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.
• 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
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
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 |
• 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 |
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
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
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)
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
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
• 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.
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.
No comments:
Post a Comment