Chapter 35.1 - Factors of Polynomials

In the previous section we saw how a second degree polynomial is split into it's factors. In this section we will see some solved examples.

Solved example 35.1
Write each polynomial below as a product of first degree polynomials. Write also the solutions of the equation p(x) = 0 in each case
(i) p(x) = x² - 7x + 12 
(ii) p(x) = x² + 7x + 12
(iii) p(x) = x² - 8x + 12 
(iv) p(x) = x² + 13x + 12
(v) p(x) = x² - 2x + 1 
(vi) p(x) = x² + x - 1 
(vii) p(x) = 2x² - 5x + 2 
(viii) p(x) = 6x² - 7x + 2

Solution:
Part (i): p(x) = x² - 7x + 12
1. From the coefficients in the given equation, we can write:
(i) (b₁+b₂) = -7
(ii) b₁b₂ = 12
Where b₁ and b₂ are the required solutions when p(x) = 0
(iii) From (ii) we get: b₂ = ¹²⁄_b1
(iv) Substituting this in (i) we get:  (b₁ + ¹²⁄_b1) = -7
• Simplifying we get: [(b₁)² + 7b₁ + 12] = 0
• This is a quadratic equation in (b₁)
Solving it we get: 
b₁ = -4  and b₂ = -3
2. So we get:
• p(x) = [x² - 7x + 12] = [(x-4)(x-3)]
• We can write: (x-4) and (x-3) are the factors of p(x) = [x² - 7x + 12]
3. When p(x) = 0, the solutions are 4 and 3. That is:
p(4) will be zero
p(3) will be zero
• In other words, we must put 4 or 3 in the place of x, to make p(x) equal to zero
The geometric meaning of this situation can be seen in the fig.35.1 below:

Fig.35.1

• The magenta curve is a part of the plot of p(x). It's 'y value' becomes zero at x = 3 and x = 4 
Part (ii): p(x) = x² + 7x + 12
1. From the coefficients in the given equation, we can write:
(i) (b₁+b₂) = 7
(ii) b₁b₂ = 12
Where b₁ and b₂ are the required solutions when p(x) = 0
(iii) From (ii) we get: b₂ = ¹²⁄_b1
(iv) Substituting this in (i) we get:  (b₁ + ¹²⁄_b1) = 7
• Simplifying we get: [(b₁)² - 7b₁ + 12] = 0
• This is a quadratic equation in (b₁)
Solving it we get: 
b₁ = 4  and b₂ = 3
2. So we get:
p(x) = [x² + 7x + 12] = [(x+4)(x+3)]
• We can write: (x+4) and (x+3) are the factors of p(x) = [x² + 7x + 12]
⟹ (x-(-4)) and (x-(-3)) are the factors of p(x) = [x² + 7x + 12]
3. When p(x) = 0, the solutions are -4 and -3. That is:
p(-4) will be zero
p(-3) will be zero
• In other words, we must put -4 or -3 in the place of x, to make p(x) equal to zero

Part (iii): p(x) = x² - 8x + 12
1. From the coefficients in the given equation, we can write:
(i) (b₁+b₂) = -8
(ii) b₁b₂ = 12
Where b₁ and b₂ are the required solutions when p(x) = 0
(iii) From (ii) we get: b₂ = ¹²⁄_b1
(iv) Substituting this in (i) we get:  (b₁ + ¹²⁄_b1) = -8
• Simplifying we get: [(b₁)² + 8b₁ + 12] = 0
• This is a quadratic equation in (b₁)
Solving it we get: 
b₁ = -2  and b₂ = -6
2. So we get:
• p(x) = [x² - 8x + 12] = [(x-2)(x-6)]
• We can write: (x-2) and (x-6) are the factors of p(x) = [x² - 8x + 12]
3. When p(x) = 0, the solutions are 2 and 6. That is:
p(2) will be zero
p(6) will be zero
• In other words, we must put 2 or 6 in the place of x, to make p(x) equal to zero

Part (iv): p(x) = x² + 13x + 12
1. From the coefficients in the given equation, we can write:
(i) (b₁+b₂) = 13
(ii) b₁b₂ = 12
Where b₁ and b₂ are the required solutions when p(x) = 0
(iii) From (ii) we get: b₂ = ¹²⁄_b1
(iv) Substituting this in (i) we get:  (b₁ + ¹²⁄_b1) = 13
• Simplifying we get: [(b₁)² - 13b₁ + 12] = 0
• This is a quadratic equation in (b₁)
Solving it we get: 
b₁ = 12  and b₂ = 1
2. So we get:
• p(x) = [x² + 13x + 12] = [(x+12)(x+1)]
• We can write: (x+12) and (x+1) are the factors of p(x) = [x² + 13x + 12]
⟹ (x-(-12)) and (x-(-1)) are the factors of p(x) = [x² + 13x + 12]
3. When p(x) = 0, the solutions are -12 and -1. That is:
p(-12) will be zero
p(-1) will be zero
• In other words, we must put -12 or -1 in the place of x, to make p(x) equal to zero
• The geometric meaning of this situation can be seen in the fig.35.2 below:

Fig.35.2

• The magenta curve is a part of the plot of p(x). It's 'y value' becomes zero at x = -12 and x = -1
■ Note that the fig.35.2 is a little distorted because, the width of PC monitor is more than it's height. If we draw manually on a graph paper, we will get a perfect curve. However the fig. shows the geometric meaning of f(x) = 0

Part (v): p(x) = x² - 2x + 1
1. From the coefficients in the given equation, we can write:
(i) (b₁+b₂) = -2
(ii) b₁b₂ = 1
Where b₁ and b₂ are the required solutions when p(x) = 0
(iii) From (ii) we get: b₂ = ¹⁄_b1
(iv) Substituting this in (i) we get:  (b₁ + ¹⁄_b1) = -2
• Simplifying we get: [(b₁)² +2b₁ + 1] = 0
• This is a quadratic equation in (b₁)
Solving it we get: 
b₁ = -1  and b₂ = -1
2. So we get:
p(x) = [x² - 2x + 1] = [(x-1)(x-1)]
• We can write: (x-1) and (x-1) are the factors of p(x) = [x² - 2x + 1]
3. When p(x) = 0, the solutions are 1 and 1. That is:
p(1) will be zero
• In other words, we must put 1 in the place of x, to make p(x) equal to zero
• The geometric meaning of this situation can be seen in the fig.35.3 below:

Fig.35.3

• The magenta curve is a part of the plot of p(x). It's 'y value' becomes zero only at x = 1

Part (vi): p(x) = x² + x - 1
1. From the coefficients in the given equation, we can write:
(i) (b₁+b₂) = 1
(ii) b₁b₂ = -1
Where b₁ and b₂ are the required solutions when p(x) = 0
(iii) From (ii) we get: b₂ = ^-1⁄_b1
(iv) Substituting this in (i) we get:  (b₁ + ^-1⁄_b1) = 1
• Simplifying we get: [(b₁)² - b₁ - 1] = 0
• This is a quadratic equation in (b₁)
Solving it we get: 
b₁ = ^[1-(√5)]⁄₂  and b₂ = ^[1+(√5)]⁄₂
2. So we get:
• p(x) = [x² + x - 1] = [(x+{^[1-(√5)]⁄₂})(x+{^[1+(√5)]⁄₂})] 
= [(x-{^-[(√5)-1]⁄₂})(x-{^[(√5)+1]⁄₂})] 
• We can write: (x-{^-[(√5)-1]⁄₂}) and (x-{^[(√5)+1]⁄₂}) are the factors of p(x) = [x² + x - 1]
3. When p(x) = 0, the solutions are {^-[(√5)-1]⁄₂} and {^[(√5)+1]⁄₂}. That is:
p({^-[(√5)-1]⁄₂}) will be zero
p({^[(√5)+1]⁄₂}) will be zero
• In other words, we must put {^-[(√5)-1]⁄₂} or {^[(√5)+1]⁄₂} in the place of x, to make p(x) equal to zero

Part (vii): p(x) = 2x² - 5x + 2
The question can be modified as: 2[x² - (⁵⁄₂)x + 1]
1. From the coefficients in the given equation, we can write:
(i) (b₁+b₂) = -⁵⁄₂
(ii) b₁b₂ = 1
Where b₁ and b₂ are the required solutions when p(x) = 0
(iii) From (ii) we get: b₂ = ¹⁄_b1
(iv) Substituting this in (i) we get:  (b₁ + ¹⁄_b1) = -⁵⁄₂
• Simplifying we get: [2(b₁)² + 5b₁ + 2] = 0
• This is a quadratic equation in (b₁)
Solving it we get: 
b₁ = -2  and b₂ = ^-1⁄₂
2. So we get:
p(x) = [x² - (⁵⁄₂)x + 1] = [(x-2)(x-(¹⁄₂))]
• We can write: (x-2) and (x-(¹⁄₂)) are the factors of p(x) = [x² - (⁵⁄₂)x + 1]
3. When p(x) = 0, the solutions are 2 and (¹⁄₂). That is:
p(2) will be zero
p(¹⁄₂) will be zero
• In other words, we must put 2 or (¹⁄₂) in the place of x, to make p(x) equal to zero

• The geometric meaning of this situation can be seen in the fig.35.4 below:

Fig.35.4

• The outer magenta curve is a part of the plot of p(x) = [x² - (⁵⁄₂)x + 1]
• The inner magenta curve is a part of the plot of p(x) = [2x² - 5x + 2]
• In both cases, 'y value' becomes zero at x = 2 and x = ¹⁄₂

Part (viii): p(x) = 6x² - 7x + 2
The question can be modified as: 6[x² - (⁷⁄₆)x + ¹⁄₃]
1. From the coefficients in the given equation, we can write:
(i) (b₁+b₂) = -⁷⁄₆
(ii) b₁b₂ = ¹⁄₃
Where b₁ and b₂ are the required solutions when p(x) = 0
(iii) From (ii) we get: b₂ = ¹⁄_3b1
(iv) Substituting this in (i) we get:  (b₁ + ¹⁄_3b1) = -⁷⁄₆
• Simplifying we get: [18(b₁)² + 21b₁ + 6] = 0
• This is a quadratic equation in (b₁)
Solving it we get: 
b₁ = ^-2⁄₃  and b₂ = ^-1⁄₂
2. So we get:
• p(x) = [x² - (⁷⁄₆)x + ¹⁄₃] = [(x-(²⁄₃))(x-(¹⁄₂))]
• We can write: (x-(²⁄₃)) and (x-(¹⁄₂)) are the factors of p(x) = [x² - (⁷⁄₆)x + ¹⁄₃]
3. When p(x) = 0, the solutions are (²⁄₃) and (¹⁄₂). That is:
p(²⁄₃) will be zero
p(¹⁄₂) will be zero
• In other words, we must put (²⁄₃) or (¹⁄₂) in the place of x, to make p(x) equal to zero



Solved example 35.2
Find a second degree polynomial p(x) such that p(1) = 0 and p(-2) = 0
Solution:
1. Suppose we have a second degree polynomial p(x)
• If we put 1 in place of x, that polynomial will become zero
    ♦ So 1 is a solution of p(x)
    ♦ So (x-1) is a factor of p(x)
• If we put -2 in place of x, that polynomial will become zero
    ♦ So -2 is a solution of p(x)
    ♦ So (x-(-2)) is a factor of p(x)
        ⟹ (x+2) is a factor of p(x)
2. We got two factors of the second degree polynomial p(x)
• Both the factors are first degree polynomials
3. Any second degree polynomial will have only two factors
• And each of those factors will be a first degree polynomial
• In our case we obtained both the factors
• If we multiply those factors, we will get the original polynomial
4. So we can write:
p(x) = (x-1)(x+2)
= x² + x - 2

Solved example 35.3
Find a second degree polynomial p(x) such that p(1+√3) = 0 and p(1-√3) = 0
Solution:
1. Suppose we have a second degree polynomial p(x)
• If we put (1+√3) in place of x, that polynomial will become zero
    ♦ So (1+√3) is a solution of p(x)
    ♦ So (x-(1+√3)) is a factor of p(x)
• If we put (1-√3) in place of x, that polynomial will become zero
    ♦ So (1-√3) is a solution of p(x)
    ♦ So (x-(1-√3)) is a factor of p(x)
2. We got two factors of the second degree polynomial p(x)
• Both the factors are first degree polynomials
3. Any second degree polynomial will have only two factors
• And each of those factors will be a first degree polynomial
• In our case we obtained both the factors
• If we multiply those factors, we will get the original polynomial
4. So we can write:
p(x) = [x-(1+√3)][(x-(1-√3)]
5. Let A = (1+√3) and B = (1-√3)
Then we get: p(x) = [x-A][x-B]
= x² - (A+B)x +  AB
6. A+B = (1+√3) + (1-√3) = 2
AB = (1+√3)(1-√3) = [1²-(√3)²] = [1-3] = -2
7. Then the result in (5) becomes:
p(x) = x² - (A+B)x +  AB 
= x² -2x - 2

Solved example 35.4
Find a third degree polynomial p(x) such that p(1) = 0, p(√2) = 0 and p(-√2) = 0
Solution:
1. Suppose we have a third degree polynomial p(x)
• If we put (1) in place of x, that polynomial will become zero
    ♦ So 1 is a solution of p(x)

    ♦ So (x-1) is a factor of p(x)
• If we put (√2) in place of x, that polynomial will become zero
    ♦ So (√2) is a solution of p(x)
    ♦ So (x-(√2)) is a factor of p(x)
• If we put (-√2) in place of x, that polynomial will become zero
    ♦ So (-√2) is a solution of p(x)
    ♦ So (x-(-√2)) is a factor of p(x) same as (x+(√2)) is a factor of p(x)
2. We got three factors of the third degree polynomial p(x)
• All the three factors are first degree polynomials
3. Any third degree polynomial will have only three factors
• And each of those factors will be a first degree polynomial
• In our case we obtained all the three factors
• If we multiply those factors, we will get the original polynomial
4. So we can write:
p(x) = [x-1][x-(√2)][(x+(√2)]
5. Let us first multiply the last two factors:
[x-(√2)][(x+(√2)] = [x²-(√2)²] = [x²-2]
Now we get:
• p(x) = [x-1][x²-2] = x³ - x² -2x + 2
• The geometric representation of p(x) can be seen in the fig.35.5 below:

Fig.35.5

■ Note that the magenta curve meets the x axis at 3 points
• One point is on the negative side of the x axis. 
    ♦ Between -1.5 and -1. It is closer to -1.5. This point is (x = -√2). 
    ♦ We know that √2 = 1.414. That is why, the point is closer to 1.5
• Two points are on the positive side of the x axis
    ♦ One is at exact (x = 1)
    ♦ The other point is between 1 and 1.5, closer to 1.5. It is (x = √2)

Solved example 35.5
Prove that the polynomial x² + x + 1 cannot be written as a product of first degree polynomials
Solution:
1. Let us first assume that the given polynomial can be written as a product of first degree polynomials
• Let those first degree polynomials be (x+b₁) and (x+b₂)
2. From the coefficients in the given equation, we can write:
(i) (b₁+b₂) = 1
(ii) b₁b₂ = 1
Where b₁ and b₂ are the required solutions when p(x) = 0
(iii) From (ii) we get: b₂ = ¹⁄_b1
(iv) Substituting this in (i) we get:  (b₁ + ¹⁄_b1) = 1
• Simplifying we get: [(b₁)² + -b₁ + 1] = 0
• This is a quadratic equation in (b₁)
3. But there is no number that will satisfy [(b₁)² + -b₁ + 1] = 0
• That means b₁ does not exist
• So we cannot use 2(i) to obtain b₂ either
• Both b₁ and b₂ does not exist.
• Consequently, there is no (x+b₁) and (x+b₂)
• Thus our assumption in (1) is wrong

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In the next section, we will see polynomial remainder.

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Chapter 34.1 - Single unknown vertex in a Parallelogram - Solved examples

In the previous section we saw how to find the unknown coordinates of a vertex in a parallelogram. We also saw a solved example. In this section we will see a few more solved examples.

Solved example 34.2
In the fig.34.4(a) below, the midpoints of the sides of the large triangle are joined to make a small triangle inside. 

Fig.34.4

Calculate the coordinates of the vertices of the large triangle.
Solution:
• Let the outer triangle be PQR and the inner triangle be ABC. This is shown in fig.b
    ♦ When the midpoints are marked and joined as in the fig.a, we get a triangle inside.
    ♦ Also we will get 3 inside parallelograms (Details here)
• So we need to find the unknown fourth vertex of those parallelograms. The steps are given below:
1. One of the 3 parallelograms is highlighted in fig.c. It is AQBC. 
Q is the unknown vertex. When we find Q of the parallelogram, we get one of the vertices of the main triangle also
(i) Group the vertices into two: [A,Q] and [C,B]
(The members of a group should not be diagonally opposite)
• Coordinates of both vertices in the second group are known. So we can write the details of the travel within that group:
(ii) To reach B from C:
• First travel 2 units horizontally to the right [∵ (5-3) = 2]
• Then travel 1 unit vertically upwards [∵ (4-3) = 1]
(iii) The same procedure of travel must be followed for the travel from A to Q
• First travel 2 units horizontally to the right
    ♦ At the end of this travel, the coordinates will be (6,2) [∵ (4+2) = 6]
• Then travel 1 unit vertically upwards
    ♦ At the end of this final lap, the coordinates will be (6,3) [∵ (2+1) = 3]
So the coordinates of Q are (6,3)
2. The next parallelogram is highlighted in fig.34.4(d) below. It is ABRC.

Fig.34.4

R is the unknown vertex.
(i) Group the vertices into two: [A,B] and [C,R]
(The members of a group should not be diagonally opposite)
• Coordinates of both vertices in the first group are known. So we can write the details of the travel within that group:
(ii) To reach B from A:
• First travel 1 unit horizontally to the right [∵ (5-4) = 1]
• Then travel 2 units vertically upwards [∵ (4-2) = 2]
(iii) The same procedure of travel must be followed for the travel from C to R
• First travel 1 unit horizontally to the right
    ♦ At the end of this travel, the coordinates will be (4,3) [∵ (3+1) = 4]
• Then travel 2 units vertically upwards
    ♦ At the end of this final lap, the coordinates will be (4,5) [∵ (3+2) = 5]
So the coordinates of R are (4,5)
3. The third parallelogram is highlighted in fig.34.4(e) below. It is PABC. 
P is the unknown vertex.
(i) Group the vertices into two: [P,A] and [B,C]
(The members of a group should not be diagonally opposite)
• Coordinates of both vertices in the second group are known. So we can write the details of the travel within that group:
(ii) To reach C from B:
• First travel 2 units horizontally to the left [∵ (3-5) = -2]
• Then travel 1 unit vertically downwards [∵ (3-4) = -1]
(iii) The same procedure of travel must be followed for the travel from A to Q
• First travel 2 units horizontally to the left
    ♦ At the end of this travel, the coordinates will be (2,2) [∵ (4-2) = 2]
• Then travel 1 unit vertically downwards
    ♦ At the end of this final lap, the coordinates will be (2,1) [∵ (2-1) = 1]
So the coordinates of P are (2,1)

Solved example 34.3
In the parallelogram OABC shown in fig.34.5 below, the coordinates of A are (x2,y2). The coordinates of C are (x1,x2). 

Fig.34.5

What are the coordinates of B?
Solution:
1. Group the vertices into two: [O,A] and [C,B]
(The members of a group should not be diagonally opposite)
• Coordinates of both vertices in the first group are known. So we can write the details of the travel within that group:
2. To reach A from O:
• First travel x2 units horizontally to the right [∵ (x2-0) = x2]
• Then travel y2 units vertically upwards [∵ (y2-0) = y2]
3. The same procedure of travel must be followed for the travel from C to B
• First travel x2 units horizontally to the right
    ♦ At the end of this travel, the coordinates will be [(x1+x2),y1]
• Then travel y2 units vertically upwards
    ♦ At the end of this final lap, the coordinates will be [(x1+x2),(y1+y2)]
So the coordinates of B are [(x1+x2),(y1+y2)]

Solved example 34.4
Prove that in any parallelogram, the sum of the squares of all sides is equal to the sum of the squares of the diagonals
Solution:
• Consider any parallelogram such as the one shown in fig.34.6(a) below:

Fig.34.6

• Any parallelogram will have:
    ♦ An unique value for it's base length
    ♦ An unique value for it's height
    ♦ An unique value for the offset
• If we know these three values of a parallelogram, we can completely define it.
• For our parallelogram, 
    ♦ Let the base length be a
    ♦ Let the height be y
    ♦ Let the offset be x
• These are shown in fig.b. We will use them to prove the result.
• We have see that, the origin of the axes can be placed any where. 
• And also, the axes can have any orientation. 
    ♦ But they must always be perpendicular to each other. (Details here)
• So let us place the origin at the bottom left vertex of the parallelogram
• Let the orientation be in such a way that, the base of the parallelogram coincides with one of the axes. This is shown in fig.c
• Now we can easily write the coordinates of the various points. Fig.c is self explanatory
• Once the coordinates are written, we can calculate the distances using the distance formula. The steps are given below. 
(Note that, Pythagoras theorem is used in those cases where it can be used with more ease than the distance formula)
1. Lengths of sides and their squares:
(i) Length of OA = a
So OA² = a²
(ii) Length of AB using Pythagoras theorem = √[x²+y²]
So AB² = [x²+y²]
(iii) Length of BC = OA = a
So BC² = a²
(ii) Length of OC = AB = √[x²+y²]
So OC² = [x²+y²]
2. Sum of the above squares = 2x²+2y²+2a² = 2(x²+y²+a²)
3. Length of diagonals and their squares:
(i) Length of OB using Pythagoras theorem = √[(a+x)²+y²] = √[a²+2ax+x²+y²]
So OB² = [a²+2ax+x²+y²]
(ii) Length of AC using distance formula =  √[(a-x)²+(0-y)²] = √[a²-2ax+x²+y²]  
So AC² = [a²-2ax+x²+y²]
4. Sum of the above squares = 2(x²+y²+a²)
Results in (2) and (4) are the same. Hence proved

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

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Chapter 33.7 - Volume of Sphere - Solved examples

In the previous section we saw surface area and volume of spheres and hemispheres. We also saw some solved examples. In this section, we will see a few more solved examples.

Solved example 33.29
The surface area of a solid sphere is 120 sq.cm. If it is cut into two halves, what would be the surface area of each hemisphere?
Solution:
1. Let r cm be the radius of the total sphere. Then it's surface area would be 4πr² cm²
2. This surface area is given as 120 cm². So we can equate the two:
4πr² = 120 ⟹ πr² = 30 ⟹ r² = ³⁰⁄_π
3. The curved surface area of each hemisphere will be ¹²⁰⁄₂ = 60 cm².
4. When the sphere is cut, each hemisphere will have a base also. This base is a circular area. 
• The radius of the base circle will be the same 'r'. So we can write:
Base area of each hemisphere = πr² = π × ³⁰⁄_π = 30 cm²
5. So total surface area of each hemisphere = (60 + 30) = 90 cm²

An easy method:
• Curved surface area of a hemisphere = ¹⁄₂ × 4πr² = 2πr².
• Base area = πr²
• So total surface area of a hemisphere = 2πr² + πr² = 3πr²
• From (2) we have: r² = ³⁰⁄_π
• So total surface area of each hemisphere = 3π × ³⁰⁄_π = 90 cm²

Solved example 33.30
The volume of two spheres are in the ratio 27 : 64. 
(i) What is the ratio of their radii?
(ii) What is the ratio of their surface areas?
Solution:
Part (i):
1. Let the volume and radius of the first sphere be V1 and r1 respectively
• Let the volume and radius of the second sphere be V2 and r2 respectively
2. (i) Then volume of first sphere = V1 = ⁴⁄₃ × [π(r1)³] 
(ii) volume of second sphere = V2 = ⁴⁄₃ × [π(r2)³] 
3. Given that ^V1⁄_V2 = ²⁷⁄₆₄
⟹ ⁴⁄₃ × [π(r1)³] ÷ ⁴⁄₃ × [π(r2)³] = ²⁷⁄₆₄
⟹ [π(r1)³] ÷ [π(r2)³] = ²⁷⁄₆₄ 
⟹ [(r1)³] ÷ [(r2)³] = ²⁷⁄₆₄
⟹ (^r1⁄_r2)³ = ²⁷⁄₆₄  
⟹ (^r1⁄_r2)³ = (³⁄₄)³ 
⟹ (^r1⁄_r2) = (³⁄₄)
Part (ii):
1. Let S1 and S2 be the surface areas. Then we get:
^S1⁄_S2 = [4π(r1)²] ÷ [4π(r2)²]
⟹ ^S1⁄_S2 = [(r1)²] ÷ [(r2)²]
⟹ ^S1⁄_S2 = (^r1⁄_r2)²
2. But from (3) in part (i), we have: (^r1⁄_r2) = (³⁄₄)
• So (^r1⁄_r2)² = (³⁄₄)² = (⁹⁄₁₆)
• Thus we get: ^S1⁄_S2 = (^r1⁄_r2)² = (⁹⁄₁₆)

Solved example 33.31
The base radius and length of a metal cylinder are 4 cm and 10 cm. If it is melted and recast into spheres of radius 2 cm, how many spheres can be made?
Solution:
1. Total volume available for melting = Volume of the cylinder
= πr²h = π×4²×10 = 160π cm³
2. Volume of one sphere = ⁴⁄₃πr³ = ⁴⁄₃ × π × 2³ = ³²⁄₃ × π cm³.
3. Number of spheres = [Total volume] ÷ [Volume of one sphere]
= [160π] ÷ [³²⁄₃ × π] 
= [160] ÷ [³²⁄₃] 
= [160] × [³⁄₃₂] 
= [16×10] × [³⁄_16×2] = 15 Nos.

Solved example 33.32
A metal sphere of radius 12 cm is melted and recast into 27 small spheres. What is the radius of each sphere?
Solution:
1. Total volume available for melting = Volume of the sphere
= ⁴⁄₃πr³ =  ⁴⁄₃×π×12³ = 2304π cm³
2. Let 'r' be the radius of one small sphere.
Then volume of one small sphere = ⁴⁄₃πr³
3. Number of spheres = [Total volume] ÷ [Volume of one sphere]
= [2304π] ÷ [⁴⁄₃πr³] 
= [2304] ÷ [⁴⁄₃×r³] 
= [2304] × [³⁄₄×(¹⁄_r³)] 
= [1728×(¹⁄_r³)]
4. But number of spheres is given as 27. So we can write:
 [1728×(¹⁄_r³)] = 27
⟹ (¹⁄_r³) = ²⁷⁄₁₇₂₈
⟹ (¹⁄_r³) = ¹⁄₆₄
⟹ r³ = 64 = 4³
⟹ r = 4 cm

Solved example 33.33
From a solid sphere of radius 10 cm, a cone of height 16 cm is carved out. What fraction of the volume of the sphere is the volume of the cone?
Solution:
1. Consider the red sphere in fig.33.32(a) below.

Fig.33.32

• Two ellipses are drawn inside it: A dotted ellipse and a dashed ellipse
[The dotted ellipse is shown just to give an emphasis to the 'spherical shape'. It does not come in any of our calculations]
• The dashed ellipse represents a circle whose centre is same as the centre of the sphere
    ♦ Also this circle is horizontal
• So this circle divides the sphere into an upper hemisphere and a lower hemisphere
• This circle is taken as the base of the cone (shown in cyan colour) in fig.b. 
• We can see that, the cone fits perfectly in the upper hemisphere. 
• This is shown more clearly in fig.c
2. From fig.c we can see that, the height of the cone will be the height of the hemisphere, which is 10 cm
• But cone given in the question has a height of 16 cm. 
• So the given cone does not fit inside the upper hemisphere alone. 
    ♦ It will occupy some portion of the lower hemisphere also
• This is shown in fig.33.33(b) below. In that fig. we can see that the, base of the new cone is below the dashed ellipse

Fig.33.33

3. In fig.33.33(c), the measurements are given
• One half of the cone is represented by the right triangle ABC
• The distance of the apex C from the centre O will be the radius of the sphere, which is 10 cm
• So the remaining distance OA will be (16-10) = 6 cm
• Distance OB will also be the radius 10 cm
• Applying Pythagoras theorem to the right triangle OAB, we get:
AB² = OB² - OA² ⟹ AB² = 10² - 6² ⟹ AB² = 100 - 36 ⟹ AB² = 64 ⟹ AB = 8 cm
4. Thus we have:
• Height of the cone, h = 16 cm
• Radius of the cone, r_c = 8 cm
• So Volume, V_c =  ¹⁄₃π(r_c)²h = ¹⁄₃×π×8²×16
• Volume of sphere, V_s = ⁴⁄₃πr³ = ⁴⁄₃×π×10³
5. Taking ratios, we get:
^V_c⁄_Vs = {¹⁄₃×π×8²×16} ÷ {⁴⁄₃×π×10³}
⟹ ^V_c⁄_Vs = {8²×16} ÷ {4×10³} 
= {8×8×16} ÷ {4×10³}
= {2×8×16} ÷ {10³}
= {256} ÷ {1000}
= ³²⁄₁₂₅
11. Thus ^V_c⁄_Vs = ³²⁄₁₂₅
⟹ Vc = ³²⁄₁₂₅ × Vs
• So 'volume of the cone' is ³²⁄₁₂₅ of the 'volume of the sphere'

Solved example 33.34
The picture shows the dimensions of a petrol tank. How many litres of petrol can it hold?

Fig.33.34

Solution:
1. The tank has two hemispherical parts and one cylindrical part
• The yellow dashed line indicates the axis of the tank
• From the fig., it is clear that radius of the hemisphere is 1 m. 
• So it's volume = Vh = ²⁄₃πr³ = ²⁄₃×π×1³= ²⁄₃×π m³
• Thus volume of two hemispheres = 2 × ²⁄₃×π = ⁴⁄₃×π m³ 
2. Height of a hemisphere will be equal to it's radius. 
So length of the cylindrical part = [6 - (2×1)] = 4 m
3. Volume of cylinder = Vc = πr²h = π×1²×4 = 4π m³ 
4. Thus total volume = ⁴⁄₃×π + 4π = ¹⁶⁄₃×π m³.
5. We know that 1 liter is the volume of a cube of edge 10 cm (Details here)
• So 1 liter = 10³ cm³ = 1000 cm³ 
• Now, (¹⁶⁄₃×π) m³ = [(¹⁶⁄₃×π) × 1000000] cm³ = 16746666.67 cm³. (∵ 1 m = 100 cm)

• Thus the no. of liters = ^16746666.67⁄₁₀₀₀ = 16746.67 liters

Solved example 33.35
A solid sphere is cut into two hemispheres. From one, a square pyramid and from the other, a cone, each of maximum possible size are carved out. What is the ratio of their volumes?
Solution:
1. Consider the red hemisphere in fig.33.35(a) below.

Fig.33.35

• A dotted ellipse and a dashed curve are drawn inside it
[The dotted curve is shown just to give an emphasis to the 'hemispherical shape'. It does not come in any of our calculations]
• The dashed ellipse represents the base of the hemisphere
• For maximum possible volume, this base is taken as the base of the cone (shown in cyan colour) in fig.33.35(b) 
• We can see that, the cone fits perfectly in the hemisphere. 
• This is shown more clearly in fig.33.35(c)
2. From fig.c, we have:
• Height of the cone, h_c = r
• Radius of the cone, r_c = r
• So Volume, V_c =  ¹⁄₃π(r_c)²h = ¹⁄₃×π×r²×r = ¹⁄₃×π×r³
3. Consider the red hemisphere in fig.33.36(a) below. It is the same hemisphere of radius r, that we saw for the cone above

Fig.33.36

• A square (seen as a rhombus in view) is drawn in the base of the hemisphere
• This square is the base of the pyramid
4. For maximum possible volume, the diagonal of the square must be equal to the diameter of the circle
• So in fig.c, we can write:
OP = OQ = half of diameter = radius = r
5. OPQ is a right triangle. We can apply Pythagoras theorem
• Then base edge = PQ = √[OP² + OQ²] = √[r² + r²] = √[2r²] = √[2]r
6. So volume of the pyramid, V_p = ¹⁄₃ × base area × height = ¹⁄₃ × √[2]r ×√[2]r × r = ²⁄₃×r³
7. Now we can take the ratio:
^V_p⁄_Vc = {²⁄₃×r³} ÷ {¹⁄₃×π×r³} = {²⁄₃} ÷ {¹⁄₃×π} = {2} ÷ {π}
• Thus we get:
 Vp : Vc = 2 : π

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We have completed this discussion on solids. In the next chapter, we will see Geometry and Algebra.

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