Given equation of lines are
3x-y=3  ...(i)                         
2x -3y =2  ...(ii)                    
and x + 2y = 8 ...(iii)     
Let lines (i), (ii) and (iii) represent the sides of Il ABC, say AB, BC and CA, respectively.
Now solve Eqs. (i) and (ii) and then Eq. (ii) and (iii) to find coordinates of point A. B. C.
= A(2, 3), B(1, 0) and C(4, 2).

Hint Let us suppose that \( \sqrt{p} \)+ \( \sqrt{q} \)is a rational
number, Again, let  \( \sqrt{p} \)+ \( \sqrt{q} \) = a, where a is rational.
Therefore, \( \sqrt{q} \)  = a - \( \sqrt{p} \)
On squaring both sides, we get
\({l} q=a^{2}+p-2 a \sqrt{p}\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right] \)
\(\\ \sqrt {p}=\frac{a^{2}+p-q}{2 a} \)
Since, p and q are primes and a is a rational number, so
\(\frac{a^{2}+p-q}{2 a}\) is rational, therefore \( \sqrt{p} \) is a rational number.
But this contradicts the fact that \( \sqrt{p} \) is irrational
number as p is prime. So, our assumption was incorrect.
Hence, \( \sqrt{p} \) + \( \sqrt{q} \)  is irrational

Given, PQRS is a trapezium in which \(PQ\parallel RS\) and PQ = 3RS.

\(\Rightarrow \frac { PQ }{ RS } =\frac { 3 }{ 1 } \) ...(i) 
In \(\triangle POQ\) and \(\triangle ROS\),
\(​​\angle SOR=\angle QOP \)  [vertically opposite angles]
\(​​\angle SRP=\angle RPQ\) [alternate angles]
\(\therefore \triangle POQ\sim \triangle ROS\)     [by AA similarity criterion]
By property of area of similar triangle,
\(\frac { ar\left( \triangle POQ \right) }{ ar\left( \triangle ROS \right) } =\frac { { \left( PQ \right) }^{ 2 } }{ { \left( RS \right) }^{ 2 } } =\left( \frac { 3 }{ 1 } \right) ^{ 2 }\)  [from Eq.(i)]
\(\Rightarrow \frac { ar\left( \triangle POQ \right) }{ ar\left( \triangle SOR \right) } =\frac { 9 }{ 1 } \)
Hence, the required ratio is 9 : 1.

Prove \(\triangle ABC\) and \(\triangle AMP\)  are similar.
then take ratio \(\frac{AC}{AP}=\frac{BC}{MP}\)

    
The correct correspondence will be \(\triangle QRP\sim \triangle SMT\) No

Diagonals of a rhombus bisect each other at right angles.

So, OA = OC = 8 cm
and OB = OD = 6 cm
Now use pythagoras theorem in  ΔAOB
10 cm.

Draw the figure according to the question and get two triangles. Then, show both triangles are similar by AAA similarity criterion and then calculate the required distance.
She is at 9 m from the base of the pole.

Given, pair oflinear equations is
x+y-2=0   ..(i)
and 2x - y - 1= 0    ..(ii)
Now, table for x + y = 2
or y=2-x is

Table for 2x - y -1 = 0
or y = 2x -1 is 

Plot the points A (0, 2) and B (2, 0) and join them to get the straight line AB. Similarly, plot the points C (0, - 1) and D (2, 3) and join them to get the straight line CD. The lines AB and CD intersect at E (1, 1). So, the solution of the given pair oflinear equations is (1, 1). 

It is clear from the graph that infinite lines can pass through the intersection point of linear equations x + y = 2 and 2x - y = 1, i.e. point E (1, 1) satisfy the many linear equations such as y = x, 2x +Y = 3, x + 2y = 3 and so on.

x=20, y=30; \(\angle A=100^{ 0 },\angle B={ 100 }^{ 0 },\angle C={ 50 }^{ 0 },\angle D=80^{ 0 }\)

Complete step-by-step answer:
Let the cost of one chair be x and the cost of one table be y.
Now from the first statement, Two chairs and three tables costs Rs.5650
The equation so formed is, 
2x+3y=5650 …eq1
From second statement, three chairs and two tables cost Rs.7100
The equation so formed is,  3x+2y=7100 …eq2
Now to solve these equations we will use elimination method,
eq1+eq2,
2x+3y+3x+2y=5650+7100
On adding the same terms,
5x+5y=12750
Dividing both sides by 5,
x+y=2550 ….eq3
Now to eliminate any one variable we need it in opposite sign, eq2-eq1,
3x+2y−(2x+3y)=7100−5650
on solving we get,
3x+2y−2x−3y=1450
subtracting the same terms,
x−y=1450….eq4
now to solve the equations eq3+-eq4,
x+y+x−y=2550+1450
Now adding same terms,
2x=40002x=4000
Dividing both sides by 2,
x=2000
This is the cost of one chair.
Now using eq3 we get,
y=x−1450
Putting the value of x,
y=2000−1450
On subtracting we get,
y=550y=550
This is the cost of one table.
Thus the cost of one chair is Rs.2000 and that of one table is Rs.550.
Note: Note that substitution method can also be used to solve the equations. But here the elimination method is easier to solve the equations. But it is important to notice the equations that are formed do not get wrong in any sign. Because that is the only mistake generally students make.

x=a², y=b²

6 sq units

Let the zeroes of q(x)=x³+2x²+a are also the zeroes of the polynomial p(x)=x⁵-x⁴-4x³+3x²+3x+b, i.e q(x) is a factor of p(x). Now, let us divide p(x) by q(x).
Then, the division process is
Since, (x³+2x²+a) is a factor
(x⁵-x⁴-4x³+3x²+3x+b), so remainder should be zero.
i.e -(1+a)x²+(3+3a)x+(b-2a)=0 or (1+a)x²+(3+3a)x+(b-2a)=0=0.x²+0.x+0
On comparing the coefficient of x² and constant term, we get
a+1=0 and b-2a=0
⇒ a=-1 and b=2a
⇒ a=-1 and b=2(-1)=-2           [∵ a=-1]
Hence, for a=-1 and b=-2, the zeroes of q(x) are also the the zeroes of the polynomial p(x).

Given, \(\sqrt { 2 } \) is one of the zeroes of the cubic polynomial. Thus, \((x-\sqrt { 2 } )\) is one of the factors of the given polynomial p(x)=\(6x^{ 3 }+\sqrt { 2x^{ 2 } } -10x-4\sqrt { 2 } \).
Then, division process is
Here, quotient \(=6x^{ 2 }+7\sqrt { 2x } +4\) and remainder =0
Now, factorise the quotient by splitting the middle term i.e. write
\(6x^{ 2 }+7\sqrt { 2x } +4=6x^{ 2 }+4\sqrt { 2x } +4\\ =2x(3x+2\sqrt { 2 } )+\sqrt { 2 } (3x+2\sqrt { 2 } )\\ =(2x+\sqrt { 2 } )(3x+2\sqrt { 2 } )\)
For other zeroes of p(x), put \(6x^{ 2 }+7\sqrt { 2x } +4=0\)
\((2x+\sqrt { 2 } )(3x+2\sqrt { 2 } )=0\)
⇒ \(x=-\frac { \sqrt { 2 } }{ 2 } ,\frac { -2\sqrt { 2 } }{ 3 } \)
Hence, other two zeroes of p(x) are \(\frac { -1 }{ \sqrt { 2 } } \) and \(\frac { -2\sqrt { 2 } }{ 3 } \) .

Given numbers are n, (n + 4), (n + 8), (n + 12) and (n + 16), where n is any positive integer. On dividing n by 5, let q be the quotient and r be the remainder.
Then, n = 5q + r, where \(0\le r<5\)
[ by Euclid's division lemma]
\(\Rightarrow \)  n = 5q + r, where r = 0, 1, 2, 3, 4
\(\Rightarrow \)  n = 5q or 5q + 1 or 5q + 2 or 5q + 3 or 5q + 4
If n = 5q, then n is only divisible by 5.
If n = 5q + 1, then n + 4 = 5q + 1 + 4 = 5q + 5 = 5(q + 1) which is divisible by 5. So, (n + 4) is only divisible by 5.
If n = 5q + 2, then n + 8 = 5q + 2 + 8 = 5q + 10 = 5(q + 2) which is divisible by 5. So, (n + 8) is only divisible by 5.
If n = 5q + 3, then n + 12 = 5q + 3 + 12  = 5q + 15 = 5(q + 3) which is divisible by 5. So, (n + 12) is only divisible by 5.
If n = 5q + 4, then n + 16 = 5q + 4 + 16 = 5q + 20 = 5(q + 4) which is divisible by 5. So, (n + 16) is only divisible by 5.
Hence, one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5, where n is any positive integer.

We have, (3 x 5 x 7) + 7 = 105 + 7 = 112
\(\therefore \)  Prime factors of 112 = 2 x 2 x 2 x 2 x 7 = 2⁴ x 7
So, it is the product of prime factors 2 and 7.
Hence, it is a composite number.

Let a be any odd positive integer, then on dividing a by b, we have a = bq + r,\( \ 0\le r ..\). (i) [by Euclid's division lemma]
On putting b = 2 in Eq. (i), we get
a = bq + r,          \(0\le\) r   \(\Rightarrow \)  r = 0 or 1
If r = 0, then a = 2q, which is divisible by 2. So, 2q is even.
If r = 1, then a = 2q + 1, which is not divisible by 2.
\(\therefore \quad \left( 2q+1 \right) \) is odd.
Now, as a is odd, so it cannot be of the form 2q. Thus, any odd positive integer a is of the form (2q + 1).
Now, consider \({ a }^{ 2 }=\left( 2q+1 \right) ^{ 2 }={ 4q }^{ 2 }+1+4q\quad \left[ \because \quad \left( x+y \right) ^{ 2 }={ x }^{ 2 }+{ y }^{ 2 }+2xy \right] \) 
= 4(q² + q) + 1 = 4 m + 1,   where m = q² + q
Hence, for some integer m, the square of any odd integer is of the form 4m + 1.

Consider 10, 20, 30, 40, ... 490. According to question
490 = 10 + (n - 1) x 10
[\(\because\) a = 10, an = 490]
and \(S_{n}=\frac{n}{2}[2 \times 10+(n-1) \times 10]\)
On solving, we get n = 49.
= 12,250

- 510

Here, 8 and 20 are the first term and common difference of an A.P.
\(\therefore\)  Sn = \({}n\over2\) [ 2(8) + ( n - 1 ) 20 ] = 8n + 10n² - 10n
= 10n² - 2n
- 30 and 8 are the first term and common difference of another A.P.
\(\therefore\)    S 2n = \({2n \over 2}\) [ 2 (-30) + ( 2n - 1 )8 ]
= - 60n +10n² - 8n = n ( - 60 + 16n - 8 )
= 16n² - 68n
As per statement of the question, we have|
16n² - 68n = 10n² - 2n
\(\Rightarrow\) 16n² - 10n² - 68n + 2n = 0
\(\Rightarrow\)  6n² - 66n = 0
\(\Rightarrow\)   6n ( n - 11 ) = 0
\(\Rightarrow\)  Either n  - 11 = 0   or n = 0
\(\Rightarrow\)        n = 11   or    n = 0  ( Rejecting )
We have   n = 11
Hence, value of n is 11.

Let the first three terms of an A.P. be a - d, a, a + d
\(\therefore\)  a - d + a + a + d = 33
\(\Rightarrow\)     3a = 33
a = 11
Now, according to the given condition|
( a - d ) ( a + d ) = a + 29
( 11 - d ) ( 11 + d ) = 11 + 29
\(\Rightarrow\)     121 - d1 = 40
\(\Rightarrow\)    d² = 81
\(\Rightarrow\)  d = \(\pm\) 9
\(\therefore\)  The required A.P is 2, 11, 20, ... or 20, 11, 2, ...

12

\(4 \pm {3 \sqrt {2}\over {2}}\)

\({1\over 5}\pm {\sqrt {51}\over {5}}\)

\(\angle \)PCA = 110°
\(\angle \)CBA = ?

\(\angle \)ACB = 90°e   [Angle in a semicircle is right agle]
\(\angle \)PCB = 110° - 90° = 20°
\(\angle \)PCB = \(\angle \)CAB = 20° [Alternate segment theorm]
In \(\triangle\)ABC
\(\angle \)ABC + \(\angle \)CAB + \(\angle \)BCA = 180°   [∵ Sum of angles of a is 180°]
⇒ \(\angle \)ABC = 180° - 110° = 70°