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.
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
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.
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.
• 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 90o, 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
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 |
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 |
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 |
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 |
• 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 90o, 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
Thank you u r great
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