 # Conversion - One Shape to Another  View Notes

## Conversion From One Shape to Another

Solids are of varying sizes and we come across them in our day to day life. Each solid that has been cast into a particular shape might not be in that shape earlier. Let’s take jewellery making for instance. How is it possible that perfectly round and circular pearls come out of the earth? It is only possible after converting a particular mould into a desired shape or size. When we convert any shape into any other shape, we are interested in the total volume of the solid. The total volume of the shape remains the same (assuming that there is no wastage in conversion). Generally, we heat the solid to convert it into the desired shape.

Let us study the conversion of solid from one shape to another in detail. We will cover the important underlying aspects of conversion. We will discuss the mathematical aspects of the conversion of solids. We will study the formulas of conversion from one shape to another.

### Conversion of Solid From One Shape to Another

The space a solid occupies or contains is defined as the volume. Two solids of different shape can have the same volume. A solid, if converted into a different shape, will keep its original volume, no matter the shape in which it is converted.

For example, if we melt one big cylindrical candle into ten smaller candles, the total volume of the candles will be equal to the larger candle.

Even if the conversion is to a different shape, the volume remains the same. If we melt a big iron cube into smaller balls of different radiuses. The total volume of the solid remains the same.

Let’s understand it with the help of an example.

An iron ball of radius 21 cm is melted into  27 smaller balls of radius 7cm.

Let’s prove this statement using the concepts of the volume of the sphere.

The volume of the sphere is given by

$Volume = \frac{4}{3} \pi r^{3}$

The volume of the iron ball of radius 21cm

$V = \frac{4}{3} \times \frac{22}{7} \times 21 \times 21 \times 21$

$V = 4 \times 22 \times 21 \times 21$

$V = 38, 808 cm^{3}$

The volume of 27 iron ball of radius 7 cm

$V = \frac{4}{3} \times \frac{22}{7} \times 7 \times 7 \times 7$

$V = 4 \times 22 \times 9 \times 7 \times 7$

$V = 38, 808 cm^{3}$

The total volume of 27 balls of radius = 7cm is equal to the volume of the iron ball of radius = 21cm.

### Conversion From One Shape to Another

When we convert a solid from one shape to another, the volume of the solid remains the same. Given below is the list of formulas pertaining to different shapes regarding their size and volumes.

In case of a cylinder of height h and radius r.

Volume = πr2h

Total surface area = 2πr(h + r)

In case of a cuboid of length l, height h and breadth b.

Volume = l * b * h

Total surface area = 2(lb+bh+ha)

In case of a cube of side a.

Volume = a3

Total surface area = 6a2

In case of a sphere of radius r.

Volume = 4/3r3

Total surface area = 4r2

Problem 1: Rainwater is collected in a flat surface of length 6m and breadth 4m. It is then transferred into a cylindrical vessel of radius 10cm. What will be the height up to which the vessel will be filled if 2cm rainfall has fallen?

Problem 2: The external dimensions of a cuboid are 20cm x 30cm x 25 cm. The width of the wood used to make the cuboid is 1.5cm. Find the amount of wood used in making the cuboidal shape.

Problem 3: A cinema hall wants to make preparations for attendance to maintain social distancing. Normally, each person requires 150cm3 of air but under new guidelines, each person needs to be given 600cm3 of space. The dimensions of the hall are 100m x 60m x 15m. Find the number of people that can sit in the hall before and after the restriction.

Problem 4: Find the total volume of 3 spheres with diameters 14cm, 21cm and 42cm. Also, these three spheres are combined to make a single sphere. Find the diameter of the sphere thus formed.

Q1. Find the Volume of a Cone with a Radius of 15 cm and Slant Height of 17cm.

Ans: This is a basic problem in the volume of cones.

The slant height of the cone = 17cm.

The radius of the cone = 15cm.

The volume of the cone = ⅓ πr2h.

The radius of the cone is given = 15cm.

However, the height of the cone isn’t given. So we have to calculate the height first using Pythagoras theorem.

l2 = r2 + h2

172 = 152 + h2

h2 = 289 - 225

h2 = 64

h = 8cm

Now we will calculate the volume of the cone using the formula

The volume of the cone = ⅓ πr2h

Volume = ⅓*22/7*15*15*8

V = *3.14*225*8

V=1884cm3

Q2. On Melting a Cylinder of Diameter 16cm and Height 2cm, we Convert Them into 12 Spheres of the Same Size. Assuming There is no Wastage in Conversion, Calculate the Diameter of each Sphere Formed. Use π = 22/7.

Ans: According to the question, the volume of the cylinder will be equal to 12 times the volume of the sphere.

Height of cylinder = 2cm.

Diameter of cylinder = 16cm.

The volume of the cylinder = πr2h

V = 22/7*8*8*2

V = 2816/7

The volume of sphere= 4/3 x π r3

The volume of 12 spheres= 12*4/3*πr3

The volume of 12 spheres is equal to the volume of the cylinder. So,

12*4/3*22/7*r3 = 2816/7

r3 = (2816 x 3)/(22 x 12 x 4)

r3 = 96/12

r3 = 8

r = 2cm

The radius of each sphere is 2cm, so the diameter of each sphere comes out to be 4cm.

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