Chapter 33.1 - Height of a Square pyramid to find it's lateral surface area

In the previous section we saw the calculation of the surface area of square pyramids. In this section, we will see a few solved examples. Later in this section, we will see how the height of a square pyramid can be used to find it's lateral surface area.

Solved example 33.1
A square of side 5 cm and four isosceles triangles of base 5 cm and height 8 cm are to be put together to make a square pyramid. How many square centimetres of paper is needed?
Solution:
1. Base area = 5 × 5 = 25 cm²
2. Area of one isosceles triangle = ¹⁄₂ × base × altitude =  ¹⁄₂ × 5 × 8 =  20 cm²
3. Area of four such isosceles triangles = 4 × 20 = 80 cm²
4. Total surface area = 25 + 80 = 105 cm² 

Solved example 33.2
A toy is in the shape of a square pyramid of base edge 16 cm and slant height 10 cm. What is the total cost of painting 500 such toys, at Rs 80 per square metre?
Solution:
1. Base area = 16 × 16 = 256 cm²
2. Area of one isosceles triangle = ¹⁄₂ × base × altitude =  ¹⁄₂ × 16 × 10 =  80 cm²
3. Area of four such isosceles triangles = 4 × 80 = 320 cm²
4. Total surface area = 256 + 320 = 576 cm²
5. Total surface area of 500 toys = 500 × 576 = 288000 cm² = 28.8 m²
6. Cost of painting 500 toys = 28.8 × 80 = Rs. 2304/- 

Solved example 33.3
The lateral faces of a square pyramid are equilateral triangles and the base edge is 30 cm. What is it's surface area?
Solution:
• Given that the lateral faces are equilateral triangles
    ♦ So the base edge of the pyramid will be equal to the lateral edge. 
    ♦ Thus we get: Lateral edge of the pyramid = 30 cm
• So the lateral faces are equilateral triangles of side 30 cm
    ♦ Area of an equilateral triangle = ^(√3×s2)⁄₄ 
    ♦ Where s is the side of the equilateral triangle (See derivation here)
1. So we get:
Area of one lateral face = ^(√3×302)⁄₄ = 225√3 cm²
2. Area of the four lateral faces = 4 × 225√3 = 900√3 cm²
3. Area of base = 30 × 30 = 900 cm²
Total surface area = 900 + 900√3 = 900(1+√3) cm²  

Solved example 33.4
The perimeter of the base of a square pyramid is 40 cm and the total length of all it's edges is 92 cm. Calculate it's surface area.
Solution:
1. Let b be the base edge and l the lateral edge. Then we can write:
4b + 4l = 92 cm
2. But 4b is given as 40 cm. So we get b = ⁴⁰⁄₄ = 10 cm
3. Substituting this value of b in (1) we get: 40 + 4l = 92 ⟹ 4l = 52 ⟹ l = 13 cm
4. Altitude of one isosceles triangle = √[13²  - 5²] = √[169  - 25] = √[144] = 12 cm
• Now we can calculate the total surface area
5. Base area = 10 × 10 = 100 cm²
6. Area of one isosceles triangle = ¹⁄₂ × base × altitude =  ¹⁄₂ × 10 × 12 =  60 cm²
7. Area of four such isosceles triangles = 4 × 60 = 240 cm²
8. Total surface area = 100 + 240 = 340 cm²

Solved example 33.5
Can we make a square pyramid with the lateral surface area equal to the base area?
Solution:
1. Consider the base of a square pyramid. Let it be a square of side a. This is shown in fig.33.10(a) below:

Fig.33.10

2. In the fig.a, P is the apex of the pyramid
• Q and R are the midpoints of two opposite sides of the base
• The diagonals of the base intersect at O
3. Consider the right triangle POQ
• Obviously, the hypotenuse of ⊿POQ is the slant height of the pyramid. Let it be l
4. Now we can calculate the areas:
(a) Base area = a²
(b) Area of one isosceles triangle on the lateral surface = ¹⁄₂ × base × altitude 
=  ¹⁄₂ × a × l = ¹⁄₂ × (al)
(c) Area of four isosceles triangles = 4 × ¹⁄₂ × (al) = 2al.
• So total lateral surface area = 2al
5. Let us assume that the two areas are equal. That is:
Base area = Lateral surface area
• Then we can write: a² = 2al
⟹ a = 2l.
6. So we can write:
• If the lateral surface area of a square pyramid is same as it's base area, then, 'twice the slant height' will have to be equal to the 'base edge'
• Let us see if such a situation is possible:
7. Consider the measurements in fig.b
• The apex P is now brought down to P'. That is., height of the pyramid is now smaller than that in fig.a
• P' is closer to O
• Consequently, l is reduced to l'.
• The length l' is now closer to ^a⁄₂.
• This is same as: 2l' is now closer to a. (∵ QR = a)
8. If we bring down P' even more closer to O, l' will become even more closer to ^a⁄₂
• If we go on lowering the apex, the stage will reach when the apex and O coincides.
• At that stage, twice the slant height will be exactly equal to a
• But such a situation is of no use to us. because, when P and O coincides, we no longer have a pyramid. It is just a plane
9. So it is clear:
• If there is to be a pyramid, P should not coincide with O and then 2l will not be equal to a
So our assumption in (5) is wrong. We can write:
• If there is to be a square pyramid, the base area can never be equal to it's lateral surface area

How to use the height of a Square pyramid to calculate lateral surface area

We have earlier seen how to determine the height of a pyramid. See fig.33.6 in the previous section.
• For a square pyramid, this height has much significance. If we know the base edge and height of a square pyramid, we can easily make that pyramid. 
• No other property of that pyramid is required 
Let us see an example:
■ It is decided to make a tent in the shape of a square pyramid. The base must be 6 × 6 m square and the height must be 4 m. How can the tent be erected?
Solution:
1. Mark a square of side 6 m on the ground. This is shown in fig.33.11(a) below.

Fig.33.11

2. Draw the diagonals. Mark the point of intersection of the two diagonals as O
3. Erect a straight pole of length 4 m at O
4. The top end of the pole is the apex. Join the apex to the four corners of the base square
• With the above four steps, we get the basic frame of the pyramidal tent. 
■ So a square pyramid can be completely defined by just two items:
• The base edge
• The height

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There are lot of technical specifications for making a tent. We are not concerned about them in our present discussion. 
• But we will surely want to know the lateral surface area. Then only we will be able to procure enough canvas sheet to cover the tent.
• Calculation of lateral surface area is easy. We do not have to take actual measurements on the tent frame. 
• We can calculate it just by using the base edge and height. Let us write the steps:
1. We have seen that, to calculate the lateral surface area, we need the slant height.
• Consider fig.33.11(b). A red triangle OAB is drawn. It is right angled at O. 
• OB is the height of the tent which is 4 m. 
• A is the midpoint of a base edge. Clearly, AB is the slant height. 
2. But to calculate AB using Pythagoras theorem, we need to know OA first.
• It can be easily seen that, OA is same as CD, the half of base edge
• Base edge = 6 m. So CD = OA = 3 m
• So we can calculate the slant height:
AB = √[4² + 3²] = √[16 + 9] = √[25] = 5 m    
3. So area of one isosceles triangle = ¹⁄₂ × base × altitude =  ¹⁄₂ × 6 × 5 =  15 m²
• So area of 4 isosceles triangles = 4 × 15 = 60  m². This is the lateral surface area. 
• But we will have to procure a little more than 60 sq.m. The extra quantity required will be mentioned in the technical specifications

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■ The above problem gives us the method to use height for calculating lateral surface area. The trick is to imagine a red triangle OAB inside the pyramid. This is shown in fig.33.11(c) 
• It should satisfy the following conditions:
    ♦ OAB must be right angled
    ♦ O must coincide with the centre of the base
    ♦ A must coincide with the midpoint of a base edge
    ♦ B must coincide with the apex
• If the above conditions are satisfied,
    ♦ The base OA of ⊿OAB will be half the base edge
    ♦ The altitude OB will be the height
   ♦ The hypotenuse AB will be the slant height, which can be easily computed using Pythagoras theorem

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Now we will see a solved example
Solved example 33.6
Part (i)
A square pyramid has the following dimensions:
(a) Side of the base is 24 cm
(b) Height of each of the four isosceles triangles is 18 cm
Calculate the height of that pyramid 
Part (ii)
A square pyramid has the following dimensions:
(a) Side of the base is 24 cm
(b) The equal sides of each of the four isosceles triangles is 30 cm
Calculate the height of that pyramid
Solution:
Part (i):
1. Imagine the ⊿OAB inside the pyramid.
• It should satisfy the following conditions (see fig.33.12.a):
    ♦ OAB must be right angled
    ♦ O must coincide with the centre of the base
    ♦ A must coincide with the midpoint of a base edge
    ♦ B must coincide with the apex

Fig.33.12

• We will get:
    ♦ OB = height of pyramid
    ♦ OA = half of base edge = 12 cm
    ♦ AB = slant height = 18 cm
2. Applying Pythagoras theorem, we get:
OB = √[18² - 12²] = √[324 - 144] = √[180] = √[5×4×9] = √5×√4×√9 = 2×3√[5] = 6√5 cm
Part (ii):
1. Imagine ⊿OCB inside the pyramid.
• It should satisfy the following conditions (see fig.33.12.b):
    ♦ OCB must be right angled
    ♦ O must coincide with the centre of the base
    ♦ C must coincide with a corner the base square
    ♦ B must coincide with the apex
• We will get:
    ♦ OB = height of pyramid
    ♦ OC = half of a diagonal of the base 
    ♦ BC = Lateral edge = 30 cm
2. Applying Pythagoras theorem, we get:
OB = √[BC² - OC²] = √[30² - OC²]
3. Now OC = half of the diagonal
Full diagonal (see fig.33.12.c)= √[24² + 24²] = 24√2 cm 
So half diagonal = OC = 12√2
4. Substituting this value of OC in (2), we get:
OB = √[30² - (12√2)²] = √[900-(144×2)] = √[612] cm

Solved example 33.7
A square pyramid of base edge 10 cm and height 12 cm is to be made of paper. What should be the dimensions of the triangles?
Solution:
1. All the four triangles must be equal and isosceles
• The base of all the triangles must obviously be 10 cm. 
    ♦ We have to find the dimension of the equal sides
2. Imagine the ⊿OAB inside the pyramid.
• It should satisfy the following conditions (see fig.33.13.a):
    ♦ OAB must be right angled
    ♦ O must coincide with the centre of the base
    ♦ A must coincide with the midpoint of a base edge
    ♦ B must coincide with the apex

Fig.33.13

• We will get:
    ♦ OB = height of pyramid = 12 cm
    ♦ OA = half of base edge = 5 cm
    ♦ AB = slant height 
3. Applying Pythagoras theorem, we get:
AB =  √[OA² + OB²] = √[5² + 12²] = √[25 + 144] = √[169] = 13 cm
4. So we have an isosceles triangle BCD with base CD as 10 cm and height AB as 13 cm (see fig.33.13.b)
• Length of it's equal sides (BC and BD) = √[5² + 13²] = √[25 + 169] = √[194] cm

Solved example 33.8
Prove that in any square pyramid, squares of the height, slant height and lateral edge are in arithmetic sequence
Solution:
1. In this problem, we will take the base edge as '2a' So half of base edge = a
Let the height of the pyramid be h
See fig.33.14.a below:

Fig.33.14

2. Then square of the slant height = l² = (a²+h²)
3. Square of diagonal = [(2a)²+(2a)²] = [4a²+4a²] = 8a²
Diagonal = √[8a²] = [(2√2)a] see fig.33.14.b
Half of diagonal = (√2)a
Square of 'half of the diagonal' = [(√2)a]² = 2a²
4. So square of 'lateral edge' = (2a²+h²). See fig.33.14.c
5. Now we can write the sequence:
square of height, square of slant height, square of lateral edge
⟹ h², (a²+h²), (2a²+h²)
5. Third term - second term = (2a²+h²) - (a²+h²) = a²
• Second term - first term = (a²+h²) - h² = a²
• So it is an arithmetic sequence with first term h² and common difference a²

Solved example 33.9
Consider an isosceles triangle of base 30 cm and equal sides 25 cm. A square pyramid is to be made with that triangle. What would be it's height? What if the base edge is 40 cm instead of 30 cm
Solution:
Part (i):
1. The given isosceles triangle has a base of 30 cm. So the base edge of the square pyramid is 30 cm. See fig.33.15(a)

Fig.33.15

2. The given isosceles triangle has equal sides of 25 cm. So the lateral edge of the square pyramid is 25 cm
3. Half of the base diagonal (see fig.33.15.b) = OC = ¹⁄₂ × {√[30² + 30²]} = ¹⁄₂ × {√[2×900]} = ¹⁄₂ × {30 ×√2} =  15√2 cm
4. Now, back in fig.a, h² = [25² + (15√2)²] =  [625 + 450] = 1075
⟹ h = √1075
Part (ii):
1. The given isosceles triangle has a base of 40 cm. So the base edge of the square pyramid is 40 cm. See fig.33.15(c)
2. The given isosceles triangle has equal sides of 25 cm. So the lateral edge of the square pyramid is 25 cm
3. Half of the base diagonal (see fig.33.15.d) = OC = ¹⁄₂ × {√[40² + 40²]} = ¹⁄₂ × {√[2×1600]} = ¹⁄₂ × {40 ×√2} =  20√2 cm
4. Now, back in fig.a, h² = [25² + (20√2)²] =  [625 + 800] = 1425
⟹ h = √1425 cm

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In the next section, we will see volume of pyramids

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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 23.3 - 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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