Again, \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}\) [From (1)]
Therefore, \(\frac{C M}{R N}=\frac{B C}{Q R}\) [From (8)]
Also, \(\frac{C M}{R N}=\frac{A B}{P Q}=\frac{2 B M}{2 Q N}\)
i.e., \(\frac{\mathrm{CM}}{\mathrm{RN}}=\frac{\mathrm{BM}}{\mathrm{QN}}\)
i.e., \(\frac{\mathrm{CM}}{\mathrm{RN}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{BM}}{\mathrm{QN}}\) [From (9) and (10)]
Therefore, Δ CMB ~ Δ RNQ (SSS similarity)

This criterion is referred to as the AAA (Angle - Angle - Angle) criterion of similarity of two triangles.
This theorem can be proved by taking two triangles ABC and DEF such that
∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F
Cut DP = AB and DQ = AC and join PQ
So, ∠ ABC ≅ ∠ DPQ
This gives ∠ B = ∠ P = ∠ E and PQ || EF
Therefore, \(\frac{\mathrm{DP}}{\mathrm{PE}}=\frac{\mathrm{DQ}}{\mathrm{QF}}\)
i.e.,  \(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{AC}}{\mathrm{DF}}\)
Similarly, \(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}} \text { and so } \frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{AC}}{\mathrm{DF}}\)

Comparing the given equations with standard forms of equations  \(a_{1} x+b_{1} y+c_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2}=0\) 
\(a 1=6, b_{1}=-3, c_{1}=10\) 
\(a_{2}=2, b_{2}=-6, c_{2}=9\) 
\(\therefore \frac{a_{1}}{a_{2}}=\frac{6}{2}=3, \frac{b_{1}}{b_{2}}=\frac{-3}{-1}=3\) 
\(\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\) 
Thus, the lines representing the pair of linear equations are parallel.

Given, a pair of linear equation is:
\(\sqrt { 2 } x+\sqrt { 3 } y=0\)   ......(i)
and \(\sqrt { 3 } x-\sqrt { 8 } y=0\quad or\quad y=\frac { \sqrt { 3 } x }{ \sqrt { 8 } } \)  ......(ii)   
On substituting y from eqn. (ii) in eqn. (i),
\(\sqrt { 2 } x+\sqrt { 3 } \times \left( \frac { \sqrt { 3 } x }{ \sqrt { 8 } } \right) =0\)
\(\Rightarrow \quad \sqrt { 2 } x\times \sqrt { 8 } +3x=0\)
\(\Rightarrow \quad \sqrt { 2 } x+\frac { 3x }{ \sqrt { 8 } } =0\)
\(\Rightarrow \quad \sqrt { 2 } x\times \sqrt { 8 } +3x=0\)
\(\\ \Rightarrow \)  4x + 3x = 0
\(\\ \Rightarrow \)  7x = 0
\(\therefore\) x = 0
On substituting x = 0 in eqn. (ii),
\(y=\frac { \sqrt { 3 } \times 0 }{ \sqrt { 8 } } =0\)
\(\therefore\)y = 0
Hence, x = 0, y = 0

Let us assume, to the contrary, that \(\sqrt 2\) is rational.
So, we can find integers r and s (≠ 0) such that \(\sqrt 2\) =\(\frac{r}{s}\) .
Suppose r and s have a common factor other than 1. Then, we divide by the common factor to get \(\sqrt 2\) = \(\frac{a}{b}\) , where a and b are coprime.
So, b\(\sqrt 2\) = a.
Squaring on both sides and rearranging, we get 2b ² = a ² .Therefore, 2 divides a ² .
Now, by  it follows that 2 divides a.
So, we can write a = 2c for some integer c.
Substituting for a, we get 2b² = 4c² , that is, b² = 2c² .
This means that 2 divides b² , and so 2 divides b (again using Theorem 1.3 with p = 2).
Therefore, a and b have at least 2 as a common factor.
But this contradicts the fact that a and b have no common factors other than 1.
This contradiction has arisen because of our incorrect assumption that \(\sqrt 2\) is rational.
So, we conclude that \(\sqrt 2\) is irrational.

Let the quadratic polynomial be ax² + bx + c, and its zeroes be \(\alpha \text { and } \beta \text { . }\) 
We have
\(\alpha+\beta=-3=\frac{-b}{a}\) 
and \(\alpha \beta=2=\frac{c}{a}\) 
If a = 1, then b = 3 and c = 2.
So, one quadratic polynomial which fits the given conditions is x² + 3x + 2.
You can check that any other quadratic polynomial that fits these conditions will be of the form k(x² + 3x + 2), where k is real.
Let us now look at cubic polynomials. Do you think a similar relation holds between the zeroes of a cubic polynomial and its coefficients
Let us consider p(x) = 2x³ – 5x² – 14x + 8.
You can check that p(x) = 0 for x = 4,\(-2, \frac{1}{2}\) .Since p(x) can have atmost three zeroes, these are the zeores of 2x³ – 5x² – 14x + 8. Now,
sum of the zeroes = \(4+(-2)+\frac{1}{2}=\frac{5}{2}=\frac{-(-5)}{2}=\frac{-\left(\text { Coefficient of } x^{2}\right)}{\text { Coefficient of } x^{3}}\) 
product of the zeroes = \(4 \times(-2) \times \frac{1}{2}=-4=\frac{-8}{2}=\frac{-\text { Constant term }}{\text { Coefficient of } x^{3}}\) 
However, there is one more relationship here. Consider the sum of the products of the zeroes taken two at a time. We have
\(\{4 \times(-2)\}+\left\{(-2) \times \frac{1}{2}\right\}+\left\{\frac{1}{2} \times 4\right\}\) 
\(=-8-1+2=-7=\frac{-14}{2}=\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{3}}\) 
In general, it can be proved that if \(\alpha, \beta, \gamma\)  are the zeroes of the cubic polynomial ax³ + bx² + cx + d, then
\(\alpha+\beta+\gamma=\frac{-b}{a}\) 
\(\alpha \beta+\beta \gamma+\gamma \alpha=\frac{c}{a}\) 
\(\alpha \beta \gamma=\frac{-d}{a}\) 

Given, the sum of first n terms,
  \({ S }_{ n }=4n-{ n }^{ 2 }\)                ...(i)
On putting n = 1 in Eq. (i), we get 
 \({ S }_{ 1 }=4\times 1-{ 1 }^{ 2 }=4-1=3\) 
Thus, sum of first term = 3
On putting n = 2 in Eq. (i), we get
  \({ S }_{ 2 }=4\times 2-{ 2 }^{ 2 }=8-4=4\) 
Thus, sum of first two terms = 4
\(\because\) the n th term of an AP, a_n = S_n - S_n-1
\(\therefore\) Second term = S₂ - S₁ = 4 - 3 = 1
On putting n = 3 in Eq. (i), we get
S₃ = 4 \(\times\) 3 - 3² = 12 - 9 = 3
\(\therefore\) third term = S₃ - S₂ = 3 - 4 = -1
On putting n = 9 in eq. (i), we get
S₉ = 4 \(\times\) 9 - 9² = 36 - 81 = -45
Again, putting n = 10 in Eq. (i), we get
S₁₀ = 4 \(\times\)10 - 10² = 40 - 100 = -60
\(\therefore\) 10th term = S₁₀ - S₉
= -60 - (-45) = -60 + 45 = -15
Now, on replacing n by n - 1 in Eq. (i), we get
\(\begin{aligned} S_{n-1} & =4(n-1)-(n-1)^2 \\ \end{aligned}\)
\(\begin{aligned} & =4 n-4-n^2+2 n-1=-n^2+6 n-5 \\ \end{aligned}\)
\(\begin{aligned} \therefore \quad n \text {th term } & =S_n-S_{n-1} \\ \end{aligned}\)
\(\begin{aligned} & =4 n-n^2-\left(-n^2+6 n-5\right) \end{aligned}\)
\(\begin{aligned} & =4 n-n^2+n^2-6 n+5=5-2 n \end{aligned}\)

Length of spiral made up of thirteen consecutive semi-circles
\(=(\pi \times 0.5+\pi \times 1.0+\pi \times 1.5+\pi \times 2.0+\ldots+\pi \times 6.5)\)
[\(\because\) circumference of semi-circle= \(\pi\)r, where, r is radius of circle]
\(\begin{aligned} & =0.5 \pi[1+2+3+4+\ldots+13] \\ \end{aligned}\)
\(\begin{aligned} & =\pi \times 0.5 \times \frac{13}{2}[2 \times 1+(13-1) \times 1] \end{aligned}\)
[\(\because\) the numbers 1, 2,3, .,.13, forms an AP with a =1, d = 2 - 1 = 1. Also,
\(\left.S_n=\frac{n}{2}\{2 a+(n-1) d\}\right]\)
\(=\frac{22}{7} \times \frac{5}{10} \times \frac{13}{2} \times 14=143 \mathrm{~cm}\)

Let the number of marbles John had be x
Then the number of marbles Jivanti had be = 45 – x (Why?).
The number of marbles left with john, when he lost 5 marbles = x – 5
The number of marbles left with Jivanti, when she lost 5 marble = 45 – x – 5 = 40 – x
Therefore, their product = (x – 5) (40 – x)
= 40x – x² – 200 + 5x
= – x² + 45x – 200
So, – x² + 45x – 200 = 124 (Given that product = 124)
i.e., – x² + 45x – 324 = 0
i.e., x² – 45x + 324 = 0
Therefore, the number of marbles John had, satisfies the quadratic equation
x² – 45x + 324 = 0
which is the required representation of the problem mathematically

(i) LHS = (x – 2)² + 1 = x² – 4x + 4 + 1 = x2 – 4x + 5
Therefore, (x – 2)² + 1 = 2x – 3 can be rewritten as
x² – 4x + 5 = 2x – 3
i.e., x² – 6x + 8 = 0
It is of the form ax² + bx + c = 0.
Therefore, the given equation is a quadratic equation
(ii) Since x(x + 1) + 8 = x² + x + 8 and (x + 2)(x – 2) = x² – 4
Therefore, x² + x + 8 = x² – 4
i.e., x + 12 = 0
It is not of the form ax² + bx + c = 0.
Therefore, the given equation is not a quadratic equation
(iii) Here, LHS = x (2x + 3) = 2x² + 3x
So, x (2x + 3) = x² + 1 can be rewritten as
2x² + 3x = x² + 1
Therefore, we get x² + 3x – 1 = 0
It is of the form ax² + bx + c = 0.
So, the given equation is a quadratic equation
(iv) Here, LHS = (x + 2)³ = x3 + 6x² + 12x + 8
Therefore, (x + 2)³ = x³ – 4 can be rewritten as
x³ + 6x² + 12x + 8 = x3 – 4
i.e., 6x² + 12x + 12 = 0 or, x² + 2x + 2 = 0
It is of the form ax² + bx + c = 0.
So, the given equation is a quadratic equation

In ΔDCA = ΔHGF,
∠DAC = ∠HFG ......(i)
\(\left[\begin{array}{l} \because \quad \Delta A B C \sim \Delta F E G \\ \therefore \angle C A B=\angle G F E \\ \Rightarrow \angle C A D=\angle G F H \text { or } \angle D A C=\angle H F G \end{array}\right]\)
and ∠DCA = ∠HGF   ........(ii)
\(\left[\begin{array}{l} \because \Delta A B C \sim \Delta F E G \\ \therefore \angle A C B=\angle F G E \\ \Rightarrow \frac{1}{2} \angle A C B=\frac{1}{2} \angle F G E \Rightarrow \angle D C A=\angle H G F \end{array}\right]\)
From Eqs. (i) and (ii),
ΔDCA - ΔHGF [by AA similarity criterion)

Let p(u) = 4u² + 8u = 4u(u+2)
To find zeroes, put p(u) = 0
\(\Rightarrow\) 4u(u+2) = 0 \(\Rightarrow\) u = 0 or u + 2 = 0 [\(\because\) 4 \(\neq\)0]
\(\Rightarrow\) u = 0 or u = -2
Hence, zeroes of the given polynonial are 0 and -2.
Verification
Here, sum of zeroes = 0 - 2 = -2 = -(8/4)
=-\(\frac{Coefficient \quad of \quad u}{Coefficient \quad of \quad u^{2}}\)
and product of zeroes
=0 \(\times\)-2 = 0 = (0/4) = \(\frac{Constant \quad term}{Coefficient \quad of \quad u^{2}}\)
so, the relationship between the zeroes and its coefficients is verified.

Proof : Let the prime factorisation of a be as follows :
a = p₁ p₂ . . . p_n , where p₁ ,p₂ , . . ., p_n are primes, not necessarily distinct. Therefore, a² = ( p₁ p₂ . . . p_n )( p₁ p₂ . . . p_n ) = p²₁ p²₂ . . . p ²_n .
Now, we are given that p divides a ² . Therefore, from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of a ² . However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only prime factors of a² are p₁ ,p₂ , . . ., p_n . So p is one of p₁ , p₂ , . . ., p_n .
Now, since a = p₁ p₂ . . . p_n ,p divides a.
We are now ready to give a proof that \(\sqrt 2\) is irrational.
The proof is based on a technique called ‘proof by contradiction’. 

Let the fixed charges for the first three days be Rs x.
Let the additional change per day be Rs y.
According to the given conditions,
Sarita paid for 7 days = Rs 27
i.e.. x + 4 \(\times\) y = 27
i.e., x + 4y = 27 ... (i)
[\(\because\) Rs 4y are to be paid for extra 4 days]
In the case of Susy,
x + 2y = 21        ...(ii)
[\(\because\) Rs 2y are to be paid for extra 2 days]
On substracting eqn. (ii) from eqn. (i),
2y = 27 - 21
\(\Rightarrow\) 2y = 6
\(\therefore\) y = 3
On substituting y = 3 in eqn. (i),
x + 4 \(\times\) 3 = 27
\(\Rightarrow\)x = 27 -12
\(\therefore\) x = 15
Hence, fixed charge for first three days = Rs 15 Additional charge per extra day = Rs 3.

The number of rose plants in the 1st, 2nd, 3rd, . . ., rows are :
23, 21, 19, . . ., 5
It forms an AP . Let the number of rows in the flower bed be n.
Then a = 23, d = 21 – 23 = – 2, a_n = 5
As, a_n = a + (n – 1) d
We have,  5 = 23 + (n – 1) (– 2)
i.e.,   – 18 = (n – 1) (– 2)
i.e.,   n = 10
So, there are 10 rows in the flower bed.

Let the two consecutive integers be x and x+1
ATQ x²+(x+1)2=365
\(\Rightarrow\) x²+x2+2x+1=365 \(\Rightarrow\) 2x²+2x-364=0
\(\Rightarrow\) x²+x-182=0 \(\Rightarrow\) x²+14x-13x-182=0
\(\Rightarrow\) x(x+14)-13(x+14)=0 \(\Rightarrow\) (x-13)(x+14)=0
\(\Rightarrow\) x=13, -14 (-14 is rejected because it is a negative integer)
Hence, the two consecutive positive integers are 13 and 13+1=14

Given, AD and CE are altitudes which intersect each other at the point P.
In ΔABD and ΔCBE,
∠ADB = ∠CEB      [each 90°]
and ∠ABD = ∠CBE      [Common angle]
\(\therefore\) ΔABD ∼ ΔCBE    [by AA similarity criterion]

The given equations can be rewritten as
2x - 3y - 8 = 0 and 4x - 6y - 9 = 0
On comparing with standard form of pair of linear equations, we get
a₁ = 2, b₁ = -3, c₁ = -8
and a₂ = 4, b₂ = -6, c₂ = -9
Now, \(\frac{a_{1}}{a_{2}}=\frac{2}{4}=\frac{1}{2},\frac{b_{1}}{b_{2}}=\frac{-3}{-6}=\frac{1}{2} and \frac{c_{1}}{c_{2}}=\frac{-8}{-9}=\frac{8}{9}\)
Thus, \(\frac{1}{2}=\frac{1}{2}\neq \frac{8}{9}\) i.e.,\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Hence, the pair of linear equations is inconsistent.

Given points are A(0,0) and B(36,15) 
Now, using distance formula, we have
AB = \(\sqrt {(36-0)^{2}+(15-0)^{2}}\)
=\(\sqrt{{36}^{2}+{15}^{2}}\) = \(\sqrt{1296+225}\)
=\(\sqrt{1521}\) =39 units

Let us take position of the town A as che origin and the position of the town B as the point B IS) in the coordinate axis as Shown in figure. 
Now, |AB|= \(\sqrt{(36^{2}+(15)^{2}}\)
=\(\sqrt{{36}^{2}+{15}^{2}}\) = \(\sqrt{1296+225}\)
=\(\sqrt{1521}\) =39 units

Here, a₁₂ = 37 and d = 3
then, a₁₂ = 37
\(\Rightarrow\) a + 11d = 37   [\(\because\) a_n = a + (n - 1) d]
\(\Rightarrow\) a + 11(3) = 37   [\(\because\) d = 3]
\(\Rightarrow\) a = 37 - 33 = 4
On putting n = 12, a = 4 and l = a₁₂ = 37 in
\(\begin{aligned} & S_n=\frac{n}{2}(a+l) \text {, we get } \\ \end{aligned}\)
\(\begin{aligned} & \qquad S_{12}=\frac{12}{2}(4+37)=6 \times 41=246 \end{aligned}\)
Hence, a = 4 and S₁₂ = 246.

Consider the equation
x² - 3x + 5 = 0
Comparing it with ax² + bx + c = 0, we get
a = 2, b = -3 and c = 5
Discriminant = b2 - 4ac
= (-3)2 - 4 (2) (5) = 9 - 40
= - 31
As b² - 4ac < 0,
Therefore, no real root is possible for the given equation.