Chapter 23.3 - Surface area of Prisms - Solved examples

In the previous section we derived the formula for surface area of prisms. In this section we will see some solved examples.
Solved example 23.4
The base of a prism is an equilateral triangle of perimeter 12 cm. It's height is 5 cm. What is the total surface area?
Solution:
1. We know that lateral surface area = perimeter × height = 12 × 5 = 60 cm²
2. We are asked to find the total surface area. So we need to find two times the 'area of the base'
3. Base is an equilateral triangle of perimeter 12 cm. In an equilateral triangle, all sides are equal. Let this side be 's'. Then: Perimeter = 3s = 12 ⇒ s = 12/3 = 4 cm
4. With this s we can find the area:
Area of the equilateral triangle = ^√3⁄₄ × 4²  = 4√3 cm². Details here.
5. Thus total surface area = 60 + (2×4√3) = (60 + 8√3) cm².

Solved example 23.5
Two identical prisms with right triangles as base are joined to form a rectangular prism as shown in fig.23.14 below. 

Fig.23.14

What is the total surface area of the rectangular prism?
Solution:
Fig.23.14(a) shows one of the two identical prisms. It's base is a right triangle. [Note that we will get a rectangle only if two right triangles are joined. We will not get a rectangle by joining other types of triangles]
1. We will first find the total surface area of one single triangular prism
(i) Lateral surface area = perimeter × height
(ii) Perimeter = (5 + 12 + hypotenuse)
(iii) hypotenuse = √[5² + 12²] = √[25 + 144] = √169 = 13
(iv) So perimeter = (5 +12 +13) = 30 cm
(v) Thus, lateral surface area = 30 × 15 = 450 cm²
(vi) Base area = 1/2 × base × height = 1/2 × 5 × 12 = 30 cm²
(vii) So total surface area of one triangular prism = 450 + (2 × 30) = 450 + 60 = 510 cm²
2. When the two prisms are joined, each of them will not contribute this whole 510 cm². Because, the hypotenuse sides will be concealed
3. Area of one hypotenuse side = 13 × 15 = 195 cm²
4. Area of two such sides = 195 × 2 = 390 cm²
5. So total area of the rectangular prism = 2 × 510 – 390 = 1020 – 390 = 630 cm².

Solved example 23.6
The lateral surface area of a wooden prism of base an equilateral triangle is 48 square centimetres. It's height is 4 cm. Six of these are put together to form a hexagonal prism. How much paper will be required to cover this hexagonal prism completely?

Solution:
1. Fig.23.15(a) given below shows the prism with base an equilateral triangle. It's lateral surface area is given as 48 cm²

Fig.23.15

2. We know that lateral surface area = perimeter × height. 
3. Height is given as 4 cm. So we can write:
48 = perimeter × 4 ⇒ perimeter = 48/4 = 12 cm
4. In an equilateral triangle, all the three sides are equal. Let this side be 's'. The we can write:
Perimeter = 3s = 12 ⇒ s = 12/3 = 4 cm
5. From fig.b it is clear that, side of the hexagonal prism is same as the side of the equilateral triangular prism. So we can write:
Side of the hexagonal prism = s = 4 cm
6. So perimeter of the hexagonal prism = 6 × 4 = 24 cm
7. Height of the hexagonal prism is same as the height of the equilateral triangular prism = 4 cm
So, lateral surface area of the hexagonal prism = perimeter height = 24 × 4 = 96 cm²
8. So we need a paper of 96 cm² area to cover the lateral faces of the hexagonal prism.
9. But we need to cover it completely. That is., top and bottom faces must also be covered. So we need to find the base area. Base area is 6 times the area of the equilateral triangle.
10. Area of the equilateral triangle = ^√3⁄₄ × 4²  = 4√3 cm². Details here.
11. So base area of the hexagonal prism = 6 × 4√3 = 24√3 cm²
12. Total area of top and bottom = 2 × 24√3 = 48√3 cm²
13. Thus the total area of paper required = (96 + 48√3) cm².
14. Taking the value of √3 as 1.73 approximately, we get: (96 + 48√3) = 179
15. We can write: 180 cm² of paper will be required to cover the hexagonal prism completely.

Solved example 23.7
A water trough is in the shape of a prism. It's base is trapezoidal. Dimensions of the trapezium are shown in the fig.23.16(a) below. It's length is 80 cm. It is to be painted inside and outside.

Fig.23.16

How much would be the cost at Rs.100 per square metre?
Solution:
The given fig.23.16(a) shows the dimensions of the base. A 3D view will give a clearer understanding of the problem. 
1. In the fig.23.17(a) below, the trough is resting on a rectangular face. 

Fig.23.17

• The length of this rectangular face is 80 cm
• Width of this rectangular face is 50 cm
• The top face of the trough is also a rectangle 
• The length of this rectangular face is 80 cm
• Width of this rectangular face is 75 cm
2. Though both the top and bottom faces are rectangles, they are not identical. So it seems that it is not a prism. But we can tilt it as shown by the green arrow. After tilting through 90^o, the position will be as shown in fig.23.17(b).
3. Now the trough is resting on a trapezium whose dimensions are those given in fig.23.16(a)
• The top face is also the same trapezium
• It is a prism. We will first find it's total surface area
4. Perimeter of the base:
(i) In fig.23.16(b), the isosceles trapezium ABCD is split into:
    ♦ two right triangles AFD and BEC 
    ♦ a rectangle ABEF
The splitting is done by drawing perpendiculars from A and B  
(ii) Consider any one right triangle, say BEC.
We have: BC = √[BE² + CE²] = √[40² + 12.5²] = √1756.25 = 41.91 cm
5. So we have BC. We can calculate the perimeter of the base:
Perimeter = AB + CD + 2BC = 75 +50 + (2 ×41.91) = 208.82 cm
6. So lateral surface area = perimeter × height = 208.82 × 80 = 16705.6 cm²
7. Now we want the area of the base. This is equal to the area of the isosceles trapezium in fig.a
= [(a+b)/2]h = [(75+50)/2]×40 =  2500 cm²
8. So total surface area of the prism = lateral surface area + 2 × area of base 
= 16705.6 + 2 × 2500 = 21705.6 cm²
9. The prism is to be painted inside and outside. Two times the area = 21705.6 × 2 = 43411.2 
But there is no top surface in fig.23.17(a). We have to deduct 2 times the area of this top surface.
10. So area to be deducted = 2 × 75 × 80 = 12000
11. So net area = 43411.2 – 12000 = 31411.2 cm²
• 1 cm² = 0.0001 m²
• So 31411.2 cm² = 3.14112 m²
12. Cost of painting = 3.14112 × 100 = Rs. 314.112
Solved example 23.8
The base length, and width of a rectangular prism are 37.5 cm and 18 cm respectively. It's height is 40 cm. It is melted and recast into a cube. What is the surface area of the cube?
Solution:
1. First we will find the volume of the rectangular prism:
Volume = base area × height = 37.5 × 18 × 40 = 27000 cm³
2. Volume of the cube will be the same. Let the side of the cube be s cm
3. Then we can write: s³ = 27000 ⇒ s = 30 cm
4. Lateral surface area of a prism = base perimeter × height 
5. For a cube, base perimeter = 4s
6. So lateral surface area = 4s × h = 4 × 30 × 30 = 4 × 30²
7. Total area of base and top face = 2 × 30 × 30 = 2 × 30²
8. So total surface area = (4 × 30²) + (2 × 30²) = 6 × 30² = 5400 cm²

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In the next section we will see cylinders.

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Chapter 23.2 - Surface area of Prisms

In the previous section we completed the discussion on volume of prisms. In this section we will see surface area of prisms.

Consider the triangular prism in fig.23.11(a) below. 
• Base of this prism is a triangle having sides 4, 5 and 6 cm. 
• Height of this prism is 8 cm

Fig.23.11

• But this prism is open at top and bottom. So it is like a tube.
• This tube is formed by three rectangles. They are:
    ♦ Rectangle with height 8 cm, width 4 cm
    ♦ Rectangle with height 8 cm, width 5 cm
    ♦ Rectangle with height 8 cm, width 6 cm

1. We know that there are 3 vertical edges between the above 3 rectangles
2. A cut is made along the edge between the first and third rectangle
3. Then the tube is opened. This is shown in fig.23.11(b)
4. Once we open it like that, we can spread it out on a flat surface. What we get is a perfect rectangle.
• The height of this rectangle is the same 8 cm of the prism
• The length of this rectangle is (4 +5 +6) = 15 cm
• So area of the rectangle = 15 × 8 = 120 cm²
5. But (4 +5 +6) is the perimeter of the base of the original prism 
6. So area of the rectangle = perimeter of the base of the original prism × height of the original prism
7. But from the fig.b, we can see that, area of this rectangle is the area of all the lateral surfaces of the original prism. So we get a formula for finding the lateral surface area of a triangular prism:
■ Lateral surface area of a triangular prism = Perimeter of the base triangle × height of the prism
[Note: In step (2) above, we made a cut along the edge between the first and third rectangle. In fact, we can make the cut along any one of the three vertical edges] 

---

• Consider the case when the prism in fig.a is not a tube. 

• In that case, it will have two extra triangles. One at top and the other at bottom. These extra triangles are shown in blue colour in the fig.23.12 below. 

• Each of these triangles have sides 4, 5 and 6 cm. The ‘spreading out’ will be as shown in fig.23.12(b).

Fig.23.12

• In such a situation, we call it: ‘total surface area of the prism’.
• Obviously, it will be equal to: Lateral surface area + (2 × Area of base triangle)

---

In the above discussion, we have considered triangular prism. We will now consider a quadrilateral prism.

• Consider the quadrilateral prism in fig.23.13(a) below. 
• Base of this prism is a quadrilateral having sides p, q, r and s cm. 
• Height of this prism is h cm

Fig.23.13

• But this prism is open at top and bottom. So it is like a tube.
• This tube is formed by 4 rectangles. They are:
    ♦ Rectangle with height h cm, width p cm
    ♦ Rectangle with height h cm, width q cm
    ♦ Rectangle with height h cm, width r cm
    ♦ Rectangle with height h cm, width s cm

1. We know that there are 4 vertical edges between the above 4 rectangles
2. A cut is made along the edge between the first and fourth rectangles
3. Then the tube is opened. This is shown in fig.23.13(b)
4. Once we open it like that, we can spread it out on a flat surface. What we get is a perfect rectangle.
• The height of this rectangle is the same h cm of the prism
• The length of this rectangle is (p +q +r +s)
• So area of the rectangle = (p +q +r +s)h
5. But (p +q +r +s) is the perimeter of the base of the original prism 
6. So area of the rectangle = perimeter of the base of the original prism × height of the original prism
7. But from the fig.b, we can see that, area of this rectangle is the area of all the lateral surfaces of the original prism. So we get a formula for finding the lateral surface area of a quadrilateral prism:
■ Lateral surface area of a quadrilateral prism = Perimeter of the base quadrilateral × height of the prism

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• Consider the case when the prism in fig.a is not a tube. 

• In that case, it will have two extra quadrilaterals. One at top and the other at bottom. These extra quadrilaterals are shown in blue colour in the fig.23.14 below.  

• Each of these quadrilaterals have sides p, q, r and s cm. The ‘spreading out’ will be as shown in fig.23.14(b).

Fig.23.14

• In such a situation, we call it: ‘total surface area of the prism’.
• Obviously, it will be equal to: Lateral surface area + (2 × Area of base quadrilateral)

---

We have seen the cases of triangle and quadrilateral. We will get the same result for any polygonal prism. We can write it in the form of a theorem.

Theorem 23.2:

• 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 lateral surface area = Perimeter of the base quadrilateral × height
■ Total surface area = Lateral surface area + 2a

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In the next section we will see a few solved examples.

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Chapter 23.1 - Volume of a Prism

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.

Fig.23.6

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 = b_n cm²
5. Let the total area of the polygon in fig.b = a cm²
Then a = (b₁ + b₂ + ... + b_n)
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 = b_n × h = b_nh 
7. So volume of the prism in fig.a = Total volume of all the prisms in fig.c = (b₁h + b₂h + ... + b_nh) 
= (b₁ + b₂ + ... + b_n) × h
8. But from (5) we have a = (b₁ + b₂ + ... + b_n). 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 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. 

Fig.23.7

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 90^o, as shown by the green arrow. The final position after tilting, is shown in fig.b

Fig.23.8

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

Fig.23.9

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

Fig.23.10

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

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In the next section we will see Surface area of prisms.

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Chapter 23 - Solids

In the previous section we completed the discussion on Absolute value. In this chapter we will see details about a few solids.

Fig.23.1 below shows a rectangular block. 

Fig.23.1

Let us try to understand it's features:
1. It sits on a base. This base is a rectangle
2. The top surface of the block is also a rectangle
3. The 'base rectangle' and the 'top surface rectangle' are exactly the same
4. The side faces of the block are all rectangles
5. All the 'side face rectangles' have the same height

Now consider the blocks shown in fig.23.2 below. 

Fig.23.2

Let us first take the block in fig.a. We will write it's features:
1. It sits on a base. This base is a triangle
2. The top surface of the block is also a triangle
3. The 'base triangle' and the 'top surface triangle' are exactly the same
4. The side faces of the block are all rectangles
5. All the 'side face rectangles' have the same height 

Block in fig.b
1. It sits on a base. This base is a quadrilateral
2. The top surface of the block is also a quadrilateral
3. The 'base quadrilateral' and the 'top surface quadrilateral' are exactly the same
4. The side faces of the block are all rectangles
5. All the 'side face rectangles' have the same height

Block in fig.c
1. It sits on a base. This base is a polygon (When we say 'polygon', it can have any number of sides)
2. The top surface of the block is also the same polygon
3. The 'base polygon' and the 'top surface polygon' are exactly the same
4. The side faces of the block are all rectangles
5. All the 'side face rectangles' have the same height

---

So, even though the solids are of different shapes, we see some common features:

• The base and top surface are always identical
• All the side faces are rectangles
    ♦ All these rectangles have the same height
■ All solids which have the above features come under the category of Prisms

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If we are given a prism, we can further classify it. This classification is based on the shape of it's base. Thus we get:

• If the base of a prism is a triangle, it is called a triangular prism
• If the base of a prism is a rectangle, it is called a rectangular prism
• If the base of a prism is a pentagon, it is called a pentagonal prism
• If the base of a prism is a hexagon, it is called a hexagonal prism
So on ...
■ The polygons at the top and bottom are called the bases
■ The rectangles on the sides are called the lateral faces
■ The bases and lateral faces are together called faces

---

Now we will try to calculate the volume of a prism:

• Of all the different types of prisms available to us, the 'rectangular prism' is the easiest one, as far as the 'calculation of volume' is concerned.
• We can derive the formula for volume of any prism, based on the rectangular prism. So we will first see the 'volume of rectangular prism' in detail.

In the fig.23.3 (a) below, a rectangular prism sits on it's base. The base is a rectangle of length 7 cm and width 3 cm.

Fig.23.3

So we can write:
Length of the rectangular prism = 7 cm
Width of the rectangular prism = 3 cm
Height of the rectangular prism = 20 cm

We know that volume of such a solid is length width height
So the volume of the prism in fig.a is 7 × 3 × 20 = 420 cm³

■ Now consider fig.b. It shows the same prism that we saw in fig.a. But it's seating has changed. Now it sits on a rectangular base, whose length is 20 cm and width is 7 cm. So we can write:
Length of the rectangular prism = 20 cm
Width of the rectangular prism = 7 cm
Height of the rectangular prism = 3 cm
So the volume of the prism in fig.b is 20 × 7 × 3= 420 cm³

■ Now consider fig.c. It shows the same prism that we saw in figs.a and b. But it's seating has changed. Now it sits on a rectangular base, whose length is 20 cm and width is 3 cm. So we can write:
Length of the rectangular prism = 20 cm
Width of the rectangular prism = 3 cm
Height of the rectangular prism = 7 cm
So the volume of the prism in fig.b is 20 × 3 × 7= 420 cm³

---

We find that the volume is the same in all the three cases. We can write this:

(7 × 3) × 20 = (20 × 7) × 3 = (7 × 3) × 20 = 420 cm³.
• In all the three cases, the quantity inside the brackets is 'area of the base of the prism'.
• The quantity outside the brackets is 'height of the prism'
■ So we get a formula for finding the volume of a rectangular prism. We can write:
Volume of a rectangular prism = Area of the base × height

We saw that, all types of prisms (triangular, quadrilateral, hexagonal so on...) will have a base and a height. Can we apply the same formula for these prisms also?

Let us find out:
• Fig.23.4 (a) shows a right triangular prism. [Note the term 'right triangular'. It means that, the base and top surface are right triangles]. We want to find the volume of this prism.

Fig.23.4

• Let us make an exact replica of this prism. It is shown in fig.b.
• If we join the two prisms together as shown in fig.c, we will get a rectangular prism
• The final rectangular prism is shown in fig.d

Now we can try to find the volume
1. Let the 'area of the base' of the triangular prism in fig.a be 'a' cm
2. Then the 'area of the base' of the final rectangular prism in fig.d = 2a
3. Let the height of the triangular prism in fig.a be 'h' cm
4. Then the height of the final rectangular prism in fig.d is also 'h' cm
5. From (2) and (4) we get:
Volume of the rectangular prism in fig.d = 2a × h = 2ah cm³
6. But this rectangular prism is made up of two identical triangular prisms. 
So volume of one triangular prism = volume of the triangular prism in fig.a = half of 2ah = ah cm³
7. But 'ah' is the product of the area of the base of the triangular prism and it's height. So we can write:
Volume of a right triangular prism = Area of the base × height

---

■ Thus we have the same formula for: 
• A rectangular prism
• A right triangular prism
Now let us check whether it will work for ‘any triangle’:
1. Fig.23.5 (a) shows a triangular prism. It’s base is shown separately in fig.b.

2. In the fig.b, the triangle is split into two right triangles. [Note that, any triangle can be split into two right triangles. All we have to do is, drop a perpendicular from the top vertex to the base] 

3. So the prism in fig.a can also be split into two. Each will have a right triangular base. This splitting is shown in fig.c
4. Let in fig.b, 
• area of one right triangle = b cm²
• area of the other right triangle = c cm²
• Let the total area of the triangle in fig.b = a cm²

• Then a = b + c
5. We can calculate the volume of the two prisms in fig.c separately. Because, each of them have right triangular base. Thus:
• Volume of one triangular prism in fig.c = area height = b × h = bh 
• Volume of the other triangular prism in fig.c = area height = c × h = ch
• Volume of the triangular prism in fig.a = Total volume of the two prisms in fig.c = bh + ch = (b+c)h
6. But from (4) we have (b+c) = a
7. So we get: Volume of triangular prism in fig.a = ah
Thus we find that, the formula can be applied to any triangular prism.

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■ So we have the same formula for: 
• A rectangular prism
• A triangular prism [The base can be any triangle]

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In the next section we will check whether it will work for ‘any polygon’.

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