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.
• 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
We can write it in the form of a theorem.
Theorem 23.4:
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
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
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:
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(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
Consider the cylinder in fig.23.21(a) below.
• Base of this cylinder is a circle.
 |
| Fig.23.21 |
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
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• Area 'a' of the base is πr2 cm2
■ Total surface area = (Lateral surface area + 2a) cm2.
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?
Solved example 23.14
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
We have completed the discussion on surface area of cylinders. We will now see some additional problems related to the whole chapter
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 |
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(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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