Let A(1, 7), B(4, 2), C(–1, –1) and D(– 4, 4) be the given points. One way of showing that ABCD is a square is to use the property that all its sides should be equal and both its digonals should also be equal. Now
\(A B=\sqrt{(1-4)^{2}+(7-2)^{2}}=\sqrt{9+25}=\sqrt{34}\) 
\(\mathrm{BC}=\sqrt{(4+1)^{2}+(2+1)^{2}}=\sqrt{25+9}=\sqrt{34}\) 
\(C D=\sqrt{(-1+4)^{2}+(-1-4)^{2}}=\sqrt{9+25}=\sqrt{34}\) 
\(\mathrm{DA}=\sqrt{(1+4)^{2}+(7-4)^{2}}=\sqrt{25+9}=\sqrt{34}\) 
\(A C=\sqrt{(1+1)^{2}+(7+1)^{2}}=\sqrt{4+64}=\sqrt{68}\) 
\(\mathrm{BD}=\sqrt{(4+4)^{2}+(2-4)^{2}}=\sqrt{64+4}=\sqrt{68}\) 
Since, AB = BC = CD = DA and AC = BD, all the four sides of the quadrilateral ABCD are equal and its diagonals AC and BD are also equal. Thereore, ABCD is a square.
Alternative Solution : We find the four sides and one diagonal, say, AC as above. Here AD² + DC² = 34 + 34 = 68 = AC². Therefore, by the converse of Pythagoras theorem, \(\angle \mathrm{D}=90^{\circ}\). A quadrilateral with all four sides equal and one angle 90° is a square. So, ABCD is a square.

Let us assume, to the contrary, that 5 - \( \sqrt{3}\) is rational.
That is, we can find coprime a and b (b \(\neq\)0) such that 5 - \( \sqrt{3}\) = \(\frac{a}{b}\)
Therefore, 5 - \(\frac{a}{b}=\sqrt{3}\)
Rearranging this equation, we get \(\sqrt{3}\) = 5 - \(\frac{a}{b}\)=\(\frac{5b-a}{b}\)
Since a and b are integers, we get 5 - \(\frac{a}{b}\) is rational, and so \( \sqrt{3}\) is rational.
But this contradicts the fact that \(\sqrt{3}\) is irrational.
This contradiction has arisen because of our incorrect assumption that 5 - \( \sqrt{3}\) is rational.
So, we conclude that 5 - \( \sqrt{3}\) is irrational.

Total number of cards in one deck of cards is 52.
\(\therefore\) Total number of outcomes =52
(i) Let E₁ = Event of getting a king of red colour
\(\therefore\) Number of outcomes favourable to E₁ =2
[\(\because\) there are four kings in a deck of playing cards out of which two are red and two are black]
Hence, probability of getting a king of red colour,
\(P\left(E_1\right)=\frac{2}{52}=\frac{1}{26}\)
(ii) Let E₂ = Event of getting a face card
\(\therefore\) Number of outcomes favourable to E₂ =12
(\(\because\) in a deck of cards, there are 12 face cards, namely 4 kings, 4 jacks, 4 queens]
Hence, probability of getting a face card,
\(P\left(E_2\right)=\frac{12}{52}=\frac{3}{13}\)
(iii) P(a red face card)  \(\frac{6}{52}\)  = \(\frac{3}{26}\)
(iv) Let E₄= Event of getting a jack of heart
\(\therefore\)Number of outcomes favourable to E₄ = 1 
[\(\because\) there are four jack cards in a deck, namely 1 of heart, 1 of club, 1 of spade and 1 of diamond] 
Hence, probability of getting a jack of heart,
\(P\left(E_4\right)=\frac{1}{52}\)
(v) Let E₅ = Event of getting a spade
\(\therefore\) Number of outcomes favourable to E₅ = 13
[\(\because\) in a deck of cards, there are 13 spades, 13 clubs, 13 hearts and 13 diamonds]
Hence, probability of getting a spade,
\(P\left(E_5\right)=\frac{13}{52}=\frac{1}{4}\)
(vi) P(the queen of diamonds) =\(\frac{1}{52}\)

Let DB is a tree and AD is the broken part of it which touches the ground at C.

Given: ㄥACB = 30\(\unicode{xb0} \) and BC = 8m
Let AB = x m and AD = y m
∴ Now, length of the tree = (x + y)m
In ΔABC,
\(\frac { AB }{ BC } \) = tan 30\(\unicode{xb0} \) ⇒ \(\frac { x }{ 8 } =\frac { 1 }{ \sqrt { 3 } } \Rightarrow \frac { 8 }{ \sqrt { 3 } } \)
and \(\frac { AB }{ AC } \)-sin 30\(\unicode{xb0} \) ⇒ \(\frac { x }{ y } =\frac { 1 }{ 2 } \)
⇒ y = 2x ⇒ y = 2 x \(\frac { 8 }{ \sqrt { 3 } } =\frac { 16 }{ \sqrt { 3 } } \)
Hence, total height of the tree
x + y = \(\frac { 8 }{ \sqrt { 3 } } +\frac { 16 }{ \sqrt { 3 } } =\frac { 24 }{ \sqrt { 3 } } =\frac { 24 }{ \sqrt { 3 } } \times \frac { \sqrt { 3 } }{ \sqrt { 3 } } =\frac { 24\sqrt { 3 } }{ 3 } \)= 8 x 1.732 = 13.856 m

\(L H S=\frac{\cot A-\cos A}{\cot A+\cos A}=\frac{\frac{\cos A}{\sin A}-\cos A}{\frac{\cos A}{\sin A}+\cos A}\)
\(=\frac{\cos A\left(\frac{1}{\sin A}-1\right)}{\cos A\left(\frac{1}{\sin A}+1\right)}=\frac{\left(\frac{1}{\sin A}-1\right)}{\left(\frac{1}{\sin A}+1\right)}=\frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1}=\operatorname{RHS}\)

Given point P(x, y) is equidistant from the points A(7, 1) and B(3, 5).
So,   AP = BP
\(\Rightarrow\)   AP² = BP²
\(\Rightarrow\) (x - 7)² + (y - 1)² = (x - 3)² + (y - 5)²
       \(\left[\because \text { distance }=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\right]\)
\(\Rightarrow\)   x² + 49 - 14x + y² + 1 - 2y
                = x² + 9 - 6x + y² + 25 - 10y
\(\Rightarrow\)  -14x - 2y + 50 = -6x - 10y + 34
\(\Rightarrow\) -6x - 10y + 14x + 2y = 50 - 34
\(\Rightarrow\)              8x - 8y = 16
\(\Rightarrow\)                 x - y = 2
                                [dividing by 8 on both sides]
Hence, the relation between x and y is x - y = 2.

510 and 92
510 = 2 x 255
= 2 x 3 x 85
= 2 x 3 x 5 x 17
= 2 x 3 x 5 x 17 x 1
Therefore 510 = 2 x 3 x 5 x 17 x 1 .....(A)
92 = 2 x 46
= 2 x 2 x 23
Hence 92 = 2 x 2 x 23 .....(B)
From (A) and (B) HCF of 510 and 92 is = 2 and their 
LCM is 2 x 2 x 3 x 5 x 17 x 23 = 23460
Product of the LCM and HCF = 2 x 23460 = 46920
Product of the two numbers = 510 x 92 = 46920
Therefore it is proved that LCM x HCF = Product of the two numbers.

8, 9 and 25
8 =  2³ x 1
9 = 3² x 1
25 = 5² x 1
From the above HCF (8, 9, 25) = 1 and LCM (8, 9, 25) = 1800

Let the quadratic polynomial be ax² + bx + c, and its zeroes be α and ß
Given \(\alpha+B=4=-\frac{b}{a}\) 
\(\alpha \beta=1=\frac{c}{a}\) 
If a = 1, then b = -4, c = 1
Therefore, the quadratic polynomial is x² – 4x +1.

(i) The number of zeroes is 1 as the graph intersects the x-axis at one point only.
(ii) The number of zeroes is 2 as the graph intersects the x-axis at two points.
(iii) The number of zeroes is 3. (Why?)
(iv) The number of zeroes is 1. (Why?)
(v) The number of zeroes is 1. (Why?)
(vi) The number of zeroes is 4. (Why?)

Let A(1, 2), B(4, y), C(x, 6) and D(3, 5) are the vertices of a parallelogram.
Since, ABCD is a parallelogram.
\(\therefore\) Diagonals AC and BD will bisect each other. So, the mid-point of AC and mid-point of BD will be same 

Thus mid-point of AC = Mid-point of BD
\(\begin{aligned} \Rightarrow \quad & \left(\frac{1+x}{2}, \frac{2+6}{2}\right)=\left(\frac{4+3}{2}, \frac{y+5}{2}\right) \\ \end{aligned}\)
\(\begin{aligned} & {\left[\because \text { coordinates of mid-point }=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\right] } \end{aligned}\)
On comparing the coordinate from both sides,we get
\(\frac{1+x}{2}=\frac{4+3}{2} \text { and } \frac{2+6}{2}=\frac{5+y}{2}\)
\(\Rightarrow\) 1 + x = 7 and 8 = 5 + y
\(\therefore\)   x = 6 and y = 3

(cos 45°)/(sec 30° + cosec 30°)
\(=\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}}+2}=\frac{\frac{1}{\sqrt{2}}}{\frac{2+2 \sqrt{3}}{\sqrt{3}}}\)
\(=\frac{\sqrt{3}}{\sqrt{2}(2+2 \sqrt{3})}=\frac{\sqrt{3}}{2 \sqrt{2}+2 \sqrt{6}}\)
\(=\frac{\sqrt{3}(2 \sqrt{6}-2 \sqrt{2})}{((2 \sqrt{6})+2 \sqrt{2})(2 \sqrt{6}-2 \sqrt{2})}\)
\(=\frac{2 \sqrt{3}(\sqrt{6}-\sqrt{2})}{(2 \sqrt{6})^{2}-(2 \sqrt{2})^{2}}=\frac{2 \sqrt{3}(\sqrt{6}-\sqrt{2})}{24-8}=\frac{2 \sqrt{3}(\sqrt{6}-\sqrt{2})}{16}\)
\(=\frac{\sqrt{18}-\sqrt{6}}{8}=\frac{3 \sqrt{2}-\sqrt{6}}{8}\) 

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’. 

Given PQ = 3 cm and PR = 6 cm.
Therefore,  \(\begin{aligned} \frac{\mathrm{PQ}}{\mathrm{PR}} & =\sin \mathrm{R} \\ \end{aligned}\)
or  \(​​​​\begin{aligned} \sin \mathrm{R} & =\frac{3}{6}=\frac{1}{2} \\ \end{aligned}\)
So, \(\begin{aligned} \angle \mathrm{PRQ} & =30^{\circ} \\ \end{aligned}\)
and therefore, \(\begin{aligned} \angle \mathrm{QPR} & =60^{\circ} \end{aligned}\)
You may note that if one of the sides and any other part (either an acute angle or any side) of a right triangle is known, the remaining sides and angles of the triangle can be determined.

Let p(x) = ax³ + bx² + cx + d
Given that, one of the zeroes of the cubic polynomial p(x) is zero.
Let \(\alpha,\beta\) and \( \gamma\) arethe zeroes of cubic polynomial p (x),
where a = 0.
We know that,
Sum of product of two zeroes at a time =\(\frac{c}{a}\)
\(​​​​\alpha\beta +\beta\gamma+\beta​​​​\alpha=\frac{c}{a}\)
\(​​​​\alpha\beta +\beta\gamma+\gamma*0=\frac{c}{a}[\because \alpha=0,given]\)
  \(0+\beta\gamma+0=\frac{c}{a}\Rightarrow\beta\gamma=\frac{c}{a}\)
Hence, product of other two zeroes =\(\frac{c}{a}\)