Chapter 34 - Geometry and Algebra

In the previous section we completed a discussion on surface area and volume of some solids. In this chapter we will see Geometry and Algebra.

• This chapter is in fact a continuation of chapter 31 - Coordinate Geometry. 
• The last section of that chapter can be seen here. In that chapter 31, we saw an interesting case:
1. We were given the coordinates of two points
• Those were the end points of a diagonal of a rectangle or a square
2. We were asked to find the coordinates of the other two corners of the rectangle
• We solved such problems very easily. (Details here)

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• Now we will see a similar case. 
    ♦ This time, instead of rectangles or squares, we look at parallelograms. 
• Coordinates of any three of the four vertices of a parallelogram will be given. 
• We will be asked to find the coordinates of the fourth vertex. 
Let us see how it is done:
1. Consider the parallelogram OABC in fig.34.1(a) below

Fig.34.1

• Coordinates of three vertices O, A and C are known. We have to find the coordinates of the vertex B
2. In fig.b, green lines are drawn. 
• These green lines have a special property: They are all parallel to the axes
    ♦ The horizontal green lines are parallel to the x axis
    ♦ The vertical green lines are parallel to the y axis
3. Consider the two green lines attached to the side CB
• They intersect at B'. 
• We know that the angle between the axes will always be 90^o
• Since the green lines are parallel to the axes, the angle at B' will also be 90^o
• So triangle BB'C is a right triangle
4. In the same way, the bottom triangle AA'O is also a right triangle
5. Now we will see the relation between the two right triangles:
• OA and CB are the opposite sides of a parallelogram. 
    ♦ So OA and CB are parallel to each other 
    ♦ and they will be inclined at the same angle with the x axis
• CB' and OA' are both parallel to the x axis. So we get: ∠BCB' = ∠AOA'
• These two equal angles are denoted as α in fig.c
6. Again, OA and CB are parallel to each other and they will be inclined at the same angle with the y axis also
• BB' and AA' are both parallel to the y axis. So we get: ∠CBB' = ∠OAA'
• These two equal angles are denoted as β in fig.c
7. We have a situation:
(i) Angles at the ends of the hypotenuse BC are α and β
(ii) Angles at the ends of the hypotenuse OA are also α and β
(iii) The two triangles are right angled. 
(iv) The two hypotenuses, being opposite sides of a parallelogram, have the same lengths
■ In this situation, we can apply the RHS criterion. The two triangles are equal. (Details here)
Let us write the correspondence:
• The 90^o angle is at A' and B'. So A'↔B'
• The α^o angle is at O and C. So O↔C
• The β^o angle is at A and B. So A↔B
• So the correspondence is A'↔B', O↔C and A↔B. This is same as A'OA↔B'CB
• So we can write: 
The two triangles are congruent to one another under the correspondence: A'OA↔B'CB
• This is same as ΔA'OA ≅ ΔB'CB
Now we can write the corresponding sides. We get: 
    ♦ A'O↔B'C
    ♦ OA↔CB
    ♦ A'A↔B'B
8. Among the above three correspondences, the middle one is already known to us. 
• Because, they are the opposite sides of a parallelogram. 
• The first and the last are of great use to us.
■ The first one indicates this:
The bases of the two triangles are of equal length
■ The last one indicates this:
The altitudes of the two triangles are of equal length
9. Now consider fig.34.2(a) below:

Fig.34.2

• AA' is parallel to the y axis. So coordinates of A' are (6,0)
• Thus length of OA' is 6 units. This is shown in fig.34.2(b)
10. We have seen that the bases are equal. 
• So CB' must also be 6 units in length
• Then coordinates of B' will be (3+6, 5). That is., (9,5)
11. Now consider the coordinates of A. They are (6,2)
• So AA' will be 2 units in length
12. We have seen that the altitudes are equal
• So BB' will also be 2 units in length
13. We already wrote the coordinates of B' as (9,5). 
• So the coordinates of B will be (9,5+2). That is., (9,7)
• This is shown in fig.c. Thus we determined the unknown coordinates at B. The problem is solved.

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• We were able to solve it because, the two bases as well as the two altitudes are equal. We proved it using the RHS criterion.
• There is an easier method to prove those equalities. In that method we use trigonometric ratios. We have already seen the basics of trigonometry here.
1. In fig.34.2(a) above, consider the right triangle BCB'. Taking trigonometric ratios, we will get:
sin α = ^{opposite side}⁄_hypotenuse = ^BB'⁄_CB.
cos α = ^{adjacent side}⁄_hypotenuse = ^B'C⁄_CB.
2. Again in fig.34.2(a) above, consider the right triangle AOA'. Taking trigonometric ratios, we will get:
sin α = ^{opposite side}⁄_hypotenuse = ^AA'⁄_OA.
cos α = ^{adjacent side}⁄_hypotenuse = ^A'O⁄_OA.
3. Now, the sine in ⊿BCB' can be equated to the sine in ⊿AOA'. Because, both are taken for the same angle α.
• For example, suppose α be 36^o. Then from the tables, we have: sin 36 = 0.5878
• So in ⊿BCB', sin α = sin 36 = ^BB'⁄_CB = 0.5878
• In ⊿AOA', ^AA'⁄_OA will also be equal to 0.5878. Because:
sin α = sin 36 = ^AA'⁄_OA= 0.5878
4. So we can write: ^BB'⁄_CB = ^AA'⁄_OA.
• But CB = OA. So we get: BB' = AA'
■ That is., altitudes are equal
5. The same is applicable to cosine also:
• Equating the cosines, we get:
^B'C⁄_CB = ^A'O⁄_OA.
• But CB = OA. So we get: B'C = A'O
■ That is., bases are equal

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• So now we are in a position to find the unknown vertex of any parallelogram. 
• The method can be applied very quickly just by doing some metal calculations. 
Let us analyse:
1. In fig.34.2(b), imagine a person standing at O
• To reach A, he must first travel 6 units horizontally to the right
    ♦ This '6 units' is determined from the x coordinate of A 
• Then he must travel 2 units vertically upwards
    ♦ This '2 units' is determined from the y  coordinate of A 
2. Another person at C must follow the same procedure to reach B. That is:
• He must first travel 6 units horizontally to the right
• Then he must travel 2 units vertically upwards
■ The direction of each travel must be clearly specified:
• Horizontal - Towards left OR Towards right
• Vertical - Upwards OR Downwards
3. We are able to use this easy method because:
    ♦ bases are equal
    ♦ altitudes are also equal
• This method will be more clear when we see a solved example

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Solved example 34.1:
What are the coordinates of the vertex C of the parallelogram shown in the fig.34.3 below:

Fig.34.3

Solution:
1. Group the vertices into two: [A,B] and [C,D]
• Coordinates of both vertices in the first group are know. So we can write the details of the travel within that group:
2. To reach B from A:
• First travel 4 units horizontally to the right [∵ (5-1) = 4]
• Then travel 1 unit vertically upwards [∵ (4-3) = 1]
3. The same procedure of travel must be followed for the travel from D to C
• First travel 4 units horizontally to the right
    ♦ At the end of this travel, the coordinates will be (6,5) [∵ (2+4) = 6]
• Then travel 1 unit vertically upwards
    ♦ At the end of this final lap, the coordinates will be (6,6) [∵ (5+1) = 6]
So the coordinates of C are (6,6)

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■ In the above steps, we chose [A,B] and [C,D]
Let us choose [A,D] and [B,C]
• Coordinates of both vertices in the first group are know. So we can write the details of the travel within that group:
2. To reach A from D:
• First travel 1 unit horizontally to the right [∵ (2-1) = 1]
• Then travel 2 units vertically upwards [∵ (5-3) = 2]
3. The same procedure of travel must be followed for the travel from B to C
• First travel 1 unit horizontally to the right
    ♦ At the end of this travel, the coordinates will be (6,4) [∵ (5+1) = 6]
• Then travel 2 units vertically upwards
    ♦ At the end of this final lap, the coordinates will be (6,6) [∵ (4+2) = 6]
So the coordinates of C are (6,6)
• We get the same answer. So we can choose groups in two different ways
■ But it is important to note that, the members of a group should not be diagonally opposite. Then the method will not work

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In the next section, we will see a few more solved examples.

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Chapter 33.5 - Volume of a Cone

In the previous section we saw how height and base radius of a cone can be used to calculate the curved surface area. We also saw some solved examples. In this section, we will learn about volume of cones.

To find the volume of a cone, we can do an experiment. It is similar to the one that we did for square pyramids. (Details here)
1. Fill a cone with sand.
2. Transfer the sand into a cylinder having the same base area and height as the cone
3. We will find that the sand fills only up to one third height of the cylinder
4. The calculations are also similar. We will get the following result:
• Volume of the cone = ¹⁄₃ × [πr_b²× h]
    ♦ Where r_b is the radius of the base of the cone and h is the height of the cone
• Note that [πr_b²× h] is the volume of a cylinder having the same base radius r_b and same height h
• We will see the actual derivation of this formula in higher classes
An example:
Volume of a cone of base radius 4 cm and height 6 cm is:
¹⁄₃ × [πr_b²× h] = ¹⁄₃ × [π × 4²× 6] = [π × 16 × 2] = 32π cm³.

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Now we will see some solved examples
Solved example 33.24
The base radius and height of a cylindrical block of wood are 15 cm and 40 cm respectively. What is the volume of the largest cone that can be carved out of it?
Solution:
1. The base radius and height of the cone will be same as those of the cylinder. See fig.33.24 below:

Fig.33.24

So we can write:
Volume of the cone = ¹⁄₃ × [πr_b²× h] = ¹⁄₃ × [π × 15²× 40] = [π × 15 × 5 × 40] = 3000π cm³. 

Solved example 33.25
The base radius and height of a solid metal cylinder are 12 cm and 20 cm. By melting it and recasting, how many cones of base radius 4 cm and height 5 cm can be made?
Solution:
1. For the cylinder: 
• Radius of base = 12 cm
• Height = 20 cm
• So volume = [πr²× height of cylinder] = [π × 12²× 20] = 2880π cm³.  
2. For the cone: 
• Radius of base = r_b = 4 cm
• Height = h = 5 cm
• So volume of one cone = ¹⁄₃ × [πr_b²× h] = ¹⁄₃ × [π × 4²× 5] =  ¹⁄₃ × [80π] cm³. 
3. So number of cones that can be obtained = ^{Volume of cylinder}⁄_{Volume of one cone} 
= {2880π} ÷ {¹⁄₃ × [80π]}= 108 Nos.

Solved example 33.26
A sector of central angle 216^o is cut out from a circle of radius 25 cm and is rolled up into a cone. What are the base radius and height of the cone? What is it's volume?
Solution:
1. The two main properties of the sector:
(i) Given that radius of the circle is 25 cm. This will be same as the radius of the sector. So we can write: r_s = 25 cm
• This r_s will be the slant height of the cone. So we can write: Slant height = 25 cm
(ii) Central angle θ = 216^o
2. From the central angle we can calculate length of arc of the sector:
• For every 1^o angle, the length of arc will be ^πr_s⁄₁₈₀. (Theorem 21.1)
• So for 216^o, the length of arc will be 216 × ^πr_s⁄₁₈₀.
• Thus we get:
Length of arc of the sector = 216 × (^π×25⁄₁₈₀) = 30π cm
3. But this is same as the circumference of the base of the cone 
• So if r_b is the radius of the base of the cone, we can write:
2πr_b = 30π ⟹ r_b = ³⁰⁄₂ = 15 cm
4. The next step is to find the slant height. Imagine a triangle OO'P. 
• It must satisfy the following conditions:
    ♦ Triangle OO'P must be right angled
    ♦ O must coincide with the apex
    ♦ O' must coincide with the base center
    ♦ P must be a point on the circumference of the base
• Then we get:
    ♦ OO' = height of the cone
    ♦ O'P = r_b = 15 cm
    ♦ OP = slant height = 25 cm
• Applying Pythagoras theorem, we get:
OO' = √[(OP)² - (O'P)²] = √[(25)² - (15)²] = √[625 + 225] = √[400] = 20 cm
5. So volume of the cone = ¹⁄₃ × [πr_b²× h] = ¹⁄₃ × [π × 15²× 20] = 1500π cm³

Solved example 33.27
The base radii of two cones are in the ratio 3:5 and their heights are in the ratio 2:3. What is the ratio of their volumes?
Solution:
1. Given that ^r_b1⁄_rb2 = ³⁄₅.
Also given that ^h₁⁄_h2 = ²⁄₃.
2. We have: V₁ ÷ V₂ = {¹⁄₃ × [πr_b1²× h₁]} ÷ {¹⁄₃ × [πr_b2²× h₂]}
= {[r_b1²× h₁]} ÷ {[r_b2²× h₂]} 
= {(^r_b1⁄_rb2)²} × {^h₁⁄_h2} 
= {(³⁄₅)²} × {²⁄₃} = ⁶⁄₂₅.
⟹ ^V₁⁄_V2 = ⁶⁄₂₅

Solved example 33.28
Two cones have the same volume and their base radii are in the ratio 4:5. What is the ratio of their heights?
Solution:
1. Given that ^r_b1⁄_rb2 = ⁴⁄₅.
Also given that V₁ = V₂
2. We have: V₁ = V₂ ⟹ {¹⁄₃ × [πr_b1²× h₁]} = {¹⁄₃ × [πr_b2²× h₂]}
⟹ {[r_b1²× h₁]} = {[r_b2²× h₂]} 
⟹ {(^r_b1⁄_rb2)²} = {^h₂⁄_h1} 
⟹ {(⁴⁄₅)²} = {^h₂⁄_h1} = ¹⁶⁄₂₅.
⟹ ^h₁⁄_h2 = ²⁵⁄₁₆

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In the next section, we will see spheres.

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Chapter 33.4 - Height and Curved surface area of a Cone

In the previous section we saw how a sector of a circle is rolled up to form a cone. We also saw some solved examples. In this section, we will learn about curved surface area of cones.

We have seen that the sector of a circle is rolled up to form a cone. So area of that sector will be the curved surface area of the cone. We have seen how to calculate the area of a sector in a previous chapter. Details here.
Let us see an example:
To make a conical hat of base radius 8 cm and slant height 30 cm, how much sq.cm of paper do we need?
Solution:
1. The two main properties of the sector:
(i) Given that slant height of the cone should be 30 cm. 
• So radius of the sector r_s = 30 cm
(ii) Central angle θ has to be calculated
2. Radius of the base r_b is given as 8 cm
• So circumference of the base = 2πr_b = 2π×8 = 16π cm
3. This circumference is equal to the length of the arc
For every 1^o angle, the length of arc will be ^πr_s⁄₁₈₀. (Theorem 21.1)
• So for θ^o, the length of arc will be (θ × ^πr_s⁄₁₈₀)
• Thus we get:
Length of arc of the sector = [θ × (^π×30⁄₁₈₀)] = [θ × (^π×1⁄₆)] cm
4. We can equate the results in (2) and (3):
16π = [θ × (^π×1⁄₆)] ⟹ θ = 96^o.
5. When we have the values for the two main properties of the sector, we can easily calculate it's area 
• Every 1^o central angle in a circle of radius r_s will give a sector of area (^πr_s2⁄₃₆₀) cm². (Theorem 21.3)
• So for 96^o, the area will be [96 × (^πr_s2⁄₃₆₀)] = [96 × 30² × (^π⁄₃₆₀)] = [96 × 30 × (^π⁄₁₂)] 
= [8 × 30 × π] = 240π cm².

Another method:
1. We know that,  every 1^o central angle in a circle of radius r_s will give a sector of area (^πr_s2⁄₃₆₀) cm².
2. So if the central angle is θ, the area of the sector will be [θ × (^πr_s2⁄₃₆₀)] cm².
3. We know that length of arc of the sector is (θ × ^πr_s⁄₁₈₀) cm 
• But length of arc of the sector is the circumference of the base of the cone, which is 2πr_b.
4. So we can write:  2πr_b = (θ × ^πr_s⁄₁₈₀)
⟹ θ =  ^360r_b⁄_rs.
5. Substituting this value of in (2), we get:
• Area of the sector = [(^360r_b⁄_rs) × (^πr_s2⁄₃₆₀)] = πr_br_s.
    ♦ But 'area of the sector' is the 'curved surface area of the cone'
    ♦ And r_s is the slant height l
■ So Area of the sector = Curved surface area of the cone =  πr_bl cm²

Height of a cone

• We have seen that a sector can be completely defined by two properties
    ♦ It's radius
    ♦ It's central angle
• In the case of square pyramids, we saw that, it can be completely defined by two properties:
    ♦ It's height
    ♦ It's base edge
• In a similar way, a cone can be completely defined by two properties:
    ♦ It's height
    ♦ It's base radius
■ That is., if we know the height and base radius of a cone, we will be able to calculate all other properties:
Slant height, circumference of the base, central angle, surface area and volume
Let us see how it is done:
Fig.33.23(a) below shows a cone. 

Fig.33.23

• The apex is marked as O. Center of it's base is marked as O'. So OO' is the height of the cone.
• Now mark any point P on the circumference of the base. Draw O'P and OP
• Obviously, O'P will be r_b and OP will be the slant height
• Also OO'P will be a right angled triangle. So we can use Pythagoras theorem to find unknown quantities
An example:
In a cone, the height is 10 cm and base radius is 5 cm. Find the height of the cone
Solution:
1. Imagine a triangle OO'P. 
• It must satisfy the following conditions:
    ♦ Triangle OO'P must be right angled
    ♦ O must coincide with the apex
    ♦ O' must coincide with the base center
    ♦ P must be a point on the circumference of the base
• Then we get:
    ♦ OO' = height of the cone = 10 cm 
    ♦ O'P = r_b = 5 cm
    ♦ OP = slant height
2. Applying Pythagoras theorem, we get:
OP = √[(O'P)² + (OO')²] = √[(5)² + (10)²] = √[25 + 100] = √[125] = 5√5 cm

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Now we will see some solved examples
Solved example 33.20
What is the area of the curved surface of a cone of base radius 12 cm and slant height 25 cm
Solution:
1. The two main properties of the sector:
(i) Given that slant height of the cone should be 25 cm. 
• So radius of the sector r_s = 25 cm
(ii) Central angle θ has to be calculated
2. Radius of the base r_b is given as 12 cm
• So circumference of the base = 2πr_b = 2π×12 = 24π cm
3. This circumference is equal to the length of the arc
For every 1^o angle, the length of arc will be ^πr_s⁄₁₈₀. (Theorem 21.1)
• So for θ^o, the length of arc will be (θ × ^πr_s⁄₁₈₀)
• Thus we get:
Length of arc of the sector = [θ × (^π×25⁄₁₈₀)] = [θ × (^π×5⁄₃₆)] cm
4. We can equate the results in (2) and (3):
24π = [θ × (^π×5⁄₃₆)] ⟹ θ = (⁸⁶⁴⁄₅)^o
5. When we have the values for the two main properties of the sector, we can easily calculate it's area 
• Every 1^o central angle in a circle of radius r_s will give a sector of area (^πr2⁄₃₆₀) cm². (Theorem 21.3)
• So for (⁸⁶⁴⁄₅)^o, the area will be [(⁸⁶⁴⁄₅) × (^πr_s2⁄₃₆₀)] = [(⁸⁶⁴⁄₅) × 25² × (^π⁄₃₆₀)] 
= [864 × 125 × (^π⁄₃₆₀)] = 300π cm².

Another method using equation:
1. We have: Curved surface area of a cone = πr_bl cm².
2. Substituting the values we get: 
Curved surface area of a cone = π×12×25 = 300π cm².

Solved example 33.21
What is the surface area of a cone of base diameter 30 cm and height 40 cm?
Solution:
1. The first step is to find the slant height. Imagine a triangle OO'P. 
• It must satisfy the following conditions:
    ♦ Triangle OO'P must be right angled
    ♦ O must coincide with the apex
    ♦ O' must coincide with the base center
    ♦ P must be a point on the circumference of the base
• Then we get:
    ♦ OO' = height of the cone = 40 cm 
    ♦ O'P = r_b = 15 cm (∵ diameter = 30 cm)
    ♦ OP = l = slant height
• Applying Pythagoras theorem, we get:
OP = l = √[(O'P)² + (OO')²] = √[(15)² + (40)²] = √[225 + 1600] = √[1825] cm
2. Curved surface area = πr_bl = π×15×√[1825] = 640.8π cm².
3. Surface area of base = πr_b² = π×15²= 225π cm².
4. Total surface area = (640.8+225)π = 865.8π cm²

Solved example 33.22
A conical fire work is of base diameter 10 cm and height 12 cm. 10000 such fire works are to be wrapped in colour paper. The price of colour paper is 2 rupees per sq.m. What is the total cost?
Solution:
1. The first step is to find the slant height. Imagine a triangle OO'P. 
• It must satisfy the following conditions:
    ♦ Triangle OO'P must be right angled
    ♦ O must coincide with the apex
    ♦ O' must coincide with the base center
    ♦ P must be a point on the circumference of the base
• Then we get:
    ♦ OO' = height of the cone = 12 cm 
    ♦ O'P = r_b = 5 cm (∵ diameter = 10 cm)
    ♦ OP = slant height
• Applying Pythagoras theorem, we get:
OP = l = √[(O'P)² + (OO')²] = √[(5)² + (12)²] = √[25 + 144] = √[169] = 13 cm
2. Curved surface area = πr_bl = π×5×13 = 65π cm².
3. Surface area of base = πr_b² = π×5²= 25π cm²
4. Total surface area = (65+25)π = 90π cm² = 90×3.14 = 282.6 cm²
5. Surface area of  10000 fire works = 282.6 × 10000 = 2826000 cm².
6. 2826000 cm² = ^2826000⁄₁₀₀₀₀  m² = 282.6 m².
7. So cost of colour paper = 282.6 × 2 = Rs 565.20

Solved example 33.23
Prove that for a cone made by rolling up a semicircle, the area of the curved surface is twice the base area
Solution:
1. The two main properties of the sector:
(i) Let the radius of the sector be r_s
(ii) Central angle θ is 180^o (∵ the sector is a semicircle)
2. Let the radius of the base be r_b
• So circumference of the base = 2πr_b 
3. This circumference is equal to the length of the arc
• We do not need to calculate the length of the arc. The arc length of a semicircle is 'half the circumference of the full circle'
• The 'circumference of the full circle' is 2πr_s. So half of it is πr_s.
4. We can equate the results in (2) and (3):
2πr_b  = πr_s ⟹ 2r_b  = r_s.
5. Now we want the area of the sector
• But the area of the sector is area of the semicircle which is ¹⁄₂ × πr_s²
• Let us substitute for r_s using the result in (4). We get:
• Area of the sector = ¹⁄₂ × πr_s² = ¹⁄₂ × π(2r_b)² = ¹⁄₂ × π × 4 × r_b² =  2πr_b²
6. Area of the sector is same as the area of curved surface. So we can write:
Area of the curved surface of the cone = 2πr_b².
7. Now we calculate the base area. 
• We have radius of the base of the cone = r_b.
• So area of the base of the cone = πr_b².
8. Comparing the results in (6) and (7), we get:
Area of the curved surface of the cone = Twice the base area

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In the next section, we will see volume of cone.

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