OA. OB = OC . OD    (Given)
So, \(\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OD}}{\mathrm{OB}}\)     (1)
Also, we have  \(\angle\)AOD = \(\angle\)COB   (Vertically opposite angles) (2)
Therefore, from (1) and (2),  \(\Delta\)AOD \(\sim\) \(\Delta\)COB (SAS similarity criterion)
So, \(\angle\)A = \(\angle\)C and \(\angle\)D = \(\angle\)B
(Corresponding angles of similar triangles)

Using distance formula
\(AB=\sqrt { ({ -3-2) }^{ 2 }+[8-({ -2)] }^{ 2 } } \)
\(=\sqrt { { 5 }^{ 2 }+10^{ 2 } } =\sqrt { 25+100 } \)
\(=\sqrt { 125 } =5\sqrt { 5 } units\)
\(BC=\sqrt { [{ -1-(-3)] }^{ 2 }+(4-8)^{ 2 } } \)
\(=\sqrt { { 2 }^{ 2 }+4^{ 2 } } =\sqrt { 4+16 } =\sqrt { 20 } \)
\(=2\sqrt { 5 } units\)
\(AC=\sqrt { ({ -1-2) }^{ 2 }+[4-(-2)]^{ 2 } } \)
\(=\sqrt { { 3 }^{ 2 }+6^{ 2 } } =\sqrt { 9+36 } =\sqrt { 45 } \)
\(=3\sqrt { 5 } units\)
\(\because \ 2\sqrt { 5 } +3\sqrt { 5 } =5\sqrt { 5 } \)
∴ Points are collinear. 

Let length be x m and breadth be y m
Perimeter =2(l+b) \(\Rightarrow \) 2(x+y)=80
\(\Rightarrow \) x+y=40 \(\Rightarrow \) y=40-x
Area = lxb \(\Rightarrow \) x(40-x) =300
\(\Rightarrow \) 40-x²=300 \(\Rightarrow \) x²-40x+300=0
D=b²-4ac=(-40)²-4x1x300=1600-1200=400 > 0
\(\therefore \) Real solution is possible 
Hence it is possible to design sides of a park.

Here  a = 17,  d = 14\(\frac{1}{2}\) - 17 =\(\frac{29}{2}\) - 17 = \(-\frac{5}{2}\)
Let number of terms in AP = n
 a_n = -38
\(\Rightarrow\) a + (n - 1)d = -38
\(\Rightarrow\)  17 + (n - 1) x \((-\frac{5}{2})\) = -38
\(\Rightarrow\)    (n - 1)\((-\frac{5}{2})\) = -55
\(\Rightarrow\)  (n - 1) = -55 x \((-\frac{2}{5})\) = 22
 n = 23

 \(\sqrt3x^2+10x+7\sqrt3=0\)
\(\sqrt3 x^2+10x+7\sqrt3=0\)
\(\sqrt3x^2+7x+3x+7\sqrt3=0\)
\(x(\sqrt3x+7)+\sqrt3(\sqrt3x+7)=0\)
\((\sqrt3x+7)(x+\sqrt3)=0\)
\(x={-7\over \sqrt3} \ or \ x=-\sqrt3\)

Let us assume that \(\frac{2\sqrt 3}{5} \)is rational number. Then, it will be
of the form \( \frac{a}{b} \) where a, b are coprime integers and b \(\neq\)0 
Now \(\frac{2\sqrt 3}{5}\) \(​​\frac{a}{b}\)
On rearranging, we get \(\frac{5a -2b}{b}\) = \( \sqrt{3} \)
Since, 5 and  are rational. So, \(\frac{5a -2b}{b}\)  Will be rational
\(\therefore\) \( \sqrt{3} \) is rational.
But given that, \( \sqrt{3} \) is irrational So, this contradicts the fact that \( \sqrt{3} \) is irrational. Therefore, our assumption is wrong.
Hence,  \(\frac{ 2+ \sqrt 3}{5}\)is Irrational

Let \(f(y)=7y^{ 2 }-\frac { 11 }{ 3 } y-\frac { 2 }{ 3 }\)
 \(=\frac { 21y^{ 2 }-11y-2 }{ 3 } \)
\(=\frac { 21y^{ 2 }-14y+3y-2 }{ 3 } \)  [by splitting the middle term]
\(=\frac { 7y(3y-2)+1(3y-2) }{ 3 } \)
\(=\frac { 1 }{ 3 } (3y-2)(7y+1)\)
So, the value of \(7y^{ 2 }-\frac { 11 }{ 3 } y-\frac { 2 }{ 3 } \) is zero whe 3y-2=0 or 7y+1=0, i.e. when \(y=\frac { 2 }{ 3 } \) or \(y=\frac { 1 }{ 7 } \).
Thus, the zeroes\(=\frac { 2 }{ 3 } -\frac { 1 }{ 7 } =\frac { 14-3 }{ 21 } =\frac { 11 }{ 21 } =-\left( \frac { -11 }{ 3\times 7 } \right) \)
\(=-(1).\left( \frac { Coefficient \ of \ y }{ Coefficient \ of \ y^{ 2 } } \right) \) and product of zeroes\(=\left( \frac { 2 }{ 3 } \right) \left( -\frac { 1 }{ 7 } \right) =\frac { -2 }{ 21 } =\frac { -2 }{ 3\times 7 } \)
\(=\left( \frac { Constant \ term }{ Coefficient \ of \ y^{ 2 } } \right) \)
Hence, the relations between the zeroes and the coefficients of the polynomials is verified.

Given numbers are 156, 208 and 260.
Here, 260 > 208 > 156
Let us first find the HCF of 260 and 208
By using Euclid's division lemma for 260 and 208, 
we get 
260 = (208 x 1) + 52
Here, remainder = 52 \(\neq \) 0
On taking 208 as new dividend and 52 as new divisor and then apply Euclid's division lemma, we get
208 = (52 x 4) + 0
Here, the remainder is zero and  the divisor is 52.
So, 52 is the HCF of 208 and 260.
Now, 156 > 52
Let us find the HCF of 52 and 156. By using Euclid's division lemma, we get
156 = (52 x 3) + 0
Here, the remainder is zero and the divisor is 52.
So, 52 is the HCF of 52 and 156.
Thus, HCF of 156, 208 and 260 is 52.
Hence, the minimum number of buses
\(=\frac { 156 }{ 52 } +\frac { 208 }{ 52 } +\frac { 260 }{ 52 } \)
= 3 + 4 + 5 = 12

1: 3

Sum of n terms of given A.P. Sn =5n² - 3n
\(\therefore\)   Sum of ( n - 1 ) terms of given AP,
Sn - 1 = 5 ( n - 1 )2 - 3( n - 1 )
Sn - 1 =5 ( n² - 2n + 1 ) - 3n + 3
= 5n² - 10n + 5 - 3n + 3 = 5n² - 13n + 8
\(\therefore\)   nth term of AP = an = Sn - Sn - 1 = ( 5n² - 3n ) - ( 5n² - 13n + 8 )
a_n = 10n - 8
\(\therefore\)    Ist term of AP = 10 x 1 - 8 = 2
2nd term of AP = 10 x 2 - 8 = 12 and 3rd term of AP = 10 X 3 - 8 = 22
\(\therefore\)   Required AP is 2, 12, 22, ...
\(\therefore\)     a₁₀ = 10 X 10 - 8 = 92

Basic Proportionality theorem It states that if a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides of the triangle in proportion.
Given Let ABC be a triangle.

In the given triangle, line I is parallel to BC which intersect AB at D and AC at E.
To prove \(\frac{A D}{D B}=\frac{A E}{E B}\)
Construction Join BE and CD. Draw perpendicular lines DM and EN to the sides AC and AB of \(\Delta\)ABC respectively, as shown in the figure.
Proof Consider the given \(\Delta\)ABC.
\(\Delta\)ADE and \(\Delta\)DEB have equal heights i.e. EN.
So, area of \(\Delta\)ADE = \(\frac{1}{2} \times A D \times E N\)
Area of \(\Delta\)DEB = \(\frac{1}{2} \times D B \times E N\)
Thus, \(\frac{\text { Area }(\triangle A D E)}{\text { Area }(\triangle D E B)}=\frac{A D}{D B}\)              ...(i)
Similarly, consider the \(\Delta\)ADE with base AE and height DM.
\(\begin{aligned} & \text { Area of } \triangle A D E=\frac{1}{2} \times A E \times D M \\ \end{aligned}\)
\(\begin{aligned} & \text { Area of } \triangle C D E=\frac{1}{2} \times E C \times D M \\ \end{aligned}\)
\(\begin{aligned} & \frac{\text { Area }(\triangle A D E)}{\text { Area }(\triangle C D E)}=\frac{A E}{E C} \end{aligned}\)             ...(ii)
It is known that triangles with the same base and between same parallel lines have an equal area
Here, \(\Delta\)BDE and \(\Delta\)CDE have the same base DE and lie between the parallel lines DE and BC
\(\Rightarrow\)    \(A(\triangle D E B)=A(\triangle D E C)\)    ...(iii)
From Eqs. (i), (ii) and (iii), we get
\(\begin{array}{rlrl} \frac{A(\triangle A D E)}{A(\triangle D E B)} =\frac{A(\triangle A D E)}{A(\triangle D E C)} \\ \end{array}\)
\(\Rightarrow \quad \frac{A D}{D B}=\frac{A E}{E C}\)             Hence proved

Δ ABC ~ Δ PQR    (Given)
So, \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{CA}}{\mathrm{RP}}\)                                  (1)
and \(\angle\) A = \(\angle\) P, \(\angle\) B = \(\angle\) Q and \(\angle\) C = \(\angle\) R           (2)
But AB = 2 AM and PQ = 2 PN 
                              (As CM and RN are medians)
So, from (1), \(\frac{2 \mathrm{AM}}{2 \mathrm{PN}}=\frac{\mathrm{CA}}{\mathrm{RP}}\)
i.e., \(\frac{\mathrm{AM}}{\mathrm{PN}}=\frac{\mathrm{CA}}{\mathrm{RP}}\)                                            (3)
Also, \(\angle\) MAC = \(\angle\) NPR  [From (2)]                         (4)
So, from (3) and (4),
Δ AMC ~ Δ PNR      (SAS similarity)                       (5)

k = 2

Let the fraction be \(\frac { x }{ 2x+1 } \)
\(\frac { x }{ 2x+1 } +\frac { 2x+1 }{ x } =\frac { 58 }{ 21 } Z\)
\(21[x^{ 2 }+(2x+)=58(2x^{ 2 }+x)\)
\(\Rightarrow 11x^{ 2 }-26x-21=0\)
\(x=3,\frac { 7 }{ 11 } \) (rejected)
Fraction = \(\frac { 3 }{ 7 } \)

We know that, in a cyclic quadrilateral, the sum of two opposite angles is 180°.
\(\therefore \quad \angle B+\angle D={ 180 }^{ 0 } and \quad \angle A+\angle C={ 180 }^{ 0 }\)
\(\Rightarrow \) 3y-5-7x+5=180 and 4y+20-4x=180
\(\Rightarrow \) 3y-7x=180   ...(i)
and 4y-4x=160 
\(\Rightarrow \) y-x=4. [dividing both sides by 4]  ...(ii)
On multiplying Eq. (ii) by 7 and then subtracting from Eq. (i), we get
-4y=180-280
\(\Rightarrow \) -4y=-100 y=25
On putting y=25 in Eq. (ii), we get
25-x=40 x=-15
On putting the values of x and y, we calculate the angles as
\(\angle A=4y+20=100+20={ 120 }^{ 0 }.\) 
\(\\ \angle B=3y-5=75-5={ 70 }^{ 0 }.\) 
\(\\ \angle C=-4x=-4(-15)={ 60 }^{ 0 }.\) 
and \( \angle D=-7x+5=105+5={ 110 }^{ 0 }.\)
Hence, the angles are \(\angle A={ 120 }^{ 0 },\angle B={ 70 }^{ 0 },\angle C={ 60 }^{ 0 }\) and \( \angle D={ 110 }^{ 0 }.\)

Let unit's digit be y and ten's digit be x. Then, x+y=8 and (10x+y)-(10y+x)=18
\(\Rightarrow\) 9x-9y=18
x=5, y=3, Required number=53

\(\because\) Points (x,y), (1,2) and (7,0) are colinear
\(\because\) Area of the triangle formed by these pointer is 0.
⇒ \(\frac{1}{2}\)[x(2-0)+1(0-y)+7(y-2)=0
⇒ [2x+1(-y)+7y-14]=0 ⇒ 2x-y+7y-14=0
⇒ 2x+6y-14=0 ⇒ x+3y-7=0

(a) Since, he pays first instalment of? 1000 and next consecutive months he pay the instalment are 1100, 1200, ..... .
Thus, we get the AP sequence,
1000, 1100, 1200, ...
Here, a = 1000, d = 1100 - 1000 = 100
Now, T₃₀ = a + (30 - 1) d
 = 1000 + 29 \(\times\) 100
= 1000 + 2900 = 3900
Hence, the amount paid by him in 30th instalment is Rs 3900.
(ii) (b) Now, \(S_{30}=\frac{30}{2}[2 a+(30-1) d]\)
= 15 (2 \(\times\) 1000 + 29 \(\times\)100)
= 15 (2000 + 2900)
= 15 \(\times\) 4900 = Rs 73500
(iii) (c) After 30th instalment, he still have to pay = 118000 - 73500= 44500
(iv) (a) The amount in last 40th instalment is
T₄₀ = a + (40 - 1)d
= 1000 + 39 \(\times\) 100
= 1000 + 3900 = Rs 4900
(v) (b) The ratio of Ist instalment to the last instalment is \(\frac{1000}{4900} \text { i.e. } \frac{10}{49}\)

(a) (i)
The given AP is 6,12,18...,
The common difference = 12-6 =6.
(b) (ii)
In the given AP, we have:a=d=6
\(\therefore a_{n}=a+(n-1) d=6+(n-1) 6=6+6 n-6=6 n\) 
(c) (ii)
The number of trees planted by the students of all the sections of class
= 8th term of the given AP
= 6n = 6X8=48
(d) (ii)  T
otal number of trees planted by class 12 students
= 6 X 12 = 72
(e) (iii)
3rd term from the end = \((n-3+1) \text { th term }\) 
= (12-3+1) th term = 10th term
= 6X10=60

(i) (a): \(x=\frac{m_{1} x_{1}+m_{2} x_{1}}{m_{1}+m_{2}}=\frac{1(8)+2(2)}{3}\)
\(=\frac{12}{3}=4\)
\(y=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}=\frac{1(3)+2(5)}{3}=\frac{13}{3}\)
co-ordinates of the point 'Q' are \(\left(4, \frac{13}{3}\right)\)
(ii) (a):  \(\mathrm{PR}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
\(=\sqrt{(8-2)^{2}+(3-5)^{2}}\)
\(=\sqrt{6^{2}+2^{2}}=\sqrt{40}\)
= \(2 \sqrt{10} \)
(iii) (c): Let A(x,0) be the point on x- axis which is equidistant from P and Q, then we have PA = QA
\(\Rightarrow \mathrm{PA}^{2}=\mathrm{QA}^{2}\) 
\((2-x)^{2}+(5-0)^{2}=(4-x)^{2}+\left(\frac{13}{3}-0\right)^{2}\)
\(\Rightarrow 4+x^{2}-4 x+25=16+x^{2}-8 x+\frac{169}{9}\)
\(\Rightarrow \quad 4 x=16+\frac{169}{9}-4-25 \Rightarrow x=\frac{13}{9}\)
Hence the point, \(\left(\frac{13}{9}, 10\right) .\)
(iv) (c): Mid-points of PR. \(=\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\)
\(=\left(\frac{2+8}{2}, \frac{5+3}{2}\right)\)
\(=\left(\frac{10}{2}, \frac{8}{2}\right)\)
= (5,4)
(v) (b): 

The co-ordinates of point R is (6,2)

(i) (b): The distance between A and C
\(=\sqrt{(8-4)^{2}+(5+3)^{2}}=\sqrt{4^{2}+8^{2}} \)
\(=\sqrt{16+64}=\sqrt{80}=4 \sqrt{5} \text { units }\)
(ii) (a): Let the coordinates of I be (x, y).

Then, by section formula
\(x =\frac{1 \times 8+2 \times 7}{1+2}=\frac{8+14}{3}=\frac{22}{3}\)
\(\text { and } y =\frac{1 \times 5+2 \times 3}{1+2}=\frac{5+6}{3}=\frac{11}{3}\)
Thus, the coordinates of I is \(\left(\frac{22}{3}, \frac{11}{3}\right)\)
(iii) (b): The mid -point of A and C
\(=\left(\frac{8+4}{2}, \frac{5-3}{2}\right)=(6,1)\)
(iv) (b): Let B divides the line segment joining A and C in the ratio k : 1. Then, the coordinates of B will be
\(\left(\frac{8 k+4}{k+1}, \frac{5 k-3}{k+1}\right)\)
\(\text { Thus, we have }\left(\frac{8 k+4}{k+1}, \frac{5 k-3}{k+1}\right)=(7,3)\)
\(\Rightarrow \frac{8 k+4}{k+1}=7 \text { and } \frac{5 k-3}{k+1}=3 \)
\(\text { Consider, } \frac{8 k+4}{k+1}=7 \Rightarrow 8 k+4=7 k+7 \Rightarrow k=3\)
Hence, the required ratio is 3 : 1
(v) (a):\(\because\) B divides AC in the ratio 3 : 4.
\(\therefore\) A, B, C lie on a straight line.

(i) (b): The shape of the path of the soccer ball is a parabola.
(ii) (c): The axis of symmetry of the given curve is a line parallel to y-axis.
(iii) (a): The zeroes of the polynomial, represented in the given graph, are -2 and 7, since the curve cuts the x-axis at these points.
(iv) (d):A polynomial having zeroes -2 and -3 is \(p(x)=x^{2}-(-2-3) x+(-2)(-3)=x^{2}+5 x+6\)
(v) (c): We have \(f(x)=(x-3)^{2}+9\)
\(\text { Now, } 9=(x-3)^{2}+9 \)
\(\Rightarrow(x-3)^{2}=0 \Rightarrow x-3=0 \Rightarrow x=3\)