Chapter 34.8 - The Equation of Circle

In the previous section we saw the slopes of perpendicular lines. We also saw some examples. In this section we will see circles.

• We have seen that y = mx + c is the equation of a line. What does it mean?
Let us analyze:
1. Consider a line whose equation is y = mx + c
2. Take any point 'P' on the line 
• In the place of 'x' in the equation, put the x coordinate of the point  
• In the place of 'y' in the equation, put the y coordinate of the point  
3. Then the two sides of the equation will become equal. That is:
■ The coordinates of any point on the line will satisfy the equation of that line.
We can write the converse also:
■ If the coordinates of a point satisfy the equation of a line, then that point will lie on that line

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Now consider a circle. Can we form a equation for that circle?
Let us try:
• There will be a large number of points in the circumference of a circle
    ♦ The coordinates of all those points should satisfy the equation of that circle
Let us see an example:
1. Consider the yellow circle in fig.34.29 below:

Fig.34.29

• It's center is at C(1,4). It's radius is 2 units
2. Consider any point P(x,y) on the circle. Then, using the distance formula, 
CP = √[(x2-x1)² + (y2-y1)]² = √[(x-1)² + (y-4)²]
3. But this distance CP is the radius of the circle
So we can write:
√[(x-1)² + (y-4)²] = r = 2
4. Squaring both sides, we get:
[(x-1)² + (y-4)²] = 4
5. The above equation can be taken as the equation of the circle because, every point on the circle will satisfy it. 
• This is because, it is simply the square of the distance between the 'center of the circle'  and 'the point on the circle'.
• And the distance from the center will be the same for which ever point we take.
An example: 
(i) The point with coordinates (1,2) lies on the circle. This is shown in the fig.
(ii) The distance between C and that point = √[(x2-x1)² + (y2-y1)²] 
= √[(1-1)² + (2-4)²] = √[(0)² + (-2)²] = √[4] = 2 units    
(iii) So 'square of the distance PO' = PO² = 2²= 4

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The general case:
1. To write the general case, we consider a circle whose center is at any point C(x1,y1), and radius 'r' units
2. Then square of the distance between the center and any point (x,y) on the circle is:
[(x-x1)² + (y-y1)²]
• But this distance is the square of the radius which is r²
3. Equating the two, we get:
[(x-x1)² + (y-y1)²] = r²
• This is the general form of the equation of a circle
• Note the pattern:
    ♦ x1, which is the x coordinate of the center is subtracted from x
    ♦ y1, which is the y coordinate of the center is subtracted from y
4. From this we get the general form of the equation of any circle whose center (x1,y1) is at the origin O
• All we need to do is, put x1 = 0 and y1 = 0 in (3)
• Then we get: [x² + y²] = r²
• This is shown in fig.34.30 below:

Fig.34.30

From this fig., we can see that:
• The distance from the center of the circle to  any point on the circle is indeed '√[x² + y²]' if the center is at the origin

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Now we will see some solved examples:

Solved example 34.16
Find the equation of the line joining (1,2) and (2,4). In this, find the sequence of y coordinates of those points with the consecutive natural numbers 3, 4, 5, . . . as the x coordinates
Solution:
1. Equation of the line joining (1,2) and (2,4):
• Slope of the line =
m = ^(y2-y1)⁄_(x2-x1) =  ^(4-2)⁄_(2-1) = ²⁄₁ = 2
• 'c' of the line = (y1-mx1) = [2 - (2 × 1)] = [2 - 2] = 0
• So equation of the line is
y = mx + c:
y = 2 × x + 0 ⟹ y = 2x
2. We know that the consecutive natural numbers form a sequence. That sequence is:
1, 2, 3, 4, 5, . . .
3. The terms of this sequence form the x coordinates of some points. We can write them as:
(1, _ ), (2, _ ), (3, _ ), (4, _ ), (5, _ ), . . .
4. The y coordinates in the above sequence are left blank. We have to find their values.
• It is given that those y coordinates form a sequence
5. It is easy to find the terms of that sequence. Because we have the equation of the line. It is: y = 2x
• From the equation it is clear that, what ever be the value of x, the value of y will be twice that x value. 
• So we can fill up the blank spaces in (3). We get:
(1,2), (2,4), (3,6), (4,8), (5,10), . . .
• Note that, the two points given to us in the question are the first two terms of the sequence.
• The line and the actual positions of the points are shown in the fig.34.31 below:

Fig.34.31

Solved example 34.17
Find the equation of the line joining (-1,3) and (2,5). Prove that if (u,v) is a point on this line, so is (u+3, v+2)
Solution:
1. Equation of the line joining (-1,3) and (2,5)
• Slope of the line =
m = ^(y2-y1)⁄_(x2-x1) =  ^(5-3)⁄_(2-(-1)) = ²⁄₃
• 'c' of the line = (y1-mx1) = [3 - (²⁄₃ × -1)] = [3 + ²⁄₃] = ¹¹⁄₃
• So equation of the line is
y = mx + c:
y = ²⁄₃ × x + ¹¹⁄₃ 
2. Given that, (u,v) is a point on the line. What does that mean?
• It simply means that, any point on the line can be taken as (u,v)
• For example, if we consider the point (-1,3), we can take u = -1 and v = 3
• If we consider the point (2,5), we can take u = 2 and v = 5
3. Now we have prove that (u+3, v+2) is also a point on the line.
• For example, we know that (-1,3) is a point on the line. 
    ♦ Then (-1+3, 3+2) should also be a point on the line. 
    ♦ That is., (2,5) should also be a point on the line
• Similarly, we know that (2,5) is a point on the line. 
    ♦ Then (2+3, 5+2) should also be a point on the line. 
    ♦ That is., (5,7) should also be a point on the line 
• We have to prove it for the general case.
4. Consider the rough sketch in fig 34.32 below:

Fig.34.32

• A possible position of (u+3, v+2) is shown.
• The slope of the line joining (u,v) and (u+3, v+2) is
m = ^(y2-y1)⁄_(x2-x1) = ^(v+2-v)⁄_(u+3-u) = ²⁄₃
• But this is same slope of the line through (-1,3), (2,5) and (u,v)
• Also, (u,v) is a common point. So (u+3, v+2) will lie in the same line

Solved example 34.18
Prove that for any number u, the point (u, 2u+3) is on the line joining (-1,1), (2,7).
Solution:
1. Consider the rough sketch in fig.34.33 below

Fig.34.33

• We can mark any point P' on the x axis. 
• If we draw a vertical line through that point it will surely meet the given line at some point. 
2. Let the marked point P' on the x axis have an x coordinate of 'u'
• Then the meeting point P will also have the x coordinate 'u'
3. The y coordinate of P is not known
■ We have to prove that the y coordinate at the meeting point is:
[(2 times the x coordinate) + 3]. 
• In other words, 
We have to prove that the y coordinate at the meeting point is:
2u+3
4. The equation of a line will help us to find the x coordinate when y coordinate is known
• The equation of a line will also help us to find the y coordinate when the x coordinate is known
5. So let us find the equation of the line:
Equation of the line joining (-1,1) and (2,7)
• Slope of the line =
m = ^(y2-y1)⁄_(x2-x1) =  ^(7-1)⁄_(2-(-1)) = ⁶⁄₃ = 2
• 'c' of the line = (y1-mx1) = [1 - (2 × -1)] = [1 + 2] = 3
• So equation of the line is
y = mx + c:
y = 2x + 3
6. From the equation it is clear that, at any point, the y coordinate will be:
[(2 times the x coordinate at that point) + 3] 
Hence proved

Solved example 34.19
The x coordinate of a point on the slanted line in fig.34.34(a) is 3

Fig.34.34

(i) What is it's y coordinate?
(ii) What is the slope of the line?
(iii) Write the equation of the line
Solution:
Part (i):
1. Drop a perpendicular from B on to the x axis. Let B' be the foot of the perpendicular. This is shown in fig.b
• So BB' will be the y coordinate of B
2. The ⊿ABB' is a 30^o 60^o triangle. We can find it's sides using trigonometric properties. (Details here)
• But for that, at least one side of the triangle should be known.
We have:
    ♦ Length OB' = 3 units
    ♦ Length OA = 1 unit (since x coordinate of A = 1)
    ♦ So AB' = 3 - 1 = 2 units
So now we have one side. We can apply the properties of a 30^o 60^o triangle 
• tan 60 = ^{opposite side}⁄_{adjacent side} = ^BB'⁄_AB' =  ^BB'⁄₂ = √3
⟹ BB' = 2√3
Thus y coordinate of B = BB' = 2√3
Part (ii):
• Slope of the line =
m = ^(y2-y1)⁄_(x2-x1) =  ^(2√3-0)⁄_(3-1) = ^(2√3)⁄₍₂₎ = √3
Another method:
• If we know the angle θ that a line makes with the horizontal, then the slope of that line is simply tanθ.
• We discussed about it earlier in this chapter. Details here.
• For our present case, slope = tan 60 = √3
Part (iii):
Equation of the line joining (1,0) and (3,2√3)
• Slope of the line = m = √3
• 'c' of the line = (y1-mx1) = [0 - (√3 × 1)] = [0 - √3] = -√3
• So equation of the line is
y = mx + c:
y = √3x - √3 ⟹ y = √3(x-1)

Solved example 34.20
In the fig.34.35(a), ABCD is a square.

Fig.34.35

Prove that for any point on the diagonal BD, the sum of x and y coordinates is zero
Solution:
1. The vertices A and C are diagonally opposite. If we are given two such points in a rectangle or a square, we can easily find the coordinates of the other two vertices. Details here.
• So the unknown coordinates at B and D are shown in fig.b
2. Now we have two points on the line BD. We can write the equation
• Equation of the line joining (2,-2) and (-2,2)
• Slope of the line =
m = ^(y2-y1)⁄_(x2-x1) =  ^(2-(-2))⁄_(-2-2) = ⁴⁄_-2 = -2
• 'c' of the line = (y1-mx1) = [-2 - (-2 × 2)] = [-2 + 4] = 2
• So equation of the line is
y = mx + c:

y = -2x + 2

Solved example 34.21
Prove that for any point on the line intersecting the axes in the fig.34.36 below, the sum of the x and y coordinates is 3

Fig.30.36

Solution:
1. We have the coordinates of two points on the line.
• Using them we can write the equation of the line
• Equation of the line joining (3,0) and (0,3)
• Slope of the line =
m = ^(y2-y1)⁄_(x2-x1) =  ^(3-0)⁄_(0-3) = ³⁄_-3 = -1
• 'c' of the line = (y1-mx1) = [0 - (-1 × 3)] = [0 - (-3)] = 3
• So equation of the line is
y = mx + c:
y = -1x + 3
2. From this equarion we get: x+y = 3
That is., for every point on this line, the sum of x and y coordinates will be 3

Solved example 34.22
Find the equation of the circle with center at the origin and radius 5 cm. Write the coordinates of 8 points on this circle
Solution:
1. We have seen at the beginning of this section that:
Equation of a circle with center O and radius r is
[x² + y²] = r². 
• So for our present case, the equation is: [x² + y²] = 5² ⟹ [x² + y²] = 25
2. The circle cuts the axes at the following four points:
(i) P at the positive side of the x axis
(ii) P' at the negative side of the x axis
(iii) Q at the positive side of the y axis
(ii) Q' at the negative side of the y axis
These are shown in the fig.34.37(a) below:

Fig.34.37

• Now we will calculate the coordinates of each of the above four points
(i) Point P: Since the point P is on the x axis, the y coordinate will be equal to zero. 
• So put y = 0 in the equation of the circle in (1)
• We get: [x² + 0] = 25  ⟹ x² = 25 ⟹ x = √25 = ±5
• There will be two roots for 25. They are +5 and -5.
• That means, the circle will intersect with the x axis at two points. 
    ♦ One point is P on the positive side of the x axis. The coordinates are (5,0)
    ♦ The other point P' is on the negative side of the x axis. The coordinates are (-5,0)
• This is shown in the fig.30.37(a) above
• So we started out to find P and got both P and P'
(ii) Point P': This is already obtained above
(iii) Point Q: Since the point Q is on the y axis, the x coordinate will be equal to zero. 
• So put x = 0 in the equation of the circle in (1)
• We get: [0 + y²] = 25  ⟹ y² = 25 ⟹ y = √25 = ±5
• There will be two roots for 25. They are +5 and -5.
• That means, the circle will intersect with the y axis at two points. 
    ♦ One point Q is on the positive side of the y axis. The coordinates are (0,5)
    ♦ The other point Q' is on the negative side of the y axis. The coordinates are (0,-5)
• This is shown in the fig.30.37(a) above
• So we started out to find Q and got both Q and Q'
(ii) Point Q': This is already obtained above
3. We want four more points. 
• For that, we draw a line at an angle of 60^o with the positive side of the x axis. 
    ♦ This line meets the circle at R. This is shown in fig.34.37(b) above
• A perpendicular is dropped from R on to the x axis. The foot of the perpendicular is R'
    ♦ Now x coordinate of R will be equal to OR'
    ♦ y coordinate of R will be equal to RR'
• This is a problem similar to solved example 34.19 that we saw above. In that problem we used the trigonometrical ratio tan. In this example we will use sine and cosine
Consider ⊿ORR'. We have:
• sin 60 = ^{opposite side}⁄_hypotenuse = ^RR'⁄_OR =  ^RR'⁄₅ 
But from the tables, sin 60 = ^√3⁄₂
So we can write: ^RR'⁄₅ = ^√3⁄₂ ⟹ RR' = ^(5√3)⁄₂
• cos 60 = ^{adjacent side} ⁄_hypotenuse = ^OR'⁄_OR =  ^OR'⁄₅ 
But from the tables, cos 60 = ¹⁄₂
So we can write: ^OR'⁄₅ = ¹⁄₂ ⟹ OR' = ⁵⁄₂ 
Thus the coordinates of R are: [⁵⁄₂ , ^(5√3)⁄₂]
Check:
• Applying Pythagoras theorem in ⊿ORR', we have:
OR = √[(OR')² + (RR')²] = √[(⁵⁄₂)² + (^(5√3)⁄₂)²] = √[²⁵⁄₄ + ⁷⁵⁄₄] = √[¹⁰⁰⁄₄] = √[25] = 5 units
• The radius is given as 5 cm. So our calculations are correct
4. Three more points can be obtained in the same way. They are: S, S'' and R''. They are shown in fig.b. They are symmetrical to the point R that we calculated above. So the coordinates can be written easily:
Point S: [^-5⁄₂ , ^(5√3)⁄₂]
Point S'': [^-5⁄₂ , ^(-5√3)⁄₂]
Point R'': [⁵⁄₂ , ^(-5√3)⁄₂]


Solved example 34.23 
Let (x,y) be a point on the circle with the line joining (0,1) and (2,3) as diameter. 
(i) Prove that x²+ y² - 2x - 4y +3 = 0. 
(ii) Find the coordinates of the points where the circle cuts the x axis
Solution:
1. The end points of a diameter are (0,1) and (2,3)
From this we get: 
• x coordinate of the midpoint = ^(x1+x2)⁄₂ = ⁽⁰⁺²⁾⁄₂ = ²⁄₂ = 1
• y coordinate of the midpoint = ^(y1+y2)⁄₂ = ⁽¹⁺³⁾⁄₂ = ⁴⁄₂ = 2
• So the coordinates of the midpoint are (Details here): (1,2)
• But the midpoint of the diameter is same as the center of the circle
2. Length of the diameter can be calculated using the distance formula:
Diameter = √[(x2-x1)² + (y2-y1)]² = √[(2-0)² + (3-1)²] = √[(2)² + (2)²] = √8 = √[2×4] = 2√2
• So radius = ^(2√2)⁄₂ = √2 units
3. The general form of the equation of any circle whose center is at any point C(x1,y1), and radius is 'r' units is: [(x-x1)² + (y-y1)²] = r².
• So we get the equation of our circle:
[(x-1)² + (y-2)²] = (√2)².
• Expanding the above expression, we get:
x² -2x + 1 + y² - 4y + 4 = 2 ⟹ x²+ y² - 2x - 4y +3 = 0
Part (ii):
1. At the points where the circle cuts the x axis, the y coordinates will be zero.
2. So we can put y = 0 in the equation of the circle:
x²+ 0² - 2x - 4×0 +3 = 0 ⟹ x² - 2x +3 = 0

3. This is of the form ax² + bx + c = 0 (Details here)

Where: a = 1, b = (-2) and c = 3

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

5. b²-4ac = (-2)²-4×1×3 = 4 - 12 = -8
• This is a negative quantity.

• So '√[b²-4ac]' cannot be calculated
• The equation does not have a solution. That means, the circle never crosses the x axis.
• The actual positions in the Cartesian plane are shown in the fig.34.38 below:

Fig.34.38

• Note that the radius is √2 units. It is difficult to mark the value of √2 precisely on the x or y axis
• But we have seen the methods to draw a line of length √2 geometrically (Details here)
• Using a graph paper, it is even more easier because, √2 is the diagonal of a square of side 1 unit

Solved example 34.24

What is the equation of the circle in the fig.34.39 below?

Fig.34.39

Solution:
• We know that, the axes are always perpendicular to each other
• In this problem, the point of intersection of the axes, that is., the origin, lies on the circle
• So AB will be a diameter of the circle (Details here)
1. The end points of a diameter are (0,2) and (4,0)
From this we get: 
• x coordinate of the midpoint = ^(x1+x2)⁄₂ = ⁽⁰⁺⁴⁾⁄₂ = ⁴⁄₂ = 2
• y coordinate of the midpoint = ^(y1+y2)⁄₂ = ⁽²⁺⁰⁾⁄₂ = ²⁄₂ = 1
• So the coordinates of the midpoint are (Details here): (2,1)
• But the midpoint of the diameter is same as the center of the circle
2. Length of the diameter can be calculated using the distance formula:
Diameter = √[(x2-x1)² + (y2-y1)]² = √[(4-0)² + (0-2)²] = √[(4)² + (-2)²] = √20 = √[5×4] = 2√5
• So radius = ^(2√5)⁄₂  = √5 units
3. The general form of the equation of any circle whose center is at any point C(x1,y1), and radius is 'r' units is: [(x-x1)² + (y-y1)²] = r².
• So we get the equation of our circle:
[(x-2)² + (y-1)²] = (√5)².
• Expanding the above expression, we get:
x² -4x + 4 + y² - 2y + 1 = 5 ⟹ x²+ y² - 4x - 2y = 0

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In the next section, we will see Polynomials.

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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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