Two lines AB and CD intersect at point O. Also, OM and ON are t6he bisectors of \(\angle AOC\) and \(\angle BOD\) respectively.

To prove: MON is a straight line/
Prove: Since the sum of all the angles around a point O is 360^o , we have
\(\angle AOC+\angle BOC+\angle BOD+\angle AOD=360^o\)
\(\Rightarrow 2\angle MOC+2\angle BOC+2\angle BON=360^o\)
\([\because \angle BOC=\angle AOD\)(ver:opp, \(\angle S\) OM is bisector of \(\angle AOC.ON\) is bisector of \(\angle BOD\)]
\(\Rightarrow \angle MOC+\angle BOC+\angle BON\)
=180^o
\(\Rightarrow \angle MON=180^o\)
\(\therefore \angle MON\) is a straight angle
Hence, MON is a straight line.

Since \(\angle1=\angle3\ and \ \angle2=\angle4\), therefore adding before equations.
\(\angle1+\angle2=\angle3+\angle4\)
\(\Rightarrow \angle BAD= \angle BCD\)
\(\Rightarrow \angle A= \angle C.\)

Given, A(-1,-1) and C(1,1)
Then, B(1,-1) and D(-1,1)
Also, equations of sides of square are,
AB: Y=-1
BC: X=1
CD: Y=1
DA: X=-1

(a)(4, 2)
(b)\(\left({36\over13},{24\over13}\right)\)

\(A\rightarrow \left( 3,4 \right) \)

\(128\sqrt { 2 } \)

\(\frac { 3 }{ 5 } =\frac { 3\times 10 }{ 5\times 10 } =\frac { 30 }{ 50 } \)
\(\\ \frac { 4 }{ 5 } =\frac { 4\times 10 }{ 5\times 10 } =\frac { 40 }{ 50 } \)
\(\\ \because 30<31<32<33<34<35\)
\(\\ \therefore \frac { 30 }{ 50 } <\frac { 31 }{ 50 } <\frac { 32 }{ 50 } <\frac { 33 }{ 50 } <\frac { 34 }{ 50 } <\frac { 35 }{ 50 } \)
Therefore, five rational numbers between \(\frac { 3 }{ 4 } \) and \(\frac { 4 }{ 5 } \) and be taken as 
\(\frac { 31 }{ 50 } ,\frac { 32 }{ 50 } ,\frac { 33 }{ 50 } ,\frac { 34 }{ 50 } \)and \(\frac { 35 }{ 50 } \)
i.e., \(\frac { 31 }{ 50 } ,\frac { 16 }{ 25 } ,\frac { 33 }{ 50 } ,\frac { 17 }{ 25 } \) and \(\frac { 7 }{ 10 } \) 

Substituting x=-1 and y=-1 in 9kx+12ky=63, we get
\(\Rightarrow\) 9k(-1)+12k(-1)=63
\(\Rightarrow\) -9k-12k=63
\(\Rightarrow\) -21k=63 k=-3

384 cm²

BC, AB

\(\frac{4}{3}\) of a right angle=\(\frac{4}{3}\times90^o=120^o\)
(Sum of supplementary angles is 180^o)
Supplement of 120^o=180^o-120^o=60^o

Let Sita contribute=Rs.x and Gita contribute=Rs.y. 
According to the question,
x+y=200
y=200-x

6 Units.

Using Identity V, we have
(4a – 2b – 3c)² = [4a + (–2b) + (–3c)]²
= (4a)² + (–2b)² + (–3c)² + 2(4a)(–2b) + 2(–2b)(–3c) + 2(–3c)(4a)
= 16a² + 4b² + 9c² – 16ab + 12bc – 24ac

\(\frac { 125 }{ 512 } \)

\(x=\angle PEQ\)
\(\angle PER+\angle QER\)
\(=(180^{ 0 }-\angle APE)+180^{ 0 }-140^{ 0 }\)

\((x-\frac{1}{x})(x+\frac{1}{x})(x^2+\frac{1}{x^2})\)
\((x^2-\frac{1}{x^2})(x^2+\frac{1}{x^2})=x^4-\frac{1}{x^4}\)

y+20o=58o(Corr angles)
y=58^o-20^o=38^o
\(\angle PRQ=180^o-(58^o+22^o)\)(Linear pair)
=180^o-80^o=100^o
x=180^o-(100^o+38^o)(Angle sum of property)
=180^o-138^o=42^o

\(\angle AGH=\angle GHD\)
\(\Rightarrow \frac{1}{2}\angle AGH=\frac{1}{2}\angle GHD\Rightarrow\angle 1=\angle2\)
\(\Rightarrow\) GM||LH
Similarly, GL||MH
GMHL is a parallelogram.
\(\angle BGH+\angle GHD=180^o\)
\(\Rightarrow\frac{1}{2}\angle BGH+\frac{1}{2}\angle GHD=90^o\)
\(\Rightarrow \angle 3+\angle 2=90^o\)
In \(\triangle GLH, \angle GLH=180^o-(\angle 2+\angle3)\)
=180^o-90^o
\(\Rightarrow \angle GLH=90^o\)
\(\angle GMH=90^o\)
So, \(\angle MGL+\angle GLH=180^o\Rightarrow \angle MGL+90^o\)
=180^o
\(\Rightarrow \angle MGL=90^o\Rightarrow\angle MHL=90^o\)

In \(\triangle\)AMB and \(\triangle\)PNQ,
AB = PQ   (Given)
AM = PN   (Given)
\(\angle\)1 = \(\angle\)2 = 90°
(AM\(\bot \)BC & PN \(\bot \)QR)
\(\Rightarrow\) \( \triangle AMP\cong \triangle PNQ\) (By RHS)
\(\Rightarrow\) \(\angle\)3 =\(\angle\)4 (By c.p.c.t.) 
Similarly, \(\triangle AMC\cong \triangle PNR\)
\(\Rightarrow\) \(\angle\)5 = \(\angle\)6
(In congruent triangles, corresponding angles are equal)
\(\therefore\) In \(\triangle\)ABC and \(\triangle\)PQR,
AB = PQ  (Given)
AC = PR (Given)
\(\angle\)A =\(\angle\)P
(\(\angle\)3 + \(\angle\)5 = \(\angle\)4 + \(\angle\)6) (proved)
\(\triangle ABC\cong \triangle PQR\)     (By SAS)

Since total number of voters who do not cast therir cotes=40%
Hence total number of voters who cast their votes =60%
Let the total numbers of voters are x and number of voters who cast their votes is y.
Then according to the question
y=60% of \(x=\frac{60}{100}x\)
Now, when c=100, then
y=60
when x=200, then y=120
when c=300, then y=180

By plotting the points (100,60), (200,120) (300,180) on the graph and by joining them, we get the graph of equation (1) as shown in fig. From the graph, we see that
(i) If 300 voters cast their votes, then total number of votes=500

(ii) If the total number of voters are 1000, then number of votes cast=600.
(iii) Everyone should cast his vote to elect an honest candidate.

\(\frac { { \left( { 3 }^{ 2 } \right) }^{ n+1 }\times { 3 }^{ n }-{ \left( { 3 }^{ 3 } \right) }^{ n } }{ { 3 }^{ 3m }\times { 2 }^{ 3 } } =\frac { 1 }{ { 3 }^{ 6 } } \)
⇒ \(\frac { { 3 }^{ 2n+2 }\times { 3 }^{ n }-{ 3 }^{ n } }{ { 3 }^{ 3m }\times 8 } =\frac { 1 }{ { 3 }^{ 6 } } \)
ஃ 3n - 3m = -6
ஃ m - n = -2

For triangular ground   a = 5m, b = 7m, c = 8m
\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 5+7+8 }{ 2 } \) m = 10 m
\(\therefore \) Area \(=\sqrt { s(s-a)(s-b)(s-c) } \)
 \(=\sqrt { 10(10-5)(10-7)(10-8) }\)
\( \\ =\sqrt { 300 } =10\sqrt { 3 } \)
=10\(\times \)1.73 =17.3 m²
\(\therefore \) Cost of levelling  = 17.3 \(\times \) 10 = Rs.173

(i) 

(ii) See above figure
(iii) The figure obtained is a parallelogram. So, Apala is correct. So, the value 'intelligent' is depicted by her statement.
(iv) ABCD is not a trapezium as in a trapezium only one pair of opposite sides are parallel.
(v) The altitude of the parallelogram ABCD corresponding to the base AB are 4 units.
(vi) The parallelogram ABCD will become rectangle ABCD when (vii) The mathematical concept covered in this problem is 'Coordinate Geometry.

Let \(f(x)={ x }^{ 3 }+2{ x }^{ 2 }-5ax-7\)
and  \(g(x)={ x }^{ 3 }-a{ x }^{ 2 }-12x+6\)
By remainder theorem,
\(f(-1)={ R }_{ 1 }\ \ \ \ ........(1)\)
\(x+1=0\Rightarrow x=-1\)
\(\Rightarrow { (-1) }^{ 3 }+2{ (-1) }^{ 2 }-5a(-1)-7={ R }_{ 1 }\) 
\(\Rightarrow \ -1+2+5a-7={ R }_{ 1 }\) 
\(\Rightarrow { R }_{ 1 }=a-6 \ \ ................(2)\)
and     \(g(2)={ R }_{ 2 }\ \ ...........(3)\)
\(x-2=0\Rightarrow x=2\)
\(\Rightarrow { (2) }^{ 3 }+a{ (2) }^{ 2 }-12(2)+6={ R }_{ 2 }\) 
\(\Rightarrow 8+4a-24+6={ R }_{ 2 }\) 
\(\Rightarrow { R }_{ 2 }=4a-10\ \ \ .....(4)\) 
According to the question,
\(2{ R }_{ 1 }+{ R }_{ 2 }=6\) 
\(\Rightarrow 2(5a-6)+(4a-10)=6\) 
\(\Rightarrow -12+4a-10=6\) 
\(\Rightarrow 14a=28\) 
\(\Rightarrow a=2\)

Proof: QO is bisector of \(\angle\)PQR
\(\angle\)OQR = \(\frac{1}{2}\)\(\angle\)PQR = \(\frac{1}{2}\) =\(\angle\)Q
RO is bisector \(\angle\)ORQ
\(\therefore\) \(\angle\)ORQ =\(\frac{1}{2}\) \(\angle\)PRQ = \(\frac{1}{2}\) \(\angle\)R
In \(\angle\)OQR
\(\angle\)QOR + \(\angle\)OQR + \(\angle\)ORQ = 180°
(Angle sum property)
\(\angle\)QOR + \(\frac{1}{2}\) \(\angle\)Q + \(\frac{1}{2}\) \(\angle\)R = 180°
\(\angle\)QOR = 180°- \(\frac{1}{2}\)(\(\angle\)Q + \(\angle\)R)
But in \(\angle\)PQR
\(\angle\)P + \(\angle\)Q + \(\angle\)R = 180°
\(\angle\)Q + \(\angle\)R = 180°- \(\angle\)P
\(\angle\)QOR = 180°- \(\frac{1}{2}\) (180°- \(\angle\)P)
= 180°-90° + \(\frac{1}{2}\)\(\angle\)P
= 90° + \(\angle\)P Hence Proved.
These type of rallies spread awareness among people for not to kill girl child and helping in equalising sex ratio.