Chapter 23.5 - Surface area of Cylinders

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

Consider the cylinder in fig.23.21(a) below.
• Base of this cylinder is a circle.

Fig.23.21

• But this cylinder is open at top and bottom. So it is like a tube.
1. A vertical cut is made on the lateral surface of the cylinder. This is indicated by the white line in fig.b
2.Then the tube is opened. This is shown in fig.c
3. 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 height of the cylinder. Let this be 'h'
• The length of this rectangle is the perimeter of the base circle. Let this be 'p'
• So area of the rectangle = p × h = ph cm²
4. So area of the rectangle = perimeter of the base of the original cylinder × height of the original cylinder
5. But from the fig.c, we can see that, area of this rectangle is the area of the lateral surface of the original cylinder. So we get a formula for finding the lateral surface area of a cylinder:
■ Lateral surface area of a cylinder = Perimeter of the base circle × height of the cylinder

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We can write it in the form of a theorem.
Theorem 23.4:

• We have a cylinder of height 'h' cm

• The radius of the base circle is 'r' cm

• Perimeter 'p' of the base is 2πr  cm
• Area 'a' of the base is πr²  cm²

■ Then the lateral surface area of the cylinder = ph cm²

■ Total surface area = (Lateral surface area + 2a) cm².

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Solved example 23.14

The inner diameter of a well is 2.5 m, and it is 8 m deep. What would be the cost of cementing it’s inside at Rs 350 per m²?
Solution:
1. Diameter of the well = 2.5 m. So radius r = 2.5/2 = 1.25 m
2. Perimeter = 2πr = 2π × 1.25 = 2.5π m
3. Surface area = perimeter × height = 2.5π × 8 = 20π = 20 × 3.14 = 62.8 m².
4. Cost of cementing = 62.8 × 350 =Rs. 21980 

Solved example 23.15
Diameter of a road roller is 80 cm, and it is 1.2 m long. What is the area of the levelled surface when it rolls once?
Solution:
1. Diameter of the roller = 80 cm. So radius r = 80/2 = 40 cm
2. Perimeter = 2πr = 2π × 40 = 80π cm

3. Area of levelled surface when the roller rolls once = Surface area of the roller
 = perimeter × length = 80π × 120 = 30144 cm² = 3.0144 m².

Solved example 23.16
The base area and the lateral surface area of a cylinder are equal. What is the ratio of the base radius and height?
Solution:
1. Let the radius be 'r' and height be 'h'
2. Then base area = πr².
3. Perimeter of base = p = 2πr.
4. Lateral surface area = ph = 2πrh.
5. Given that base area is equal to the lateral surface area. So we get: πr² = 2πrh.
⇒ r = 2h ⇒ ^r⁄_h = 2 ⇒ ^r⁄_h = ²⁄₁
6. So the ratio r:h = 2:1
7. This result can be used as a general case:
Whenever the radius of the base of a cylinder is two times it's height,
Base area = Lateral surface area

Solved example 23.17
The base area and curved surface area of a cylinder are equal. If the base area is 314 m²,  What is it’s height?
Solution:
1. Let the radius be 'r' and height be 'h'
2. Then base area = πr².= 314 m²

⇒ 3.14 × r² = 314 ⇒ 3.14 × r² = 3.14 × 100 ⇒ r² = 100 ⇒ r = 10 m

3. Perimeter of base = p = 2πr = 2π × 10 = 20π.
4. Lateral surface area = ph = 20πh.
5. Given that base area is equal to the lateral surface area. So we get: 314 = 20πh
⇒ 3.14 × 100 = 20 ×3.14 × h ⇒ 100 = 20h ⇒ h = 5 m
Just like in the previous example, here also we get r = 2h when base area is equal to lateral surface area

Solved example 23.18
There are 18 cylindrical pillars in an auditorium. Each pillar has a diameter of 20 cm and height 2.5 m. Calculate the cost of painting all the pillars at Rs.80 per m²
Solution:
1. Diameter of one pillar = 20 cm. So radius r = 20/2 = 10 cm
2. Perimeter = 2πr = 2π × 10 = 20π cm
3. Surface area = perimeter × height = 20π × 250 = 5000π = 5000 × 3.14 = 15700 cm² = 1.57 m².
4. Cost of painting one pillar = 1.57 × 80 = Rs. 125.6
5. Cost of painting 18 pillars = 18 × 125.6 = Rs. 2260.8

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We have completed the discussion on surface area of cylinders. We will now see some additional problems related to the whole chapter

Solved example 23.19
A rectangular sheet of paper 8 × 6 cm, is bent along it’s longer side to make a cylinder. What is the radius of the circle required to close it’s base?
Solution:
The method of bending is shown in the fig.23.22 below:

Fig.23.22

Based on the fig., we can write the steps
1. The rectangle is bent along the longer edge. So the 8 cm long edge becomes the perimeter of the base. We can write:
2. 2πr = 8 cm same as πr = 4 cm same as r = ⁴⁄_π cm
3. So area of the base = πr² = π(⁴⁄_π)² = ¹⁶⁄_π cm².

Solved example 23.20
The volume of a square prism is 2420 cm³. It’s height is 20 cm. Find it’s total surface area
Solution:
1. Let the side of the base of the cube be 's' cm
2. Then volume = s² × h = s² × 20 =  2420
⇒ s² = 2420/20 = 121 ⇒ s = √121 = 11 cm
3. Perimeter of base = 4s = 4 × 11 = 44 cm
4. Lateral surface area = perimeter × height = 44 × 20 = 880 cm².
5. Area of base = s² = 121 [From (2)]
6. So total surface area = 880 + 2 × 121 = 1122 cm². 

Solved example 23.21
Crushed rocks for the construction of a road are heaped in the form of a trapezoidal prism. It’s two end faces are isosceles trapeziums, with upper width 2 m and lower width 3 m. The height of the heap is 1.5 m. The length of the prism is 15 m. What is the volume of the crushed rocks in the heap?
Solution:
In the presentation about shape of water troughs, we saw two types of troughs. If those troughs are placed upside down, we will get the shape of 'crushed rock heap'. The upside down positions of the two troughs are shown in the fig.23.23 below:

Fig.23.23

• Fig.23.23(a) shows the upside down position of type 1 trough in the presentation 
• Fig.23.23(b) shows the upside down position of type 2 trough in the presentation 
■ But there is a problem. We will not be able to heap crushed rock as shown in the fig.23.23(a). This is because, the shorter sides of the heap are exactly vertical. We cannot heap with those sides vertical. The crushed rocks will roll downwards. So we can heap only as in fig.23.23(b).
■ But in this problem, type 1 shown in fig.23.23(a) is specified. Because it can be considered as a prism. The type 2 shown in fig.23.23(b) cannot be considered as a prism. We will learn to calculate it's volume in higher classes.

So let us calculate the volume of the heap in fig.23.23(a):
1. Volume = base area × height
2. Base area = area of trapezium = ¹⁄₂ × (Upper width + lower width) × height of trapezium
=  ¹⁄₂ × (2 + 3) × 1.5 = 2.5 × 1.5 = 3.75 m²
3. Height of prism = length of heap = 15 m

4. So volume = Base area × height = 3.75 × 15 = 56.25 m³ 

Solved example 23.22
The sum of the base areas of a rectangular prism is equal to it’s lateral surface area. If it’s base has a length of 28 cm and width 22 cm, find it’s height
Solution:
1. Length of base = 28 cm. Width of base = 22 cm
2. Area of base = 28 × 22 = 616 cm².
3. Sum of base areas = 2 × 616 = 1232 cm²
4. Given that sum of base areas = lateral surface area. So we can write:
lateral surface area = perimeter of base × height = 1232 cm².
5. Perimeter of base = 2(28+22) = 2 × 50 = 100 cm
6. Substituting this value of perimeter in (4) we get:
7. 100 × height = 1232 ⇒ height = 1232/100 = 12.32 cm

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In the next chapter we will see Proportions.

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Chapter 23.4 - Properties and Volume of Cylinders

In the previous section we completed the discussion on surface area of prisms. In this section we will see cylinders.
We have seen the properties of prisms: 
• They have identical polygons at the base and top surface. 
• They have rectangles on sides.
Now we consider a special type of the above properties. In this type,
• Instead of polygons at base and top, we have circles
• Instead of rectangles on sides, we have one smooth curved surface
Such a solid is called a cylinder. Some examples can be seen here.

We will now try to find the volume of a cylinder. Consider fig.23.18(a) below. 
1. A triangular prism is inscribed inside a cylinder. 

Fig.23.18

• The base of this triangular prism is an equilateral triangle. An equilateral triangle is a regular polygon.
• We can see that the volume of the triangular prism is very less than the volume of the cylinder
2. Consider fig.23.18(b). The number of lateral sides of the inscribed prism is now increased to 5. 
• So the base is a regular pentagon. 
• We can see that the volume of the inscribed prism in fig.b is closer to the volume of the cylinder.
3. Consider fig.23.18(c). The number of lateral sides of the inscribed prism is now increased to 10. 
• So the base is a regular decagon. 
• We can see that the volume of the inscribed prism in fig.c is very close to the volume of the cylinder.
■ So we can make a conclusion:
If the number of lateral sides (denoted as 'n') of the regular polygonal prism is very large, it's volume will be very equal to the volume of the cylinder of same height

This information can be used to calculate the volume of a cylinder. So let us take a closer look at the fig.23.18(c). This is shown in fig.23.19(a) below

Fig.23.19

1. The cylinder and inscribed prism in fig.23.19(a) is the same in fig 23.18(c)
2. The regular polygon at the base, can be split into triangles. All we need to do is, join the centre to each vertex. This is shown in fig.23.19(b)
3. All triangles are equal. So they have the same area. Let this area be denoted as 'p'
4. If the height of cylinder and the prism is 'h', the volume of one triangular prism = ph
5. If there are n sides, there will be n triangular prisms. So total volume = nph
6. If n is very large, two things will happen:
(i) The total volume of the triangular prisms will become equal to the volume of the cylinder
    ♦ That means volume of the cylinder = nph
(ii) The 'total area of all the base triangles' will become equal to the 'area of the base circle' of the cylinder. [We have seen this situation when we discussed area of circles. See fig.21.19]
7. But total area of all the base triangles = np
8. So np = area of the base circle of the cylinder 
9. Substituting this value 'np' in 6(i) we get:
Volume of the cylinder = nph = area of base circle × h

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We can write it in the form of a theorem:

Theorem 23.3:

• We have a cylinder of height 'h' cm

• The base is a circle of area 'a' cm²

• Then the volume of the cylinder is ah cm³

Note: If the radius of the base circle is 'r' cm, then we know that 'a' =  πr² cm²

Then volume of the cylinder = πr²h cm³

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We will now see some solved examples

Solved example 23.9

The base radius of an iron cylinder is 15 cm and it's height is 32 cm. It is melted and recast into a cylinder of base radius 20 cm. What is the height of this cylinder?

Solution:

1. First cylinder:

base radius r1 = 15 cm,

height h1 = 32 cm

So volume v1= πr²h = π × 15² × 32 = 7200π cm³

2. Second cylinder:

Volume will be the same. So v2 = v1 = 7200 cm³

base radius r2 = 20 cm

• 7200π = π × 20² × h2 ⇒ 7200 = 400 × h2 ⇒ h2 = ⁷²⁰⁰⁄₄₀₀ = 18 cm

Solved example 23.10

The base radii of two cylinders are in the ratio 3:4. Their heights are equal. What is the ratio of their volumes?

Solution:

1. First cylinder:

base radius = r1

height = h1

So volume v1= πr²h = π(r1)²h1

2. Second cylinder:

base radius = r2

height = h2

So volume v2= πr²h = π(r2)²h2

3. v1:v2 = π(r1)²h1 : π(r2)²h2

But h1 = h2. So we can write: 

4. v1:v2 = π(r1)²h1 : π(r2)²h1 ⇒ v1:v2 = (r1)² : (r2)²

⇒ ^v1⁄_v2 = [^r1⁄_r2]² ⇒ ^v1⁄_v2 = [³⁄₄]² (∵ ^r1⁄_r2 = ³⁄₄)

⇒ ^v1⁄_v2 = [⁹⁄₁₆]

So the ratio v1 : v2 is 9:16

Solved example 23.11

The base radii of two cylinders are in the ratio 2:3. Their heights are in the ratio 5:4.

(i) What is the ratio of their volumes?

(ii) The volume of the first cylinder is 720 cm³. What is the volume of the second?

Solution:
Part (i)

1. First cylinder:

base radius = r1

height = h1

So volume v1= πr²h = π(r1)²h1

2. Second cylinder:

base radius = r2

height = h2

So volume v2= πr²h = π(r2)²h2

3. v1:v2 = π(r1)²h1 : π(r2)²h2

⇒ ^v1⁄_v2 = [^r1⁄_r2]² × [^h1⁄_h2] ⇒ ^v1⁄_v2 = [²⁄₃]² × [⁵⁄₄] = [⁴⁄₉] × [⁵⁄₄] = ⁵⁄₉.

So the ratio v1 : v2 is 5:9
Part (ii)
1. Ratio of the volumes is 5:9
2. So ^v1⁄_v2 = ⁵⁄₉. 
3. Given that v1 = 720. So we can write:
⁷²⁰⁄_v2 = ⁵⁄₉. same as v2 = ^{(720 × 9)}⁄₅ = 1296 cm³.
Another method:
1. Ratio of the volumes = 5:9. So if the total volume of the two cylinders is split into 14 equal part, the first cylinder will take up 5 such parts. Also, the second cylinder will take up 9 such parts.
2. Let 'k' be the total volume. Then each of the 14 equal parts will be ^k⁄₁₄.
3. 5 such parts = ^(5k)⁄₁₄.
4. So = ^(5k)⁄₁₄ = 720 ⇒ k = ^{(720 × 14)}⁄₅ = 2016 cm³.
5. So volume of second cylinder = 2016 - 720 = 1296 cm³

Solved example 23.12

A barrel in the form of a cylinder is full of oil. When 70.65 litres of oil was taken from it, it's level was decreased by 25 cm. 

(i) Find the radius of the barrel. 

(ii) If the total height of the barrel is 80 cm, what is the volume of the remaining oil? 

Solution:

Part (i):

1. Let the radius of the barrel be 'r' cm

2. 70.65 litres of oil will form a cylinder of height 25 cm at the top of the barrel

3. So volume of a cylinder with radius r and height 25 cm is 70.65 litres

4. 1 litre = 1000 cm³. So 70.65 litres = 70.65 × 1000 = 70650 cm³

5. Thus we can write: 70650 = π × r² × 25 ⇒ r²= ⁷⁰⁶⁵⁰⁄_{(3.14 × 25)} = 900

⇒ r = √900 = 30 cm

Part (ii)

1. Given that total height = 80 cm

2. So remaining height = 80 – 25 = 55 cm

So volume of remaining oil = π × 30² × 55 = 155430 cm³ = 155.43 litres

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We will now see another type of problem:

• Fig.23.20 shows a cylinder inscribed in a square prism. 

Fig.23.20

• Note that, the prism has a square base. Every square will have a corresponding inscribed circle. This is the largest possible circle that can be drawn inside that square. So, every square prism will have the largest possible cylinder that can be inscribed inside it. 
• Obviously, the diameter of the base of the cylinder will be equal to the side 's' of the square. So radius is half the side of the square. 
• Once we have the radius, we can calculate the area of the base of the cylinder. 
• And once we calculate that area, we can calculate the volume of the cylinder.

1. If, in the fig.23.20 above, side of the square is 10 cm, radius of the circle = ¹⁰⁄₂ = 5 cm
2. Area of the circle = πr² = π× 5² = 25π

3. If the height h = 20 cm,
Volume of the cylinder = area × height = 25π × 20 = 500π cm².

Solved example 23.13

The base length and height of a wooden square prism are 12 cm and 70 cm respectively. What is the volume of the cylinder of maximum size that can be made from it?
Solution:
1. Side of the square is 10 cm, radius of the circle = ¹²⁄₂ = 6 cm
2. Area of the circle = πr² = π× 6² = 36π

3. Height h = 70 cm,
Volume of the cylinder = area × height = 36π × 70 = 2520π cm³.

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In the next section we will see surface area of cylinders.

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