In the previous section we saw that the same formula can be used to calculate the volume of a rectangular prism and a triangular prism. In this section we will see prisms of any polygonal shape.
1. Fig.23.6 (a) shows a prism. It’s base is a polygon. Let it’s height be ‘h’ cm. It’s base is shown separately in fig.b.
2. In the fig.b, the polygon is split into triangles. [Note that, any polygon can be split into triangles. All we have to do is this: Choose any one vertex. Join this vertex to all other vertices]
3. So the prism in fig.a can also be split into triangular prisms. Each will have a triangular base. This splitting of the prism is shown in fig.c
4. Let in fig.b,
area of the first triangle = b1 cm2
area of the second triangle = b2 cm2
area of the third triangle = b3 cm2
- - - -
- - - -
area of the nth triangle = bn cm2
5. Let the total area of the polygon in fig.b = a cm2
Then a = (b1 + b2 + ... + bn)
6. We can calculate the volume of each of the prisms in fig.c separately. We will get:
Volume of first triangular prism = b1 × h = b1h
Volume of first triangular prism = b2 × h = b2h
Volume of first triangular prism = b3 × h = b3h
- - - -
- - - -
Volume of nth triangular prism = bn × h = bnh
7. So volume of the prism in fig.a = Total volume of all the prisms in fig.c = (b1h + b2h + ... + bnh)
= (b1 + b2 + ... + bn) × h
8. But from (5) we have a = (b1 + b2 + ... + bn). So we can rewrite (7):
Total volume of all the prisms in fig.c = ah cm3.
Thus we find that, the formula can be applied to any polygonal prism.
■ So we have the same formula for:
• A rectangular prism
• A triangular prism [The base can be any triangle]
• A polygonal prism [The base can be any polygon]
But when we say 'any polygon', triangles and rectangles are also included
A sample calculation:
A prism has it’s base in the shape of an equilateral triangle. The side of this equilateral triangle is 4 cm. Height of the prism is 10 cm. What is the volume of the prism?
Solution:
1. We have: Volume of any prism = area of the base × height
2. In our case, area of the base = area of the equilateral triangle of side 4 cm
= √3⁄4 × 42 = 4√3 cm2. Details here.
3. So volume = 4√3 × 10 = 40√3 cm3
Now we will see an interesting case:
In fig.23.7 below, a water trough is resting on it's base.
We want to calculate the volume of the trough. So that we can know how many litres of water it will hold.
Solution:
1. We can see that the base on which the trough is resting, is a rectangle of size 1.4 × 0.7 m
2. The top face is also a rectangle of size 2.0 × 0.7 m
3. Even though the base and top face are rectangles, they are not identical. So it seems that we can not use the 'formula for volume of prisms' in this case. But by making a change to the 'seating', we can make it into a prism.
4. In fig.23.8(a) below, the trough is tilted through 90o, as shown by the green arrow. The final position after tilting, is shown in fig.b
5. Now the base is an isosceles trapezium. This isosceles trapezium has the following measurements:
•Longer edge = 2 m •Shorter edge = 1.4 m •height = 0.4 m
[At this stage, it is better to have a good understanding about the shapes of such troughs. A presentation can be seen here]
6. The top face is also the same isosceles trapezium
7. The sides are all rectangles. Height of all these rectangles are the same. That is., 0.7 m
8. So all the conditions for a prism are satisfied. We can use the formula.
[Note that we do not need to consider the other dimensions of the side rectangles. Only the height is required]
9. Volume = base area × height
Base area = (a+b)⁄2 × height of isosceles trapezium = (2+1.4)⁄2 × 0.4 = (2+1.4)×0.2 = 3.4×0.2 = 0.68 m2.
So volume = 0.68 × 0.7 = 0.476 m3.
10. We want this volume in litres. We know that, 1 litre = 1000 cm3.
But 1000 cm3 = 0.001 m3.
So 1 litre = 0.001 m3. same as 1 m3 = 1000 litres
Thus we get: Volume of the trough = 0.476 m3 = 0.476 × 1000 = 476 litres
[Moral: A prism need not be always placed with it's base at the bottom]
Now we will see some solved examples:
Solved example 23.1
The base of a prism is an equilateral triangle of perimeter 15 cm and it's height is 5 cm. Calculate it's volume
Solution:
1. Perimeter of the equilateral triangle = 15 cm
2. So one side = 15⁄3 = 5 cm
3. Base area = Area of the equilateral triangle = √3⁄4 × 52 = 6.25√3 = 6.25×1.73 = 10.8125 cm2. Details here.
4. So volume = base area × height = 10.8125 × 5 = 54.0625 cm3.
Solved example 23.2
A hexagonal hole of each side 2 m is dug in the school ground to collect rain water. It is 3 m deep. Some water is now collected in it. The top surface of water is at a depth of 2 m from the top of the pit. How many litres of water is in it?
Solution:
1. The water collected in the pit is in the form of a hexagonal prism.
2. Height of this prism = 3 – 2 = 1 m. This is shown in fig.23.9 below
3. Base area of this prism = Area of a regular haxagon of side 2 m
= 6 times the area of an equilateral triangle of side 2 m = 6 × √3⁄4 × 22 = 6√3
= 6 ×1.73 = 10.38 m2. Details here.
4. So volume = base area × height = 10.38 × 1 = 10.38 m3.
5. We want this volume in litres. We know that, 1 litre = 1000 cm3.
But 1000 cm3 = 0.001 m3.
So 1 litre = 0.001 m3. same as 1 m3 = 1000 litres
Thus we get: Volume of the trough = 10.38 m3 = 10.38 × 1000 = 10380 litres
Solved example 23.3
A hollow prism of base a square of side 16 cm contains water 10 cm high. If a solid cube of side 8 cm is immersed in it, by how much would the water level rise?
Solution:
1. The water present is in the form of a square prism.
(i) Height of this prism = 10 cm
(ii) Base area of this prism = 16 × 16 = 256 cm2
(iii) So volume of water present = base area × height = 256 × 10 = 2560 cm3
2. Increase in volume (see fig.23.10 below) = volume of cube = 8 × 8 × 8 = 83 = 512 cm3
3. Total volume after immersing the cube = 2560 + 512 = 3072 cm3
4. This is the volume of the new prism
5. Base area of the new prism is same as before, which is 256 cm2
6. Let the height of the new prism = h
7. Then 256 × h = 3072 cm3
8. So h = 3072⁄256 = 12 cm
9. Thus, rise in water level = 12 – 10 = 2 cm
1. Fig.23.6 (a) shows a prism. It’s base is a polygon. Let it’s height be ‘h’ cm. It’s base is shown separately in fig.b.
 |
| Fig.23.6 |
3. So the prism in fig.a can also be split into triangular prisms. Each will have a triangular base. This splitting of the prism is shown in fig.c
4. Let in fig.b,
area of the first triangle = b1 cm2
area of the second triangle = b2 cm2
area of the third triangle = b3 cm2
- - - -
- - - -
area of the nth triangle = bn cm2
5. Let the total area of the polygon in fig.b = a cm2
Then a = (b1 + b2 + ... + bn)
6. We can calculate the volume of each of the prisms in fig.c separately. We will get:
Volume of first triangular prism = b1 × h = b1h
Volume of first triangular prism = b2 × h = b2h
Volume of first triangular prism = b3 × h = b3h
- - - -
- - - -
Volume of nth triangular prism = bn × h = bnh
7. So volume of the prism in fig.a = Total volume of all the prisms in fig.c = (b1h + b2h + ... + bnh)
= (b1 + b2 + ... + bn) × h
8. But from (5) we have a = (b1 + b2 + ... + bn). So we can rewrite (7):
Total volume of all the prisms in fig.c = ah cm3.
Thus we find that, the formula can be applied to any polygonal prism.
■ So we have the same formula for:
• A rectangular prism
• A triangular prism [The base can be any triangle]
• A polygonal prism [The base can be any polygon]
But when we say 'any polygon', triangles and rectangles are also included
We can write it in the form of a theorem.
Theorem 23.1:
• We have a prism of height 'h' cm
• The base can be of any polygonal shape
• Area of the base is 'a' cm2
• Then the volume of the prism is ah cm3
A prism has it’s base in the shape of an equilateral triangle. The side of this equilateral triangle is 4 cm. Height of the prism is 10 cm. What is the volume of the prism?
Solution:
1. We have: Volume of any prism = area of the base × height
2. In our case, area of the base = area of the equilateral triangle of side 4 cm
= √3⁄4 × 42 = 4√3 cm2. Details here.
3. So volume = 4√3 × 10 = 40√3 cm3
Now we will see an interesting case:
In fig.23.7 below, a water trough is resting on it's base.
 |
| Fig.23.7 |
Solution:
1. We can see that the base on which the trough is resting, is a rectangle of size 1.4 × 0.7 m
2. The top face is also a rectangle of size 2.0 × 0.7 m
3. Even though the base and top face are rectangles, they are not identical. So it seems that we can not use the 'formula for volume of prisms' in this case. But by making a change to the 'seating', we can make it into a prism.
4. In fig.23.8(a) below, the trough is tilted through 90o, as shown by the green arrow. The final position after tilting, is shown in fig.b
 |
| Fig.23.8 |
•Longer edge = 2 m •Shorter edge = 1.4 m •height = 0.4 m
[At this stage, it is better to have a good understanding about the shapes of such troughs. A presentation can be seen here]
6. The top face is also the same isosceles trapezium
7. The sides are all rectangles. Height of all these rectangles are the same. That is., 0.7 m
8. So all the conditions for a prism are satisfied. We can use the formula.
[Note that we do not need to consider the other dimensions of the side rectangles. Only the height is required]
9. Volume = base area × height
Base area = (a+b)⁄2 × height of isosceles trapezium = (2+1.4)⁄2 × 0.4 = (2+1.4)×0.2 = 3.4×0.2 = 0.68 m2.
So volume = 0.68 × 0.7 = 0.476 m3.
10. We want this volume in litres. We know that, 1 litre = 1000 cm3.
But 1000 cm3 = 0.001 m3.
So 1 litre = 0.001 m3. same as 1 m3 = 1000 litres
Thus we get: Volume of the trough = 0.476 m3 = 0.476 × 1000 = 476 litres
[Moral: A prism need not be always placed with it's base at the bottom]
Now we will see some solved examples:
Solved example 23.1
The base of a prism is an equilateral triangle of perimeter 15 cm and it's height is 5 cm. Calculate it's volume
Solution:
1. Perimeter of the equilateral triangle = 15 cm
2. So one side = 15⁄3 = 5 cm
3. Base area = Area of the equilateral triangle = √3⁄4 × 52 = 6.25√3 = 6.25×1.73 = 10.8125 cm2. Details here.
4. So volume = base area × height = 10.8125 × 5 = 54.0625 cm3.
Solved example 23.2
A hexagonal hole of each side 2 m is dug in the school ground to collect rain water. It is 3 m deep. Some water is now collected in it. The top surface of water is at a depth of 2 m from the top of the pit. How many litres of water is in it?
Solution:
1. The water collected in the pit is in the form of a hexagonal prism.
2. Height of this prism = 3 – 2 = 1 m. This is shown in fig.23.9 below
 |
| Fig.23.9 |
= 6 times the area of an equilateral triangle of side 2 m = 6 × √3⁄4 × 22 = 6√3
= 6 ×1.73 = 10.38 m2. Details here.
4. So volume = base area × height = 10.38 × 1 = 10.38 m3.
5. We want this volume in litres. We know that, 1 litre = 1000 cm3.
But 1000 cm3 = 0.001 m3.
So 1 litre = 0.001 m3. same as 1 m3 = 1000 litres
Thus we get: Volume of the trough = 10.38 m3 = 10.38 × 1000 = 10380 litres
Solved example 23.3
A hollow prism of base a square of side 16 cm contains water 10 cm high. If a solid cube of side 8 cm is immersed in it, by how much would the water level rise?
Solution:
1. The water present is in the form of a square prism.
(i) Height of this prism = 10 cm
(ii) Base area of this prism = 16 × 16 = 256 cm2
(iii) So volume of water present = base area × height = 256 × 10 = 2560 cm3
2. Increase in volume (see fig.23.10 below) = volume of cube = 8 × 8 × 8 = 83 = 512 cm3
 |
| Fig.23.10 |
4. This is the volume of the new prism
5. Base area of the new prism is same as before, which is 256 cm2
6. Let the height of the new prism = h
7. Then 256 × h = 3072 cm3
8. So h = 3072⁄256 = 12 cm
9. Thus, rise in water level = 12 – 10 = 2 cm
No comments:
Post a Comment