High school Mathematics Lessons: Base edge

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Wednesday, January 24, 2018

Chapter 33.2 - Volume of Square pyramids

In the previous section we saw how to use the height of a square pyramid to calculate it's lateral surface area. We also saw some solved examples. In this section, we will see volume of a square pyramid.

We know how to calculate the volume of prisms. Simply multiply the base area of the prism with it's height. To find the volume of a square pyramid, we will do an experiment:
1. Make a square pyramid using thick paper. See fig.33.16(a) below:

Volume of a square pyramid will be one third of that of a square prism having same base area and height.

Fig.33.16

2. Make a square prism having the same height and same base area. (fig.33.16.b)
• At this stage, both the pyramid and prism are empty. This is shown in fig.33.17(a) below:

Fig.33.17

3. Fill the pyramid completely with sand
• Now the pyramid is full with sand. But the prism is empty. This is shown in fig.33.17(b)
4. Transfer this sand into the prism.
• Now the pyramid is empty. But the prism contains some sand. This is shown in fig.33.17(c)
• Some thing is not right. The pyramid was completely full. All the sand was transferred. But the prism is only partially full
• But this is natural. It happens because the volume of a square pyramid will be less than the volume of a square prism having same base area and same height.
• Let us try to find out how much the pyramid is smaller
5. Measure the height of the sand in the prism. We will see that it is one third of the total height of the prism. This is shown in fig.c
6. Now let us calculate the volumes:
(i) Let V_s be the volume of sand. From fig.c, we will get: V_s = base area × height = a² × ^h⁄₃.
(ii) Let V_pr be the volume of the prism. Obviously, V_pr = base area × height = a² × h.
7. Taking ratios, we get:

• This is same as: V_s = ¹⁄₃× V_pr
8. So volume of the sand is only one third of the volume of the prism
• But volume of the sand is same as the volume of the pyramid. So we can write:
• Volume of the square pyramid = ¹⁄₃× V_pr = ¹⁄₃× (a²h)
♦ Where a is the base edge of the square pyramid
♦ And h is the height of the square pyramid
• We will see the actual derivation of this formula in higher classes
An example:
Find the volume of a square pyramid of base edge 10 cm and height 8 cm
Solution:
1. Given: a = 10 cm, h = 8 cm
2. Then volume = ¹⁄₃× (a²h) = ¹⁄₃× 10²× 8 = 266²⁄₃ cm³
Another example:
A metal cube of side 15 cm is melted and recast into a square pyramid of base edge 25 cm. What is it's height?
Solution:
1. Volume of the given cube = 15³ cm³
2. Volume of the square pyramid will also be 15³ cm³
3. So we can write: ¹⁄₃× (a²h) = ¹⁄₃× 25²× h = 15³
⟹ h = 3 × ^15²⁄_25³ = 16.2 cm

Now we will see some solved examples
Solved example 33.10
What is the volume of a square pyramid of base edge 10 cm and slant height 15 cm
Solution:
In this problem, we are not given h. But h is essential to calculate volume. So we will find it first.
1. Imagine the ⊿OAB inside the pyramid. See fig.33.18(a) below:

Fig.33.18

• 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
• We will get:
♦ OB = height of pyramid
♦ OA = half of base edge = 5 cm
♦ AB = slant height = 15 cm
2. Applying Pythagoras theorem, we get:
h = OB = √[15² - 5²] = √[225 - 25] = √[200] = 10√2 cm
3. Then volume = ¹⁄₃× (a²h) = ¹⁄₃× 10²× 10√2 = ¹⁄₃× (1000√2) cm³

Solved example 33.11
Two square pyramids have the same volume. The base edge of one is half that of the other. How many times the height of the second pyramid is the height of the first?
Solution:
1. Let the volume, base edge and height of the first pyramid be V1, a1 and h1 respectively
• Let the volume, base edge and height of the second pyramid be V2, a2 and h2 respectively
2. Then volume of first pyramid = V1 = ¹⁄₃× [(a1)²h1]
• Volume of second pyramid = V2 = ¹⁄₃× [(a2)²h2]
3. Given that V1 = V2
• So we can write: ¹⁄₃× [(a1)²h1)] = ¹⁄₃× [(a2)²h2]
⟹ [(a1)²h1] = [(a2)²h2]
4. But given that a2 = ^a1⁄₂
• Substituting this in (3), we get: [(a1)²h1] = [(^a1⁄₂)²h2]
⟹ [(a1)²h1] = ¹⁄₄× [(a1)²h2] ⟹ [h1] = ¹⁄₄× [h2] ⟹ 4[h1] = [h2]
• So height of the second pyramid (whose base edge is half) is four times the height of the first pyramid

Solved example 33.12
The base edge of two square pyramids are in the ratio 1:2 and their heights are in the ratio 1:3. The volume of the first is 180 cm³. What is the volume of the second?
Solution:
1. Let the volume, base edge and height of the first pyramid be V1, a1 and h1 respectively
• Let the volume, base edge and height of the second pyramid be V2, a2 and h2 respectively
2. (i) Then volume of first pyramid = V1 = ¹⁄₃× [(a1)²h1]
(ii) Volume of second pyramid = V2 = ¹⁄₃× [(a2)²h2]
3. Given that ^a1⁄_a2= ¹⁄₂ . So we get: a1 = ^a2⁄₂
4. Given that ^h1⁄_h2= ¹⁄₃ . So we get: h1 = ^h2⁄₃
5. Substituting (3) and (4) in 2(i), we get:
• V1 = 180 = ¹⁄₃× [(^a2⁄₂)²× ^h2⁄₃] ⟹ 180 × 3 = [(^a2⁄₂)²× ^h2⁄₃] ⟹ 540 = ¹⁄₄×¹⁄₃× [(a2)²h2]
⟹ 540 × 4 = ¹⁄₃× [(a2)²h2]
• So we get the value of ¹⁄₃× [(a2)²h2]. But this is V2
• Thus, the volume of the second pyramid = V2 = 540 × 4 = 2160 cm³.

Solved example 33.13
All edges of a square pyramid are 18 cm. What is it's volume?
Solution:
In this problem, we are not given h. But h is essential to calculate volume. So we will find it first.
1. Imagine ⊿OCB inside the pyramid. See fig.33.18.b above
• It should satisfy the following conditions:
♦ 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 = 18 cm
2. Applying Pythagoras theorem, we get:
OB = √[BC² - OC²] = √[18² - OC²]
3. Now OC = half of the diagonal
• Full diagonal = √[18² + 18²] = 18√2 cm. See fig.33.18.c
• So half diagonal = OC = 9√2
4. Substituting this value of OC in (2), we get:
• OB = √[18² - (9√2)²] = √[324-(81×2)] = √[162] cm
5. Then volume = ¹⁄₃× (a²h) = ¹⁄₃× 18²× √162 = ¹⁄₃× 18²× √[9×9×2] = 972√2 cm³.

Solved example 33.14
The slant height of a square pyramid is 25 cm and it's surface area is 896 cm². What is it's volume?
Solution:
1. Let the base edge be (2a) cm
2. Then area of one isosceles triangle = ¹⁄₂ × base × altitude = ¹⁄₂ × (2a) × 25 = 25a cm².
• Area of four isosceles triangles = 4×25a = 100a cm².
3. Base area = (2a)²= 4a²cm².
4. Total surface area = (100a + 4a²) cm²
5. But the total surface area is given as 896 cm². So we can write: (100a + 4a²) = 896
6. Dividing both sides by 4 we get: (25a + a²) = 224
⟹ a²+ 25a - 224 = 0
7. To avoid any confusion, we will use 'x' instead of 'a'. Then we get:
x²+ 25x - 224 = 0
8. This is a quadratic equation of the form ax² + bx + c = 0 (Details here)

Where: a = 1, b = 25 and c = (-224)

9. So we can use the general formula to solve the equation

10. b²-4ac = 25²-4×1×(-224) = 625 + 896 = 1521

• So √[b²-4ac] = √1521 = ±39
• This '±' sign is already present in the numerator in the formula
• The numerator is: -b±√[b²-4ac] = -(25)±39 = -25+39 OR -25-39
⟹ numerator = 14 OR -64
• The denominator = 2a = 2×1 = 2
• Thus x = ¹⁴⁄₂OR ^-64⁄₂ ⟹ x = 7 OR -32
♦ The base edge cannot be negative value. So we get x = 7
• That is., half the base edge = a = 7 cm
11. Now we have the base edge and slant height. We want the height h
Imagine the ⊿OAB inside the pyramid. See fig.33.19(a) below:

Fig.33.19

In the next section, we will see cones.

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Monday, January 22, 2018

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

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

■ 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

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

In the next section, we will see volume of pyramids