\(P R^{2}-P Q^{2}=Q R^{2} \Rightarrow P R^{2}=P Q^{2}+Q R^{2}\)
\(\Rightarrow\) ΔPQR is right angled triangle, right angled at Q.

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

Altitude AD which is perpendicular to BC, then D is the mid-point of BC.
BD = CD = \(\frac{1}{2}\) BC = \(\frac{8}{2}\) = 4 cm.
Apply, Pythagoras theorem to find AD.
AD = \(4\sqrt{3}\) cm.

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.

Given, \(\triangle ABC\sim \triangle EDF\)
Also, AB = 5 cm, AC = 7 cm, DF = 15 cm
and DE = 12 cm                         .... (i)
Since, \(\triangle ABC\sim \triangle EDF\)
\(\therefore \frac { AB }{ ED } =\frac { AC }{ EF } =\frac { BC }{ DF } \)
[ ∵ Corresponding sides of similar triangles are proportions]
\(\Rightarrow \frac { 5 }{ 12 } =\frac { 7 }{ EF } =\frac { BC }{ 15 } \)  [from Eq. (i)]

On taking first and second terms, we get
\(\frac { 5 }{ 12 } =\frac { 7 }{ EF } \Rightarrow EF=\frac { 7\times 12 }{ 5 } \) = 16.8 cm
On taking first and third terms, we get
\(\frac { 5 }{ 12 } =\frac { BC }{ 15 } \Rightarrow BC=\frac { 5\times 15 }{ 12 } \) = 6.25 cm
Hence, lengths of the remaining sides of the triangles are EF = 16.8 cm and BC = 6.25 cm.

Draw the figure according to the given conditions and use Pythagoras theorem to find the value of x, then required saved distance will be equal to the difference of (AC + BC) and 26.
8 km.

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

x=a², y=b²

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 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.

No, because HCF does not divide LCM.

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.

Given, edge of the cube=22cm
Volume of the cube=(Edge)³=(22)³=10648cm³
Also, given that diameter of a marble=0.5cm
Radius of a marble\(=\frac { 0.5 }{ 2 } =0.25cm\)  \(\left[ \because \quad radius=\frac { diameter }{ 2 } \right] \)
Volume of one spherical marble
\(=\frac { 4 }{ 3 } \times \frac { 22 }{ 7 } \times \left( 0.5 \right) ^{ 3 } \ \ \ \ \left[ \because \ volume \ of \ sphere=\frac { 4 }{ 3 } \pi r^{ 3 } \right] \\ =\frac { 88 }{ 21 } \times 0.015625=\frac { 1375 }{ 21 } =0.0655cm^{ 3 }\)
Filled space of the cube with marble
=Volume of cube\(-\frac { 1 }{ 8 } \times \)Volume of the cube
\(=10648-\frac { 1 }{ 8 } \times 10648=10648-1331\\ =9317cm^{ 3 }\)
Required number of marbles
\(=\frac { Total \ space \ filled \ by \ marbles \ in \ a \ cube }{ Volume \ of \ one \ marble } \\ =\frac { 9317 }{ 0.0655 } =142244\)
Hence, the number of marbles that the cube accommodate is 142244.

To draw a rhombus AB' C' D' similar to rhombus ABCD, we use the following steps of construction
1. First, draw a rhombus ABCD, in which AB = 4 cm and \(\angle ABC=60°\)  and join its diagonal ACwhich divides it into two triangles ABC and ADC
2. Now, construct MB' C' similar to MBC with scale factor 2/3
(i) First, draw a ray AX making an acute angle with AB downwards.
(ii) Mark 3 points A₁ A₂  and A₃ on AX such that
 AA₁ = A₁A₂ = A₂A₃
(iii) Join A₃B and then draw A ₂B' II A₃B which intersects AB at B'.
(iv) Draw B^' C^' parallel to BC which intersects AC at C^' . Thus,\(\Delta A{ B }^{ ' }{ C }^{ ' }\) similar, to \(\Delta ABC\) is constructed.
3. Draw the line segment D^'C^' II DC. Thus, AB' C' D' is the required rhombus similar to given rhombus ABCD.
   

Given, S₅ + S₇ = 167 and S₁₀ = 235
\(\Rightarrow 12 a+31 d=167 \text { and } 2 a+9 d=47\)
Now, find a and d, and then S₂₀
S₂₀ = 970

\((i)\frac { -3 }{ 2 } \) and \(\frac { 2 }{ 3 } \)  \((ii)\frac { 1 }{ 2 } \) and 3 

Let her pocket money be Rs x. Now, she takes Rs 1 on day 1, Rs 2 on day 2, Rs 3 on day 3 and so on till the end of the month, from this money.
i.e.  1 + 2 + 3 + 4 + ..... + 31
which form an AP, in which number of terms is 31 and first term 
\(\therefore\) Sum of first terms = S ₃₁
Sum of n terms,  \({ S }_{ n }=\frac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] \)
\(\therefore\)  \({ S }_{ 31 }=\frac { 31 }{ 2 } \left[ 2\times 1+\left( 31-1 \right) \times 1 \right] \)
\(=\frac { 31 }{ 2 } \left( 2+30 \right) \)
\(=\frac { 31\times 32 }{ 2 } \)
\(=31\times 16=496\)
So, Kanika takes Rs 496 till the end of the month from this money.
Also, she spent Rs 204 of her pocket money and found that at the end of the month, she still has Rs 100 with her.
Now, according to the condition
\(\left( x-496 \right) -204=100\)
\(\Rightarrow\) x - 700 = 100
\(\Rightarrow\)   x = Rs 800
Hence, Rs 800 was her pocket money for the month.

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

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

Here a = \(\frac{a-b}{a+b},d=\frac{3a-2b}{a+b}-\frac{a-b}{a+b}\)
\(={2a-b\over a+b}\) and n = 11.
S_n  = \({n\over2}[2a+(n-1)d]\)
\(\Rightarrow\) S₁₁ = \(\frac{11}{2}\left[ 2\left( a-b\over a+b \right)+(11-1)\left( 2a-b \over a+b \right)\right]\)
\(\Rightarrow\)  S₁₁ = \({11\over2}\times2\left[ {a-b\over a+b }+{5(2a-b)\over a+b} \right]\)
\(\Rightarrow\)   S₁₁ = \(11\left[ {a-b+10a-5b\over a+b} \right]\)
\(\Rightarrow\)   S₁₁ = \(11\left[ 11a-6b\over a+b \right]\)

In \(\triangle\)ABC
AC is diameter, B = 90°    ......(i)
[In a semicircle there is always a right angle]
So, ACB + CAB = 90°
[∵ sum of angles of a\(\triangle\) is 180°]
OA ⊥ AT    [Radius and tangent are ⊥ to each other at the point of contact]
\(\angle \)OAT = 90°
\(\angle \)OAB + \(\angle \)BAT = 90°   ......(ii)
From (i) and (ii),
\(\angle \)ACB = \(\angle \)BAT     Hence proved.

\(\angle \)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°