \( = 2\pi \times {\left({3.5} \right)^2} + \left[ \pi \right]3.5 \times \sqrt {\left\{{{{3.5}^2} +{{12}^2}} \right\}} \)
\( = 214.5\,{\text{sq}}.{\text{cm}}\)
Thus, the surface area of the toy is \(214.5\,{\text{sq}}.{\text{cm}}{\text{.}}\)

Summary

The conversion and combination of solids are discussed in this article. We have given numerous formulas relating to volume and surface area since the formulae of fundamental solids are important. Then we looked at some examples of solids being converted from one shape to another.

We’ve also talked about combining two or more solids and calculating the surface area and volume of the resulting solid.  Solved examples are also provided to students on this page to help them better understand the topic.

FAQs on Conversion and Combination of Solids

Q.1. What will happen to the volume during the conversion of a solid from one shape to another?
Ans: When you convert one solid shape to another, its volume remains the same, no matter how different the new shape is. If you melt one big cubical candle into \({\text{six}}\) small cylindrical candles, the sum of the volumes of the smaller candles is equal to the volume of the big cubical candle.

Q.2. What is the combination of solid?
Ans: By combining solids, we mean a solid shape consisting of two or more regular solid shapes.

Q.3. Write examples of combined solids.
Ans: Some examples of combined solids are ice cream cones, toys, tents, etc.

Q.4. What is the volume of the solid?
Ans: Solid is the \(3\)-dimensional figure. The volume of a solid is the measure of how much space an object takes.

Q.5. Write down the formula o calculate the volume of a capsule.
Ans: We can imagine that a capsule is a combination of a cylinder and two hemispheres on both ends of the cylinder.
The volume of a capsule \( = \) volume of a cylinder \( + 2 \times \) volume of a hemisphere
Thus, the volume of a capsule \( = \pi {r^2}h + 2 \times \frac{2}{3}\pi{r^3}.\)

We hope this detailed article on the conversion and combination of solids helped you in your studies. If you have any doubts, queries or suggestions regarding this article, feel to ask us in the comment section and we will be more than happy to assist you. Happy learning!