MATHS MCQ

$ \Rightarrow \,\,2x(1 + 3) + 2y(3 - 5) = (1 + 9) - (9 + 25)$

$ \Rightarrow \,\,2x - y + 6 = 0$ .....$(i)$

Perpendicular bisector of $A\,(1,\,\,3)$ and $C\,(5,\,\, - 1)$ is

$2x\,(1 - 5) + 2y(3 + 1) = (1 + 9) - (25 + 1)$

$ \Rightarrow \,\,x - y - 2 = 0$ .....$(ii)$

Point of intersection of $(i)$ and $(ii)$ is $P = ( - 8,\,\, - 10)$

Then $PA = \sqrt {{{(1 + 8)}^2} + {{(3 + 10)}^2}} = \sqrt {81 + 169} $

$ = \sqrt {250} = 5\sqrt {10} $.

$(i)$ Point $B \,(x, y)$ divides $AD$ in $1 : 2$

$\therefore \,\,\,\,x = \frac{{0 + 9}}{3} = 3$ and $y = \frac{{0 + 12}}{3} = 4$

$(ii)$ Now point $C \,(x, y)$ divides $AD$ in $2 : 1$,

Then $x = \frac{{0 + 18}}{3} = 6$ and $y = \frac{{0 + 24}}{3} = 8$.

${x_1} = 7,\,\,{x_2} = - 3,\,\,{x_3} = 1$

Similarly ${y_1},\,\,{y_2},\,\,{y_3}$ can be found.

so coordinates of $C$ will be $\left( { - \frac{1}{3},\,\frac{5}{3}} \right)$,

which is the required point.

Obviously $C$ divides the line $AB$ in $1 : 2$, therefore coordinates of $C$ are
$\left( {\frac{{1( - 3) + 2(3)}}{{1 + 2}},\,\frac{{1( - 4) + 2( - 2)}}{{1 + 2}}} \right) = \left( {1,\,\, - \frac{8}{3}} \right)$.

Similarly $D$ divides the line $AB$ in $2 : 1$, hence coordinates of $D$ are
$\left( {\frac{{2( - 3) + 1(3)}}{{2 + 1}},\,\frac{{2( - 4) + 1( - 2)}}{{2 + 1}}} \right) = \left( { - 1,\,\, - \frac{{10}}{3}} \right)$.

Now, $A{C^2} = A{B^2} + B{C^2}$

$A{C^2} = 2A{B^2} \Rightarrow 8 = A{B^2}$

$AB = BC = 2\sqrt 2 $

Now, let $B\,(x,y)$; $\therefore A{B^2} = B{C^2}$

==>${(x - 0)^2} + {(y + 1)^2} = {(x - 0)^2} + {(y - 3)^2}$

${x^2} + {y^2} + 2y + 1 = {x^2} + {y^2} - 6y + 9$

==> $y = 1;\,\,\,\therefore {x^2} + {(2)^2} = 8; \Rightarrow {x^2} = 4 \Rightarrow x = \pm \,2$

$\therefore $ other vertices are $(2, 1)\,,(-2, 1)$.

$BC = \sqrt {{{(\sqrt 3 a)}^2} + {a^2}} = 2a$

$CA = \sqrt {{{(\sqrt 3 a)}^2} + {{( - a)}^2}} = 2a$

Since $AB = BC = CA,$ hence triangle is equilateral.

Therefore, it is an acute angled triangle.

and ${x^2} + {y^2} = {(x + 2)^2} + {(y + 2)^2}\,\, \Rightarrow \,\,4x + 4y + 8 = 0$

On simplification, we get $y = - 8$ and $x = 6$.

Hence incentre is

$\left( {\frac{{11 \times 0 + 20 \times 5 + 13 \times 16}}{{11 + 20 + 13}},\,\,\frac{{11 \times 0 + 20 \times 12 + 13 \times 12}}{{11 + 20 + 13}}} \right)$i.e. $(7, 9)$.

Therefore, area $PQRS$ $ = 2.\frac{1}{2}\,\left| {\,\begin{array}{*{20}{c}}2&1&1\\4&{ - 1}&1\\3&2&1\end{array}\,} \right| = 4$.

then $ A,\, B,\, C$ are collinear if area of $\Delta ABC = 0$

$ \Rightarrow \,\,\left| {\,\begin{array}{*{20}{c}}{x + 1}&2&1\\1&{x + 2}&1\\{\frac{1}{{x + 1}}}&{\frac{2}{{x + 1}}}&1\end{array}\,} \right| = 0$ $ \Rightarrow \,\,\left| {\,\begin{array}{*{20}{c}}x&{ - x}&0\\1&{x + 2}&1\\{\frac{1}{{x + 1}}}&{\frac{2}{{x + 1}}}&1\end{array}\,} \right| = 0$

$({R_1} \to {R_1} - {R_2})$

$ \Rightarrow \,\,\left| {\,\begin{array}{*{20}{c}}x&0&0\\1&{x + 3}&1\\{\frac{1}{{x + 1}}}&{\frac{3}{{x + 1}}}&1\end{array}\,} \right| = 0$ $({C_2} \to {C_2} + {C_1})$

$ \Rightarrow \,\,x\,\left( {x + 3 - \frac{3}{{x + 1}}} \right) = 0\,\, $

$\Rightarrow \,\,x({x^2} + 3 + 4x - 3) = 0$

$ \Rightarrow \,\,{x^2}(x + 4) = 0\,\, $

$\Rightarrow \,\,x = 0,\,\, - 4$.

i.e., given triangle is equilateral.

(In centre of a triangle are same as the centriod when triangle is equilateral)

.Hence, incentre = $\left( {\frac{{7 - 1 + 3 + 2\sqrt 3 }}{3},\frac{{1 + 5 + 3 + 4\sqrt 3 }}{3}} \right)$

$= \left( {3 + \frac{2}{{\sqrt 3 }},3 + \frac{4}{{\sqrt 3 }}} \right)$.

Slope of $BC = \frac{{3 - 4}}{{1 + 2}} = \frac{{ - 1}}{3}$

Slope of $ \bot $ to $BC = 3$

Equation of altitude through $A$ is

$y + 6 = 3(x + 2)$ ==> $y + 6 = 3x + 6$ ==> $y = 3x$ .....$(i)$

Slope of $CA = \frac{{3 + 6}}{{1 + 2}} = \frac{9}{3} = 3$

Slope of $ \bot $ to $CA = \frac{{ - 1}}{3}$

Equation of altitude through $B$ is

$y - 4 = \frac{{ - 1}}{3}(x + 2)$ ==> $3y - 12 = - x - 2$

==> $x + 3y - 10 = 0$ .....$(ii)$

Solving $(i)$ and $(ii)$ $x = 1,y = 3$

Hence, orthocentre $(1, 3).$

and $x + y + 2 = 0$.....$.(ii)$

from $(i)$ and $(ii)$, $xy = 0 \Rightarrow x = y = 0$

 Vertices of triangle are $( - 2,0)\,\,(0,0)\,\,(0, - 2)$

(In a right angled triangle circumcentre is mid point of hypotenuse)

 $( - 1, - 1)$ is the circumcircle.

Co-ordinates of point of intersection of the $x = 2$ and $y + 1 = 0$ is $(2, -1)$

Co-ordinates of point of intersection of $x = 2$ and $x + 2y = 4$ is $(2, 1)$

Co-ordinates of point of intersection of $y + 1 = 0$ and $x + 2y = 4$ is $(6, -1)$

Let Co-ordinates of circumcentre is $(x, y)$

${(x - 2)^2} + {(y + 1)^2} = {(x - 2)^2} + {(y - 1)^2}$

${(y + 1)^2} = {(y - 1)^2}$; ${y^2} + 2y + 1 = {y^2} - 2y + 1$

$4y = 0$, $y = 0$ and ${(x - 2)^2} + {(y - 1)^2} = {(x - 6)^2} + {(y + 1)^2}$

In this equation put $y = 0$

${(x - 2)^2} + {(0 - 1)^2} = {(x - 6)^2} + {(0 + 1)^2}$

${(x - 2)^2} + 1 = {(x - 6)^2} + 1$; ${(x - 2)^2} - {(x - 6)^2} = 0$

$(x - 2 + x - 6)(x - 2 - x + 6) = 0$

$4(2x - 8) = 0$==> $8(x - 4) = 0$; $x - 4 = 0$ ==> $x = 4$

 Co-ordinates of circumcentre is $(4, 0)$.

Hence incentre is $\left( {\frac{{0 + 0 + 3 \times 4}}{{5 + 4 + 3}},\frac{{0 + 4 \times 3 + 0}}{{5 + 4 + 3}}} \right)$= $(1, 1)$.

Thus $c = AB = 6,\,\,a = BC = 8$ and $b = AC = 10$

Hence incentre

$ = \left( {\frac{{8 \times 0 + 10 \times 6 + 6 \times 6}}{{8 + 10 + 6}},\,\frac{{8 \times 0 + 10 \times 0 + 6 \times 8}}{{8 + 10 + 6}}} \right) = (4,\,\,2)$.

ase said to be cerlineas if Area $(\Delta A B C)=0$

$=\frac{S}{2}\left|\begin{array}{lll}2 & y & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right|=0$

$\therefore \quad\left|\begin{array}{lll}\lambda+1 & 1 & 1 \\ 2 \lambda+1 & 3 & 1 \\ 2 \lambda+2 & 2 \lambda & 1\end{array}\right|=0$

$\Rightarrow(\dot{x}+1)(3-2 \lambda)-1(82+1-2 \lambda-2)$

$+1\left(4 x^{2}+2 x-6 x-6\right)=0$

$\Rightarrow 3 \lambda+3-2\left(\lambda^{2}-2 \lambda+1+4 x^{2}-4 \lambda-6=0\right.$

$\Rightarrow 2 \lambda^{2}-3 \lambda-2=0$

$\Rightarrow 2 \lambda^{2}-4 \lambda+\lambda-2=0$

a) $2 x(x-2)+1(x-2)=0$

$\Rightarrow \lambda=-1 / 2$ or 2

" polsible values of

$\lambda \operatorname{are} 2$

To eliminate linear terms, we should have

$2h - 4 = 0,\,\,2k + 6 = 0\,\,\, \Rightarrow \,\,h = 2,\,\,k = - 3$

i.e., $(h,\,\,k) = (2,\,\, - 3)$.

So,${(h - a)^2} + {(k - 0)^2} = {h^2}$$ \Rightarrow \,\,\,{h^2} + {a^2} - 2ah + {k^2} = {h^2}$

Hence locus is ${y^2} - 2ax + {a^2} = 0$.

$\sqrt {{{(x - 1)}^2} + {y^2}} - \sqrt {{{(x + 1)}^2} + {y^2}} = \pm 1$

On squaring both sides, we get

$2{x^2} + 2{y^2} + 1 = 2\sqrt {{{(x - 1)}^2} + {y^2}} .\sqrt {{{(x + 1)}^2} + {y^2}} $

Again on squaring, we get $12{x^2} - 4{y^2} = 3$.

This is required locus of the point $(x, y)$.

$ \Rightarrow \,\,(1 - {k^2})\,({x^2} + {y^2}) - 2akx + 2akx + {a^2}{k^2} - {a^2} = 0$

$ \Rightarrow \,\,{x^2} + {y^2} - {a^2} = 0$.

Putting the value of t in $k = u\,\sin \alpha \,.\,t - \frac{1}{2}g{t^2},$ we get $k = h\,\tan \alpha - \frac{1}{2}g\frac{{{h^2}}}{{{u^2}{{\cos }^2}\alpha }}$

$\therefore \,\,$Locus of (h, k) is $y = x\tan \alpha - \frac{1}{2}g\frac{{{x^2}}}{{{u^2}\,{{\cos }^2}\alpha }}$, which is a parabola.

$h = \frac{{\cos \alpha + \sin \alpha + 1}}{3}$ and $k = \frac{{\sin \alpha - \cos \alpha + 2}}{3}$

$ \Rightarrow \,\,3h - 1 = \cos \alpha + \sin \alpha $ and $3k - 2 = \sin \alpha - \cos \alpha $

$ \Rightarrow \,\,{(3h - 1)^2} + {(3k - 2)^2} = 2$, (squaring and adding)

$ \Rightarrow \,9\,({h^2} + {k^2}) - 6h - 12k + 3 = 0$

$ \Rightarrow \,\,3\,({h^2} + {k^2}) - 2h - 4k + 1 = 0$

$\therefore $ Locus of $(h,\,\,k)$ is $3\,({x^2} + {y^2}) - 2x - 4y + 1 = 0$.

Given that $PA - PB = 4$

$\sqrt {{{(h - 3)}^2} + {k^2}} - \sqrt {{{(h + 3)}^2} + {k^2}} = 4$

$ \Rightarrow \,\,\sqrt {{{(h - 3)}^2} + {k^2}} = 4 + \sqrt {{{(h + 3)}^2} + {k^2}} $

Squaring both sides, we get

${(h - 3)^2} + {k^2} = 16 + {(h + 3)^2} + {k^2} + 8\sqrt {{{(h + 3)}^2} + {k^2}} $

$ \Rightarrow \,\,\,{h^2} + 9 - 6h + {k^2} = 16 + {h^2} + 9 + 6h + {k^2}$

$ + \,8\sqrt {{{(h + 3)}^2} + {k^2}} $

$ \Rightarrow \,\, - 6h = 16 + 6h + 8\sqrt {{{(h + 3)}^2} + {k^2}} $

$ \Rightarrow \,\, - 8\,\sqrt {{{(h + 3)}^2} + {k^2}} = 12h + 16$

Again, squaring both sides, we get

$64\,({h^2} + 9 + 6h + {k^2}) = 144{h^2} + 256 + 2.16.12h$

$ \Rightarrow \,\,4\,({h^2} + 9 + 6h + {k^2}) = 9{h^2} + 16 + 24h$

$ \Rightarrow \,\,4{h^2} + 36 + 24h + 4{k^2} = 9{h^2} + 16 + 24h$

$ \Rightarrow \,\,5{h^2} - 4{k^2} = 20\,\, \Rightarrow \,\,\frac{{{h^2}}}{4} - \frac{{{k^2}}}{5} = 1$

Hence, the locus of point $P$ is $\frac{{{x^2}}}{4} - \frac{{{y^2}}}{5} = 1$.

In right angled triangle $CDA,\,\,\,\tan A = \frac{k}{{a - h}}$

and similarly in triangle $CDB,\,\,\tan B = \frac{k}{{a + h}}$

Also from $(i)$, $\frac{{\tan A - \tan B}}{{1 + \tan A\,.\,\tan B}} = \tan \theta $

Substituting the values of $\tan A$ and $\tan B,$ we get

${h^2} - {k^2} + 2hk\cot \theta = {a^2}$

Hence the locus is ${x^2} - {y^2} + 2xy\cot \theta = {a^2}$.

$D\,\left( {\frac{{2 \times 6 + 0}}{3},\,\frac{{ - 6 + 3}}{3}} \right)\, \equiv \,(4,\, - 1)$.

$AC$ is bisector of $\angle DAB$

$\therefore \angle DAC =\angle CAB$

and $DC \| AB$

$\therefore \angle CAB =\angle ACD \quad \ldots(2)$

$\therefore \angle DAC =\angle ACD$

Hence $AD = DC$

similarly $BC = DC$

Hence Exactly three sides are equal.

In $\triangle ABC$ and $\triangle DCB$

$AB = DC$

$BC = CB$

$AC = DB$

$\triangle ABC \cong \triangle DCB$

$\therefore \angle BAC =\angle CDB =70^{\circ}$

$\angle ABC =\angle DCB =60^{\circ}$

$\angle ACB =180^{\circ}-\angle BAC -\angle ABC$

$=180^{\circ}-70^{\circ}-60^{\circ}$

$=50^{\circ}$

$\angle DCO =60^{\circ}-\angle ACB$

$=60^{\circ}-50^{\circ}=10^{\circ}$

$\angle DOC =180^{\circ}-70^{\circ}-10^{\circ}=100^{\circ}$

Acute angle $\angle DOC =80^{\circ}$

$\frac{\operatorname{ar} \triangle ABC }{\operatorname{ar} \triangle ADE }=\frac{\frac{1}{2} \cdot AB \cdot AC \sin \theta}{\frac{1}{2} \cdot AD \cdot AE \sin \theta}$

$=\frac{ AB }{ AD } \times \frac{ AC }{ AE }$

$=\frac{5}{3} \times \frac{3}{2}=\frac{5}{2}$

correct option is $\frac{5}{2}$

Let the remaining two sides be $a$ and $b$, then

.Since, $A B C D$ is a rectangle, then

$\frac{9}{\sqrt{2}}+6+\frac{b}{\sqrt{2}}=\frac{7}{\sqrt{2}}+10+\frac{5}{\sqrt{2}}$

$\quad \frac{b}{\sqrt{2}}=4+\frac{3}{\sqrt{2}}$

$\text { Similarly, } \frac{9}{\sqrt{2}}+8+\frac{7}{\sqrt{2}}=\frac{b}{\frac{\sqrt{2}}{}}+a+\frac{5}{\sqrt{2}}$

$\quad a=4+\frac{8}{\sqrt{2}}$

$\therefore \quad a+b=7+8 \sqrt{2}=18.3$

We have, $\cos \frac{B}{2}=\frac{\frac{9}{2}+4-1}{6 \sqrt{2}}=\frac{5}{4 \sqrt{2}}$

$\therefore$ Length of angle bisector,

$B D=\frac{2 a c}{a+c} \cos \frac{B}{2}$

$\frac{3}{\sqrt{2}} =\left(\frac{4 c}{e+2}\right) \cdot \frac{5}{4 \sqrt{2}}$

$c =3$

We know that, $\frac{A B}{B C}=\frac{A D}{C D}$

$A D=\frac{3}{2}$

$\therefore$ Perimeter of $\triangle A B C=1+\frac{3}{2}+3+2=\frac{15}{2}$

If in a $\triangle A B C, A D$ is median, then by Apollonius Theorem

$A B^2+A C^2=2\left(A D^2+B D^2\right)$

Here, by using Apollonius theorem, we get

$x^2+(x-2)^2=2\left[6^2+4^2\right]$

$2 x^2-4 x+4=104$

$x^2-2 x-50=0 \Rightarrow x=1+\sqrt{51}$

The nearest integer to $b$ is $8$ .

Area of parallelogram whose diagonals are 10 and 4 and if angle between adjacent side is $\theta$ is

$A=\frac{10 \times 4}{\sin \theta}$

The area will be maximum if $\theta=\frac{\pi}{2}$, so the parallelogram must be rhombus

Perimeter of rhombus

$=4 \sqrt{2} \overline{9}=\sqrt{46} \overline{4} \text { and } \sqrt{46} \overline{4} \in(21,22]$

For an equilateral triangle $A B C$ having side length $a$. If $R$ and $r$ are radii of the circumcircle and the incircle of triangle $A B C$ respectively, then

$R=\frac{a}{2} \sec 30^{\circ}=\frac{a}{2}\left(\frac{2}{\sqrt{3}}\right)=\frac{a}{\sqrt{3}}$

and $r=\frac{a}{2} \tan 30^{\circ}=\frac{a}{2} \times \frac{1}{\sqrt{3}}=\frac{a}{2 \sqrt{3}}$

$\therefore \frac{R}{r}=\frac{\frac{a}{\sqrt{3}}}{\frac{a}{2 \sqrt{3}}}=2$, which is independent of $a$ and it is constant.

It is given that in triangle $A B C$, $\angle B A C=90^{\circ}, A D$ is the altitude from $A$ on to $B C$.

Since, $A B=15$ and $B C=25$

$\therefore \quad A C=\sqrt{B C^2-A B^2}=\sqrt{625-225}$

$=\sqrt{400}=20$

Now, since area of $\triangle A B C=\frac{1}{2}(B C)(A D)$

$=\frac{1}{2}(A B)(A C)$

$\Rightarrow \frac{1}{2}(B C)(A D)=\frac{1}{2} \times 15 \times 20$

$\Rightarrow \quad 25 \times A D=300$

$\Rightarrow \quad A D=12$

$\because A E D F$ is a rectangle, then

$E F=A D=12$

Let a convex quadrilateral $A B C D$ and $K, L, M, N$ be the mid-point of $A B$, $B C, C D, D A$ respectively.

Now, as area $\Delta A K P=$ area $\Delta B K P=x$ (let) Similarly

$\triangle B L P =\Delta C L P=y$

$\Delta C P M =\triangle D P M=z$

And  $\Delta D N P =\triangle A N P=w$

It is given that Area $(P K A N)=x+w=25$ area $(P L B K)=x+y=36$

and area $(P M D N)=z+w=41$

So $\operatorname{area}(P L C M)=y+z$

$=(x+y)+(z+w)-(x+w)$

$=\operatorname{area}(P L B K)+$ area $(P M D N)-$ area $(P K A N)$

$=36+41-25=77-25=52$

It is given that circumference of circle $C$ is $l$ and the perimeter of triangle $T$ is $l$.

Now, let the radius of circle $C$ is $r$, so

$2 \pi r=l \Rightarrow r=\frac{l}{2 \pi}$

$\therefore$ area of circle $C$ is $A_1=\pi r^2=\frac{l^2}{4 \pi}$

Now, as we know that area of triangle will be maximum for given perimeter if it is an equilateral triangle, let the length of side of equilateral triangle is ' $a$ ', then

$3 a=l \Rightarrow a=\frac{l}{3}$

and area of equilateral triangle is

$A_2=\frac{\sqrt{3}}{4} a^2$

So, $A_2=\frac{\sqrt{3}}{4}\left(\frac{l^2}{9}\right)=\frac{l^2}{12 \sqrt{3}}$

Since, as we took an equilateral triangle, which has maximum area. But we can take a triangle $T$ such that the ratio area $(C)$ is greater than any positive real area $(T)$ number $\alpha$.

Given, $(a-8)^2-(b-7)^2=5$

$\Rightarrow (a-8+b-7)(a-8-b+7)=5$

$\Rightarrow (a+b-15)(a-b-1)=5$

There are four case

$a+b-15=5 ; a-b-1=1$

$a+b-15=1 ; a-b-1=5$
$a+b-15=-5 ; a-b-1=-1$

$a+b-15=-1 ; a-b-1=-5$

On solving, we get

$(i)$ $a=11, b=9$

$(ii)$ $a=11, b=5$

$(iii)$ $a=5, b=5$

$(iv)$ $a=5, b=9$

$\therefore$ Perimeter $=2(4+6)=20$

In $\triangle P Q R$

Given, $Q R^2+P R^2=5 P Q^2$

Median $P M$ and $Q N$ intersect of $G$

$Q G=\frac{2}{3} Q N, G M=\frac{1}{3} P M$

$Q G^2+G M^2=\left(\frac{2}{3} Q N\right)^2+\left(\frac{1}{3} P M\right)^2$

$=\frac{4}{9} Q N^2+\frac{1}{9} P M^2$

$=\frac{\frac{4}{9}\left(\frac{2 P Q^2+2 Q R^2-P R^2}{4}\right) +\frac{1}{9}\left(\frac{2 P Q^2+2 P R^2-Q R^2}{4}\right)}{8 P Q^2+8 Q R^2-4 P R^2}$

$ \frac{1}{9}+\frac{2PQ^2+2PR^2-QR^2}{4} $

$=\frac{1}{9}\left[\frac{10 P Q^2+7 Q R^2-2 P R^2}{4}\right]$

$=\frac{1}{9}\left[\frac{2\left(5 P Q^2-P R^2\right)+7 Q R^2}{4}\right]$

$=\frac{1}{9}\left[\frac{2 Q R^2+7 Q R^2}{4}\right]=\frac{1}{4} Q R^2=Q M^2$

$\therefore O G^2+G M^2 =Q M^2$
$\therefore \angle Q G M =90^{\circ}$

Let $\triangle A B C$

$B C=a$

$A C=b$

$A B=c$

and median of $\triangle A B C$

$A D=l$

$B E=m$

$C F=n$

$A D$ is median,

$l < \frac{b+c}{2}$

$m < \frac{a+b}{2}$ and $n < \frac{a+c}{2}$

$l+m+n < a+b+c$

$=\frac{l+m+n}{a+b+c}$

Also in $\triangle B G C, B G+G C > B C$

$\therefore \quad \frac{2}{3}(m+n) > a$

Similarly, $\frac{2}{3}(n+l) > b$ and $\frac{2}{3}(m+l) > c$

$\because \quad \frac{4}{3}(l+m+n) > a+b+c$

$=\frac{l+m+n}{a+b+c} > \frac{3}{4}$

From Eqs. $(i)$ and $(ii)$, we get

$\frac{l+m+n}{a+b+c} \in\left(\frac{3}{4}, 1\right)$

Given, in $\triangle P Q R$

$P Q=3$

$Altitude\,\,R S=\sqrt{3}$

$P S=Q R$

In $\triangle S Q R, \quad Q R^2=S R^2+S Q^2$

$P S^2=(\sqrt{3})^2+(Q P-P S)^2$

$[\because S Q=P Q-P S]$

$P S^2=3+(3-P S)^2$

$P S^2=3+9-6 P S+P S^2 \Rightarrow P S=2$

In $\triangle P R S$,

$P R^2=P S^2+R S^2=(2)^2+(\sqrt{3})^2=4+3$

$P R=\sqrt{7}$

Given, $A B C D$ is a square.

$\angle C A P=\angle M A P=45^{\circ}$

$A M=M P=Q N$

$P Q R S$ is a square,

$\because \quad M N=P Q=P S$

$P S =2 P M=2 A M$

$A N =A M+M N=3 A M$

In $\triangle A B Q$,

$A Q^2=A N^2+Q N^2$

$1=(3 A M)^2+A M^2[\because A Q=1]$

$10\,A M^2 =1$

$A M^2 =1$

$A M^2 =\frac{1}{10}$

Area of square $P Q R S=P S^2$

$=4\,A M^2=\frac{4}{10}=\frac{2}{5}$

Area square $P Q R S$

Area of square $A B C D=\frac{2}{5}$

We have,

$A B C D$ is a trapezium.

$A B$ is parallel to $C D$.

Area of trapezium $=12$

$\frac{1}{2} \times h(A B+C D)=12$

$A B+C D=\frac{24}{h}$

Sides and height of trapezium are integer.

$\therefore h$ is a factor of $24$

$h=1,2,3,4,6,8,12,24$

$A B+C D=24,12,8,6,4,3,2,1$

But $A B+C D > h$

$A B+C D=24,12,8,6$

In $\triangle B E C$,

$B E C$ is a right angled triangle.

$\therefore h$ must be $3$ and $4$

When $h=3, B E=4, C E=5$

$A B+C D=8$

$A E+B E+A E=8$

$2 A E=8-B E=8-4$

$A E=2$

$\therefore A B=4+2=6, C D=2$

$\therefore|A B-C D|=|6-2|=4$

We have,

Four triangle having sides are $(5,12,9),(5,12,11),(5,12,13),(5,12,15)$ A right triangle has maximum area.

$\therefore$ Among these the triangle whose sides $(5,12,13)$ form a right angled triangle.

$\therefore$ It has maximum area.

Given,

$A=\left( a _1, a_2\right), B=\left(b_1, b_2\right)$, where $a_1, a_2, b_1, b_2$ are integers.

Distance between $A B$

$=\sqrt{\left(b_1-a_1\right)^2+\left(b_2-a_2\right)^2}$

$\therefore A B=\sqrt{\text { sum of squares of two number }}$

$\sqrt{65}=\sqrt{64+1}$, it is possible.

$\sqrt{74}=\overline{\sqrt{49+2}}$, it is also possible.

$\sqrt{97}=\sqrt{81+16}$, it is also possible.

$83$ is not a sum of squares of two number.

Hence, option $(c)$ is correct.

Given in $\triangle A B C, B C$ is fixed, $A$ is variable and $A B C$ is an isosceles triangle.

Case $I$ In $\triangle A B C$,

If $\quad A B=A C$

Then, locus of $A$ is perpendicular bisector of $B C$.

$\therefore$ Locus of $A$ is a straight line.

Case $II$ When $A B=B C$

$B C$ fixed $B(a, 0), C(0, a)$, locus of $A$ is $(x-a)^2+y^2=2 a^2$, which represents the equation of circle.

Similarly, when $A C=B C$ also locus of $A$ is circle.

$\therefore$ Locus of $A$ is the union of two circles and the lines.

$\triangle A B C \sim \triangle A N M$

$\frac{\text { Area of } \triangle A B C}{\text { Area of } \triangle A N M}=\frac{A C^2}{A M^2}$

$\triangle A B C \sim M P C$

$\frac{\text { Area of } \triangle A B C}{\text { Area of } \triangle M P C}=\frac{A C^2}{M C^2}$

From Eqs.$(i)$ and $(ii)$,we get

$\frac{\text { Area of } \triangle A N M}{\text { Area of } \triangle M P C}=\frac{A M^2}{M C^2}$

Area of $\triangle A N M+$ Area of $\triangle M P C$

Area of $\triangle M P C=\frac{AM^2+MC^2}{MC^2}$

Now, Area of $\triangle A N M+$ Area of $\triangle M P C$

$=$ Area of $\triangle A B C-$ Area of $B N M P$

$\therefore \quad \frac{13 \text { (Area of } \triangle A B C)}{18 \text { (Area of } \triangle M P C)}=\frac{A M^2+M C^2}{M C^2}$

From Eq. $(iii)$, $\frac{13}{18}\left(\frac{\left(A C^2\right)}{M C^2}\right)=\frac{A M^2+M C^2}{M C^2}$

$13(A M+M C)^2=18\left(A M^2+M C^2\right)$

$\frac{A M}{M C}=5$

Given, $A B C$ is right angled triangle

$\angle C=90^{\circ}$

$C D$ is perpendicular on $A B, D N$ and $D M$ are parallel to $A C$ and $B C$, respectively. $D N=4$ and $D M=5$

In $\triangle D M C$ and $\triangle D N B$

$\Delta D M C \sim \Delta D N B$

$\frac{D M}{D N} =\frac{M C}{N B}$

$\frac{5}{4} =\frac{4}{N B} \Rightarrow N B=\frac{16}{5}$

In $\triangle D N C$ and $\triangle D M A$

$\Delta D N C \sim \Delta D M A$

$\frac{D N}{D M} =\frac{N C}{M A}$

$\frac{4}{5} =\frac{5}{M A} \Rightarrow M A=\frac{25}{4}$

$\therefore A C=M C+A M=4+\frac{25}{4}=\frac{41}{4}$

$\operatorname{In} \triangle A M D$

$\quad \cos \theta=\frac{A M}{30} \Rightarrow A M=30 \cos \theta$

$\quad \sin \theta=\frac{D M}{30}=D M=30 \sin \theta$

$\text { Area of trapezium }=\frac{1}{2}(A B+C D) D M$

$\Rightarrow \quad A=\frac{1}{2}(60+60 \cos \theta) 30 \sin \theta$

$\Rightarrow \quad \quad \quad \quad \quad d A=900(\sin \theta+\sin \theta \cos \theta)$

$\Rightarrow \quad d \theta=900\left(\cos \theta-\sin ^2 \theta+\cos ^2 \theta\right)$

$\text { For maximum or minimum, put } \frac{d A}{d \theta}=0$

$\therefore \quad \cos \theta-\sin 2 \theta+\cos ^2 \theta=0$

$\Rightarrow \quad \cos \theta+\cos 2 \theta=0$

$\Rightarrow \quad 2 \cos \frac{3 \theta}{2} \cos \frac{\theta}{2}=0$

$\frac{3 \theta}{2} =\frac{\pi}{2} \text { or } \frac{\theta}{2}=\frac{\pi}{2}$

$\theta =\frac{\pi}{3} \text { or } \theta=\pi$

For maximum $\theta=\frac{\pi}{3}$

Given, area of $\triangle E A B=$ area of square $A B C D$

$E B=E D=\sqrt{130}$

Let side of square $=x$

$B M=\frac{x}{\sqrt{2}}=A M$

Area of $\triangle A E B=$ Area of $\triangle B E M$ - area of

$=\frac{1}{2} E M \times B M-\frac{1}{2} A M \times B M$

$=\frac{1}{2} B M(E M-A M)$

$=\frac{1}{2} \frac{x}{\sqrt{2}}\left(\sqrt{\left.130-\frac{x^2}{2}-\frac{x}{\sqrt{2}}\right)}\right.$

$=\frac{1}{2}+\frac{x}{\sqrt{2}}\left(\sqrt{\left.130-\frac{x^2}{2}-\frac{x}{\sqrt{2}}\right)=x^2}\right.$

$=\sqrt{130-\frac{x^2}{2}}=2 \sqrt{2} x+\frac{x}{\sqrt{2}}$

$130-\frac{x^2}{2}=\left(\begin{array}{c}5 x \\ \sqrt{2}\end{array}\right)^2$

$130-\frac{x^2}{2}=\begin{array}{c}25 x^2 \\ 2\end{array}$

$13 x^2=130 \Rightarrow x^2=10$

$\therefore$ Area of square $=10$

Given $A B C D$ is a square $A$ and $B . A$ and $B$ slide along theirs respective axes.

Let $A B=B C=C D=A D=a$ $\ln \Delta B M C$

$\sin \theta=\frac{B M}{B C}$

$\sin \theta=\frac{h}{a}$

$h=a \sin \theta$

$k=O B+N B$

$k=a \sin \theta+a \cos \theta$

$[\because O B=a \sin \theta, N B=a \cos \theta]$

$k=h+a \cos \theta$

$k-h=a \cos \theta$

On squaring and adding Eqs.$(i)$ and $(ii)$, we get

$h^2+(k-h)^2=a^2$

$\because$ Locus of $C B$

$x^2+(y-x)^2 =a^2$

$2 x^2+y^2-2 x y =a^2$

which is ellipse not a circle.

We have, $l_1, l_2, l_3 \ldots, l_n$ be the lengths of the side of arbitrary $n$ sided non-degenerate polygon $P$ and $\frac{l_1}{l_2}+\frac{l_2}{l_3}+\frac{l_3}{l_4}+\ldots+\frac{\ln -1}{l_n}+\frac{\ln }{l_1}=n, n \geq 4$

Using AM $\geq$ GM, we get

$\therefore$ The length of sides of $P$ are equal and $P$ is regular polygon of it is cyclic.

We have, side of quadrilateral has distinct integer second largest size has length $10.$

Let $a=8, b=9, c=10$, (All are distinct) We know, in quadrilateral Sum of three sides is greater than fourth side

$\therefore a+b+c>d \Rightarrow 8+9+10 > d \Rightarrow d < 27$

$\therefore$ Maximum length of $4$th side is $26.$

We have,

four shape of tiles

$(1)$ Equilateral triangle

$(2)$ Square

$(3)$ Regular pentagon

$(4)$ Regular hexagon

Cover the $X Y$-plane with four shapes such that no two tiles are overlaps.

$\therefore$ We use cquilateral triangular, regular hexagon is also made of equilateral triangle but pentagon cannot cover the plane because of its shape.

$\therefore$ Exactly three of four shapes, i.e. equilateral triangle, square and regular hexagon.

$A B C D$ is a trapezium.

$A B$ is parallel to $C D$.

$A B=11, B C=4, C D=6$ and $D A=3$

Construct $C E$ is parallel to $D A$.

$C E=3$

$B C=4$

$B E=5$

$\therefore \angle B C E$ is a right angled triangle.

$\therefore$ Area of $\triangle B C E=\frac{1}{2} E C \times B C$

$=\frac{1}{2} \times 3 \times 4$

Also, area of

$\triangle B C E =\frac{1}{2} \times B E \times h$

$=\frac{1}{2} \times 5 \times h$

From Eqs.$(i)$ and $(ii)$,

$\frac{1}{2} \times 3 \times 4 =\frac{1}{2} \times 5 \times h$

$\Rightarrow \quad h =2.4$

$A B C$ is a right angled triangle, $\angle A B C=90^{\circ}$

Circumcentre of $\triangle A B C$ is mid-point of $A C$ i.e. $M$.

Circumcentre of $\Delta P_1 P P_2$ is mid-point of $P_1 P_2$

$A B$ is perpendicular bisector of $P P_1$ and $B C$ is perpendicular bisector of $P P_2$.

Perpendicular bisector of $P P_1$ and $P P_2$ intersect at $B$.

$\therefore B$ is circumcentre of $\Delta P_1 P P_2$.

$\therefore$ Distance between

$B M=A M=M C=\frac{A C}{2}$

Given, a cuboid has all edges are integers and base is square.

Let the length, breadth and height of cuboid is $x, x, y$.

Sum of all edges of cuboid $=4 x+4 x+4 y$

Sum of area of all faces $=2 x^2+2 x y+2 x y$

Given,

Sum of all edges of cuboid = Sum of area of all faces

$\therefore \quad 4 x+4 x+4 y=2\left(x^2+x y+x y\right)$

$\Rightarrow \quad 4(2 x+y)=2\left(x^2+2 x y\right)$

$\Rightarrow \quad x^2+2 x y-4 x-2 y=0$

$\Rightarrow \quad x^2+2 x(y-2)-2 y=0$

$\Rightarrow \quad x=\frac{-2(y-2) \pm \sqrt{4(y-2)^2+4(2 y)}}{2}$

$\Rightarrow \quad x=y-2 \pm \sqrt{y^2-2 y+4}$

$x \text { is integer, when } y=2$

$\therefore \quad \quad y=2 x=2$

Hence, sum of edges $=8 x+4 y=16+8=24$

Attime $t=0, k$ th ant starts at point $k^2$ and reaches at time $t=0$ at the point $(11-k)^2$.

$\therefore$ Velocity of $k$ th ant

$=\frac{x_2-x_1-(11-k)^2-k^2}{t_2-t_1}$

$u=121-12\,h$

Now, two ants are at the same location

$x_i=x_j$

$x_i=x+u t$

$\Rightarrow \quad k_i^2-22 k_i t+121 t=k_j-22 k_j t+121 t$

$\Rightarrow \quad t=\frac{k_j^2-k_i^2}{22\left(k_j-k_i\right)}=\frac{k_j+k_i}{22}$ $\quad\left[k_i \neq k_j\right]$

Now, for $i=1$,

values of $t$ will be $\frac{3}{22}, \frac{4}{22}, \frac{5}{22}, \ldots, \frac{11}{22}$

($9$  values)

When $i=2$

value of $t$ will be $\frac{4}{22}, \frac{5}{22}, \ldots, \frac{11}{22}, \frac{12}{22}$

Only one distinct value.

Similarly, for $i=3,4,5,6,7,8,9$, we get only $1$ distinct value.

So, in all there are $17$ distinct value of $t$.

Given,

Area of both figures are equal

Area of fig. $(i)$ $=2 x y+\frac{1}{2} \cdot x(3 y)$

$=2 x y+\frac{3 x y}{2}=\frac{7 x y}{2}$

Area of fig.$(ii)$

Area of $A B C G+$ Area of $D E F G$

$=2 x y+(2 x-y) y$

$=2 x y+2 x y-y^2=4 x y-y^2$

$\therefore \frac{7 x y}{2}=4 x y-y^2 \Rightarrow y^2=\frac{1}{2} x y \Rightarrow 2 y=x$

$A B C$ is a triangle. $P$ be interior point of a $\triangle A B C, Q$ and $R$ be the reflections of $P$ in $A B$ and $A C$ respectively.

$Q A R$ are collinear

$\therefore \angle Q A R=180^{\circ}$

$Q \text { is reflection of } P \text { on } A B$

$\therefore \angle Q A B=\angle P A B$

$R \text { is reflection of } P \text { on } A C$

$\therefore \angle R A C=\angle P A C$

$\therefore 2(\angle P A B+\angle P A C)=180^{\circ}$

$\Rightarrow \angle P A B+\angle P A C=90^{\circ}$

$\Rightarrow \angle B A C=90^{\circ}$

We have,

A rectangular corner is cut form a rectangular piece of paper.

Area of rectangle

$\quad=12 \times 9=108 \text { sq units }$

$\text { Area of pentagon }$

$\qquad \quad=\text { Area of rectangle }-\text { Area of triangle }$

$\quad=108-6=102$

$\therefore \text { Ratio }=\frac{102}{108}=\frac{17}{18}$

Given,

$A B C D$ is a trapezium.

$A D$ is parallel to $B C$

$M$ is point on $B C$

such that $A B=A M$ and $D C=D M$

In $\triangle A M D$

Area of $\triangle A M D=$ Area of $\triangle A M N$

$+$ Area of $\triangle D M N$

Area of $\triangle A M N=$ Area of $\triangle A M P$

$=$ Area of $\triangle A B P$

Area of $\triangle D M N=$ Area of $\triangle D Q M$

$=$ Area of $\triangle D Q C$

$\therefore$ Area of trapezium $A B C D$

$=$ Area of $\triangle A B M+$ Area of $\triangle A M D$

$+$ Area of $\triangle M D C$

$=3[$ Area of $\triangle A P M+$ Area of $\triangle D M N)$

$=3$ Area of $\triangle A M D$

Area of trapezium $A B C D$

$\quad$ Area of $\triangle A M D$

$=\quad 3[$ Area of $\triangle A D M]=\frac{3}{1}=3: 1$

Given,

Perimeter of rectangle is $76$ units.

Let $x$ and $y$ are sides of each rectangles.

$\therefore$ Perimetre of rectangle $=6 x+5 y=76$

and $\quad 4 x=3 y \quad \ldots (i)$

On solving Eqs.$(i)$ and $(ii)$,we get

$x=6, y=8$

$\therefore$ Perimeter of each rectangle

$=2(x+y)=2(6+8)=28 \text { units }$

$A B C D$ is a quadrilateral

$\angle D A B=\angle A B C=60^{\circ}$

and $\quad \angle C A B=\angle C B D$

Construction, $A D$ and $B C$ produced to meet at such that

$\triangle A E B$ is an equilateral.

$\because \quad A B=B E=A E$

In $\triangle B D E$ and $\triangle A B C$,

$\angle B E D=\angle A B C$

$\angle D B E=\angle C A B$ given,

$[\because \angle D B E=\angle D B C]$

$\triangle B E D \sim \triangle A B C$

$\frac{B E}{A B}=\frac{B D}{A C}=\frac{E D}{B C} \Rightarrow \frac{B E}{A B}=\frac{E D}{B C}$

$\frac{A B}{A B}=\frac{A E-A D}{B C}$

$\Rightarrow \quad A E-A D=B C \Rightarrow A B=A D+B C$

Given,

In $\triangle A B C$

The angle bisector $B D$ and $C E$ are divided by incentre $I$ in the ratio $3: 2$ and $2: 1$ respectively.

$\frac{B I}{I D}=\frac{3}{2}$

and$\frac{C I}{I C}=\frac{2}{1}$

We know, $\frac{A I}{I F}=\frac{b+c}{a}, \frac{B I}{I D}=\frac{a+c}{b}$.

$\frac{C I}{I E}=\frac{a+b}{c}$

$\because \quad \frac{B I}{I D}=\frac{a+c}{b}=\frac{3}{2} \Rightarrow 2(a+c)=3 b \quad \ldots (i)$

and $\frac{C I}{I E}=\frac{a+b}{c}=\frac{2}{1} \Rightarrow a+b=2 c$

On solving Eqs.$(i)$ and $(ii)$, we get

$b=\frac{3}{2} a \text { and } c=\frac{5}{4} a$

$\because \quad \frac{A I}{I F}=\frac{b+c}{a}=\frac{\frac{3}{2} a+\frac{5}{4} a}{a}=\frac{11}{4}$

Hence, ratio $=11: 4$.

In $\triangle A B C, \angle A < \angle B < \angle C$

In $\triangle A B D, \quad \angle D > \angle C$

So, $\triangle A B D$ not similar to $\triangle A B C$.

Given, in $\triangle A B C$

$A(0,0), B(1,1) C(9,1)$

$\text { Area of } \triangle C D E=\frac{1}{2} \text { Area of } \triangle A B C$ $\Rightarrow \quad \frac{1}{2} \times C D \times D E=\frac{1}{4} \times B C \times A P$ $\Rightarrow \quad \frac{1}{2}(9-a) \times\left(1-\frac{a}{9}\right)=\frac{1}{4} \times 8 \times 1$

$\Rightarrow \quad (9-a)(9-a)=36$

$\Rightarrow \quad 9-a=6 \Rightarrow a=3$

We have,

$G$ is the centroid of $\triangle A B C, M$ and $N$ are interior point on sides $A B$ and $A C$ respectively.

$M G N$ are collinear.

If $M$ and $B$ are coincide.

$\therefore M G N$ are median of $\triangle A B C$.

$\therefore$ Area of $\triangle A M N=\frac{1}{2}$ area of $\triangle A B C$

Given, $\frac{\text { area of } \triangle A M N}{\text { area of } \triangle A B C}=r$

$\therefore \quad r_{\max }=\frac{1}{2}$

Case I When MGN are parallel to $B C$

$A G: A D=2: 3$

$\triangle A M N \sim \triangle A B C$

$\frac{\operatorname{ar}(\triangle A M N)}{\alpha r(\triangle A B C)}=\left(\frac{2}{3}\right)^2=\frac{4}{9}$

$r \geq \frac{4}{9}$

From Eqs.$(i)$ and $(ii)$, we get

$\frac{4}{9} \leq r < \frac{1}{2}$

We have sides of triangle are,

$b+5,3 b-2,6-b$

Triangle are isosceles.

$\therefore$ Two sides are equal.

Case $I$ $b+5=3 b-2$

$\therefore \quad b=\frac{7}{2}$

So, sides are $\frac{17}{2}, \frac{17}{2}, \frac{5}{2}$

Case $II$ $3 b-2=6-b \Rightarrow b=2$

$\therefore$ Sides are 7, 4, 4

Case $III$ $b+5=6-b \Rightarrow b=\frac{1}{2}$

Sides are $\frac{11}{2},-\frac{1}{2}, \frac{11}{2}$ which is not possible.

$\therefore$ Only for two values of $b$, triangles are isosceles.

Given, $A B C D$ is rectangle.

$\therefore \quad A B=C D, B C=A D$

$X$ and $Y$ are mid-point of $A D$ and $C D$ recnertivels

Let $A B=2 x, B C=2 y$

$\therefore A X=X D=y$

$D Y=Y C=x$

Area of rectangle $A B C D=4 x y=60$ $\Rightarrow x y=15$

In $\triangle A B X$ and $\triangle D E X$

$\triangle A B X \simeq \Delta D E X$

$D E =A B=2 x$

Similarly, $\triangle C B Y \simeq \triangle D F Y$

$\therefore F D=B C=2 y$

$\therefore$ Area of $\triangle B E F$

$=$ Area of $\triangle E F X+$ Area of $\triangle B F X$

$=\frac{1}{2} F X \cdot D E+\frac{1}{2} F X \cdot A B$

$=\frac{1}{2} \times 3 y \times 2 x+\frac{1}{2} \times 3 y \times 2 x$

$=6 x y$

$=6 \times 15=90 \quad[\because x y=15]$

Given,

$A B C D E F$ is a regular hexagon of side length $1.$

$A B Q R$ and $A F P S$ is a square of each side length also $1 .$

$A D C D E F$ is a regular hexagon

$\therefore \quad \angle F A B=120^{\circ}$

$\text { In square } A B Q R,$

$\qquad A B=B Q=1$

$A Q$ is a diagonal of square

$\therefore \quad A Q=\sqrt{A B^2+B Q^2}=\sqrt{2}$

$\Rightarrow \angle B A S=\angle F A B-\angle F A S$

$\quad=120^{\circ}-90^{\circ}=30^{\circ}$

$\Rightarrow \angle S A R=\angle B A R-\angle B A S$

$\quad=90^{\circ}-30^{\circ}=60^{\circ}$

$\Rightarrow \angle A S R=60^{\circ}$

$\qquad \quad[\because A R S \text { is an equilateral triangle }]$

$\Rightarrow \angle R S P=\angle A S P-\angle A S R$

$\quad=90^{\circ}-60^{\circ}=30^{\circ}$

$\Rightarrow \angle F A B=\angle F A P+\angle P A Q+\angle Q A B$

$\Rightarrow \quad 120^{\circ}=45^{\circ}+\angle P A Q+45^{\circ}$

${\left[\because \angle F A P=\angle Q A B=45^{\circ}\right.}$

$F A=F P$ and $A B=B Q]$

$\therefore \angle P A Q=30^{\circ}$

$\therefore \frac{\text { Area of } \triangle P A Q}{\text { Area of } \triangle R S P}=\frac{\frac{1}{2} \times A Q \times A P \times \sin 30^{\circ}}{\frac{1}{2} \times R S \times P S \times \sin 30^{\circ}}$

$=\frac{\sqrt{2} \times \sqrt{2}}{1}=2$

$[\because A Q=A P=\sqrt{2}, R S=P S=1]$

Let the length, breadth and height of cuboid be $x, y$ and $z$ respectively. $\therefore$ Perimeter of face $P Q R S=2(x+y)$

Area of $P Q R S=x y$

$2(x+y)+x y=16$

Similarly, for face $A P S D$,

$\qquad 2(y+z)+y z=24$

$\text { and for face } A P Q B$

$2(x+z)+x z=31$

$\text { From Eqs. $(ii)$ and $(iii)$, we get }$

$(x-y)(2+z)=7$

From Eqs. $(ii)$ and $(iv)$, we get

$4 x=2+5 y$

On solving Eqs.$(i)$ and $(v)$, we get

$x=3, y=2, z=5$

$\therefore$ Volume of cuboid $=x y z=3 \times 2 \times 5=30$

Hence, option $(d)$ is correct.

Given, $A B C D$ is a square.

Let $A B=B C=C D=A D=3 x$

$P D: P C =1: 2$

$P D =x$

$P C =2 x$

In $\triangle D A P$ and $\triangle Q B A$,

$\angle D A P =\angle Q B A$

$\angle D =\angle Q=90^{\circ}$

$\triangle D A P \sim \triangle Q B A$

$\frac{D A}{Q B}=\frac{A P}{B A}=\frac{D P}{Q B}$

$\frac{3 x}{Q B}=\frac{\sqrt{10} x}{3 x}=\frac{x}{Q A}$

$\quad\left[\because A P=\sqrt{\left.9 x^2+x^2=\sqrt{10} x\right]}\right.$

$\therefore \quad Q B=\frac{9}{\sqrt{10}} x \Rightarrow Q A=\frac{3}{\sqrt{10}} x$

Area of quadrilateral $B Q P C=$ area of square $A B C D-($ area of $\triangle A P D+$ area of $\triangle A B Q)$

$=(3 x)^2-\left(\frac{1}{2} \times 3 x \times x+\frac{1}{2} \times \frac{9}{\sqrt{10}} x \times \frac{3}{\sqrt{10}} x\right)$

$=9 x^2-\left(\frac{3}{2} x^2+\frac{27}{20} x^2\right)=\frac{123 x^2}{20}$

$\frac{\text { Area of quadrilateral } P Q B C}{\text { Area of square } A B C D}=\frac{\frac{123 x^2}{20}}{9 x^2}$

$=\frac{41}{60}$

Given, $H$ is orthocentre of $\triangle A B C$ and $O$ is circumcentre of $\triangle A B C$.

Let centroid of $\triangle A B C$ is $G$.

In acute angle $\triangle A B C$,

We know, $H G: G O=2: 1$

$L =\frac{ A + B + C }{3}$

Now, $G =\frac{2 O + H }{3}$

$\Rightarrow \quad 3 G =2 O + H$

$\therefore HA + HB + HC$

$= A - H + B - H + C - H$

$=( A + B + C )-3 H$

$=3 G -3 H$

$=2 O + H -3 H$

$=2 O -2 H$

$=2 HO$

We have,

$A H K F, F K D E$ and $H B C K$ is a unit square. $A D$ and $B F$ intersect at $X$. In $\triangle A B F, \quad A B=2$

$A F=1$

$\therefore F B^2=A B^2+A F^2=4+1=5$

In $\triangle A X F$ and $\triangle B A F$,

$\angle F =\angle F$

$\angle X =\angle A=\left(90^{\circ}\right)$

$\therefore \quad \triangle A X F \sim \Delta B A F$

$\therefore \quad \frac{\operatorname{ar}(\triangle A X F)}{\operatorname{ar}(B A F)}$ $=\frac{A F^2}{B F^2}=\frac{1}{5}$

Given,

$A B C$ is right angled triangle with $B$ is $90^{\circ} .$

$A D$ is angle bisector of $\angle A$.

$\therefore \quad \frac{A B}{A C}=\frac{B D}{D C}$

$\Rightarrow \quad A B \cdot C D=B D \cdot A C$

Area of $\triangle A D C=10$

$\Rightarrow \quad \frac{1}{2} \times A B \cdot C D=10$

$\Rightarrow \quad \frac{1}{2} \times B D \cdot A C=10$

$\Rightarrow \quad B D=\frac{20}{A C}$

$\Rightarrow \quad B D=\frac{20}{6} \quad[\because A C=6]$

$\Rightarrow \quad B D=\frac{10}{3}$

Given, $A(4,0)$ and $B(0,12)$.

Let $C(x, y)$

$\begin{array}{l} \text { Area of } \triangle A B C=\frac{1}{2}\left|\begin{array}{ccc} x & y & 1 \\ 4 & 0 & 1 \\ 0 & 12 & 1 \end{array}\right|\end{array}$

$\Rightarrow 18=\frac{1}{2}|\{x(0-12)-y(4-0)+1(48-0)\}|$

$\Rightarrow 18=\frac{1}{2}|(-12 x-4 y+48)|$

$\Rightarrow 18=\frac{-4}{2}|(3 x+y-12)|$

$\Rightarrow \quad(3 x+y-12)^2=81$

$\Rightarrow \quad(y+3 x-12)^2=81$

We have, initial position of frog $=(0,0)$

After $1st$ jump position of frog at $(4,3)$.

At $2 nd$ jump position of frog at $(0,6)$.

At $3rd$ jump position of frog at $(0,1)$.

$\therefore$ Minimum number of jumps required for the frog to go from $(0,0)$ to $(0,1)$ and each distance is $5$ units is $3$.

Given $A B C D$ is square

$A B=B C=12$ units

$M$ is mid-point of $A B$,

$N$ is mid-point of $C D$,

$P$ is point of $M N$ and $C P=S$

$\because A P B$ is an isosceles triangle,

$A P=P B=r$

$O P=M B=6$ units

Area of triangle whose sides $r, s$ and $12$ are the area of $\triangle P B C$

$=\frac{1}{2} \times$ Base $\times$ Height $=\frac{1}{2} \times B C \times O P$

$=\frac{1}{2} \times 12 \times 6=36$ sq units

Let the length, breadth and height of cuboid is $l, b$ and $h$ respectively.

$Given, l^2+h^2=39^2$

$\Rightarrow b^2+h^2=40^2$

$\Rightarrow \quad l^2+b^2=41^2$

$\Rightarrow \quad 2\left(l^2+b^2+h^2\right)=39^2+40^2+41^2$

$\Rightarrow \quad l^2+b^2+h^2=2401$

$\therefore$ Length of longest diagonal

$=\sqrt{l^2+b^2+h^2}$

$=\sqrt{2401}=49$

Given,

$A B C$ is an equilateral triangle.

$\therefore \quad A B=B C=A C$

$K L M N$ be a rectangle.

$\therefore \quad K L=M N$

and $N K=L M$

$\frac{A N}{N B}=2$

$\therefore \quad A N=2 N B=A M=M N$

$A B=A N+N B=3 N B$.

Area of $\triangle B K N=6$

$\ln \triangle B K N$,

$\begin{array}{l}\quad \sin 60^{\circ}=\frac{N K}{B N} \Rightarrow N K=\frac{\sqrt{3}}{2} B N \\\text { Area of } \triangle B K N=\frac{1}{2} \cdot B N \cdot N K \sin 30^{\circ} \\\Rightarrow \quad 6=\frac{1}{2} \cdot B N \cdot \frac{\sqrt{3}}{2} \cdot B N \cdot \frac{1}{2} \Rightarrow B N^2=\frac{48}{\sqrt{3}} \\ \therefore \text { Area of } \triangle A B C=\frac{\sqrt{3}}{4} \times A B^2 \\ =\frac{\sqrt{3}}{4} \times 9 B N^2=\frac{\sqrt{3}}{4} \times 9 \times \frac{48}{\sqrt{3}}=108 \end{array}$

$A B C D$ is a cyclic quadrilateral $\angle A=120^{\circ}$

$A B C D$ has maximum area possible.

When $\angle B=\angle D=90^{\circ}$

$\therefore$ Area of quadrilateral

$=2 \times \frac{1}{2} \times A D \times D C$

$=2 \times \frac{1}{2} \times R \times \sqrt{3} R$

$=\sqrt{3} R^2 \quad[\because A D=R, D C=\sqrt{3} R]$

Area of rectangles are

$a c=10$

$b c=4$

$b d=12$

$d e=15$

$e f=25$

$\therefore \frac{a c}{b c} \times \frac{b d}{d e} \times e f =\frac{10}{4} \times \frac{12}{15} \times 25$

$\Rightarrow a f =50$

$P_1, P_2, P_3, P_4, P_5$ be five equally spaced points on the circumference of circle of radius $1$ .

Let $R$ which is near to point $O$. $\therefore O R$ is lie between the pentagonal region $Q_1, Q_2, Q_3, Q_4, Q_5$

$\because O(2, z)$ is circumcentre of $\triangle A B C$.

$O A^2=O B^2$

$\Rightarrow (2-z)^2+(z-3)^2=(4-2)^2+(z)^2$

$\Rightarrow z^2-6 z+9=4+z^2$

$\Rightarrow z=\frac{5}{6}$

$\Rightarrow O A=r=\sqrt{\left(\frac{5}{6}-3\right)^2}=\frac{13}{6}$

$t_n$ denotes the number of integral sided triangle with distincts sides from $\{1,2,3, \ldots, n\} . t_{19}$ is the number of triangle formed by the sides from $\{1,2,3, \ldots, 19\}$ and $t_{20}$ is the number of triangle formed by the distinct sides from $\{1,2,3, \ldots, 20\}$.

Any triangle counted in $t_{19}$ is also counted in $t_{20}$, but $t_{20}-t_{19}$ is the number of triangle counted in $t_{20}$ but not in $t_{19+}$ A triangle is counted in $t_{20}$ but no $t_{19}$ if and only if its largest side is $20$.

The middle side of is $a$ and the smallest side can be $21-a$ to $a-1$

So, the number of triangle with largest side 20 and middle side.

$a=11$, then other sides are 21-11, 11-1

i.e. $10,10(11,10,10) 1$ triangle. Similarly

$a=12$, (smallest sides are $(9,10,11)=3$

triangle $a=13$, smallest sides are

$(8,9,10,11,12)=5$ triangle

$\therefore$ Total number of triangles on

$1+3+5+7+\ldots .+17=81$

$ A O=B O $

$ (a-5)^2+\left(\frac{a}{4}+2\right)^2=(a-5)^2+\left(\frac{a}{4}-6\right)^2 $

$ a=8 $

$ A B=8, A C=6, B C=10 $

$ \alpha=5, \beta=24, \gamma=24$A

$2^{\mathrm{n}}\left(2^{\mathrm{m}-\mathrm{n}}-1\right)=2^3 \times 7 $

$ 2^{\mathrm{n}}=2^3 \text { and } 2^{\mathrm{m}-\mathrm{n}}-1=7 $

$ \Rightarrow \mathrm{n}=3 \text { and } 2^{\mathrm{m}-\mathrm{n}}=8 $

$ \Rightarrow \mathrm{n}=3 \text { and } \mathrm{m}-\mathrm{n}=3 $

$ \Rightarrow \mathrm{n}=3 \text { and } \mathrm{m}=6 $

$ \mathrm{P}(6,3) \text { and } \mathrm{Q}(-2,-3) $

$ \mathrm{PQ}=\sqrt{8^2+6^2}=\sqrt{100}=10$

Hence option $(1)$ is correct

Slope of $\mathrm{BC}=-\frac{1}{3}$

equation of $\mathrm{BC}=3 \mathrm{y}+\mathrm{x}-17=0$

slope of $\mathrm{BE}=1$

Slope of $A C=-1$

equation of $A C$ is $x+y-3=0$

point $C$ is $(-4,7)$

$ x+2 y-1=0 $

$ a x+b y-1=0$

$Image$

$ \left(\frac{6-6}{3}, \frac{8-8}{3}\right) $

$ =(0,0)$

$Image$

$ a x+b y-1=0 $

$ \left(\frac{1-0}{-1-0}\right)\left(\frac{-a}{b}\right)=-1 $

$ \Rightarrow-a=b $

$ \Rightarrow \quad a x-a y-1=0 $

$ a x-a\left(1-\frac{2 x}{3}\right)-1 $

$ x\left(a+\frac{2 a}{3}\right)=\frac{a}{3} $

$ x=\frac{a+3}{5 a} $

$ 2\left(\frac{a+3}{5 a}\right)+3 y-1=0 $

$ y=\frac{1-\frac{2 a+6}{5 a}}{3}=\frac{3 a-6}{3 \times 5 a} $

$ y=\frac{a-2}{5 a} $

$ \left(\frac{a-2}{5 a}\right) $

$ \left(\frac{a+3}{5 a}\right) $

$ a=-8 $

$ b=8 $

$ -8 x+8 y-1=0 $

$ |a-b|=16$

So centroid : $(\alpha, \beta)$ is

$\alpha=\frac{6+1+5}{3}=4$

$\beta=\frac{8+2-7}{3}=1$

$\alpha+2 \beta=6$

$2 \alpha-\beta=7$

$\text { Ans. } x^2-13 x+42=0$

$\left(\frac{\beta-3}{\alpha-2}\right)\left(\frac{1}{-2}\right)=-1$

$\beta-3=2 \alpha-4$

$\beta=2 \alpha-1$

$m _{ AH } \times m _{ HC }=-1$

$\left(\frac{\beta-2}{\alpha-1}\right)(-2)=-1$

$2 \beta-4=\alpha-1 $$2(2 \alpha-1)=\alpha+3$

$3 \alpha=5$

$\alpha=\frac{5}{3}, \beta=\frac{7}{3} \Rightarrow H \left(\frac{5}{3}, \frac{7}{3}\right)$

$\alpha+4 \beta=\frac{5}{3}+\frac{28}{3}=\frac{33}{3}=11$

$\beta+4 \alpha=\frac{7}{3}+\frac{20}{3}=\frac{27}{3}=9$

$x ^2-20 x +99=0$

$4 x+3 y=69$

$25 y=175$

$y=7, x=12$

$A(12,7)$

$-3 x+4 y=17$

$3 x+21 y=183$

$25 y=200$

$y=8, x=5$

$B (5,8)$

$\therefore \text { Circumcenter }$

$\alpha=\frac{17}{2} \beta=\frac{15}{2}$

$\left(\frac{17}{2}, \frac{15}{2}\right)$

$(\alpha-\beta)^2+\alpha+\beta$

$1+16=17$

$3 y+21=-5 x+15$

$5 x+3 y+6=0$

Altitude of $AC : y -2=\frac{-1}{12}( x +1)$

$12 y-24=-x-1$

$x+12 y=23$

$\alpha=\frac{-47}{19}, \quad \beta=\frac{121}{57}$

$9 \alpha-6 \beta+60=25$

Let $M$ be the mid-point of $A B$.

$M ( h , k ) \equiv\left(\frac{\alpha}{2}, \frac{\beta}{2}\right)$

$\frac{10-\frac{1}{2} \alpha \beta}{\frac{1}{2} \alpha \beta}=4$

$\frac{5}{2} \alpha \beta=10 \Rightarrow \alpha \beta=4$

$(2 h )(2 K )=4$

Locus of $M$ is $x y=1$

Which is a hyperbola.

$\frac{\alpha+\beta+3+1}{3}=2 \quad \text { also } \frac{-2 \alpha-2+\beta}{3}= a$

$\begin{array}{lc}\Rightarrow \alpha+\beta=2 & -2 \alpha-2+\beta=3 a \\\Rightarrow \beta=2-\alpha & -2 \alpha- 2+ 2-\alpha=3 a \Rightarrow \alpha=- a\end{array}$

Now both $B$ and $C$ lies as given line

$\alpha- p \cdot 2 \alpha=21 a$

$\alpha(1-2 p )=21 a$

$-\alpha(1-2 p )=21 a \Rightarrow p =11$

$\beta+3+ p \beta=21 a$

$\beta+3+11 \beta=21 a$

$21 \alpha+12 \beta+3=0$

Also $\beta=2-\alpha$

$21 \alpha+12(2-\alpha)+3=0$

$21 \alpha+24-12 \alpha+3=0$

$9 \alpha+27=0$

$\alpha=-3, \beta=5$

So $BC =\sqrt{122}$ and $( BC )^2=122$

$k=\frac{\frac{-2 \lambda}{\sqrt{2}} 2 \cos \theta+\frac{3 \lambda}{\sqrt{2}} \sin \theta}{5}$

$h^2+k^2=\frac{19 \lambda^2}{5}$

$r=\frac{\sqrt{19} \lambda}{5}$

$AC \equiv 2 x - y -1=0$

So $A (1,1)$

Altitude from $B$ is $BH = x +2 y -7=0 \Rightarrow B (3,2)$

Altitude from $C$ is $CH =2 x + y -7=0 \Rightarrow C (2,3)$

Centroid of $\triangle ABC = E (2,2) OE =2 \sqrt{2}$

$\sin 60^{\circ}=\frac{5 / \sqrt{2}}{a}$

$a=\frac{5 \sqrt{2}}{3}$

Area of $\triangle PQR =\frac{\sqrt{3}}{4} a ^{2}=\frac{25}{2 \sqrt{3}}$

$y=2 x$

Slope of $AB =2$

Slope of given diameter $=2$

So the diameter is parallel to $AB$

Distance between diameter and line $AB$ $=\left(\frac{4}{\sqrt{2^{2}+12}}\right)=\frac{4}{\sqrt{5}}$

Thus $BC =2 \times \frac{4}{\sqrt{5}}=\frac{8}{\sqrt{5}}$

$AB =\sqrt{(1-3)^{2}+(2-6)^{2}}=\sqrt{20}=2 \sqrt{5}$

$\text { Area }= AB \times BC =\frac{8}{\sqrt{5}} \times 2 \sqrt{5}=16 \text { Ans. }$

$B =\left(\frac{-3}{\sqrt{ a }}, \sqrt{ a }\right)$

$C =\left(-\frac{3}{\sqrt{ a }},-\sqrt{ a }\right)$

Area of $ACD$

$\frac{1}{2}\left|\begin{array}{cc}\frac{3}{\sqrt{a}} & \sqrt{a} \\ -\frac{3}{\sqrt{a}} & -\sqrt{a} \\ 3 \cos \theta & a \sin \theta \\ \frac{3}{\sqrt{a}} & \sqrt{a}\end{array}\right|$

$\frac{1}{2} 6 \sqrt{ a }(\cos \theta-\sin \theta)$

$3 \sqrt{ a }(\cos \theta-\sin \theta)$

max values of function is $3 \sqrt{ a } \sqrt{2}$

$3 \sqrt{a} \sqrt{2}=12$

$2 a=16$

$a=8$

Now let $C$ be $(h, 4-2h)$

(As $C$ lies on $2 x+y=4$ )

$\because \Delta$ is isosceles with base $B C$

$\therefore A B=A C$

$\sqrt{25+1}=\sqrt{(6-h)^{2}+(2 h-3)^{2}}$

$\sqrt{26}=\sqrt{36+h^{2}-12 h+4 h^{2}+9-12 h}$

$26=5 h^{2}-24 h+45 \Rightarrow 5 h^{2}-24 h+19=0$

$\Rightarrow 5 h^{2}-5 h-19 h+19=0$

$h=\frac{19}{5}$ or $h =1$

Thus $C\left(\frac{19}{5}, \frac{-18}{5}\right)$

Centroid $\left(\frac{6+1+\frac{19}{5}}{3}, \frac{1+2-\frac{18}{5}}{3}\right)$

$\left(\frac{35+19}{15}, \frac{15-18}{15}\right)$

$\left(\frac{54}{15}, \frac{-3}{15}\right)$

$\alpha=\frac{54}{15} ; \beta=\frac{-3}{15}$

$15(\alpha+\beta)=51$

orthocentre $H (2$, a)

Slope of $AH = p$

$a +2= p$

Slope of $BH =-1$

$31 a -2 ab =15 a +4 p -2$

From $(1)$ and $(2)$

$a =1$ and $p =3$

$x+y=5$

Take image of $A$

$\frac{ x -1}{1}=\frac{ y +2}{1}=\frac{-2(-6)}{2}=6$

$(7,4)$

$7+4 p =39$

$p =8$

$\text { solving } x +8 y =39 \text { and } y =-2 x$

$x =\frac{-39}{15} \quad y =\frac{78}{15}$

$AC =72=9\,p4$

$AC ^{2}+ p ^{2}=72+64=136$

$\Delta ABC =\frac{1}{2}\left|\begin{array}{ccc}1 & -2 & 1 \\ 7 & 4 & 1 \\ \frac{-39}{15} & \frac{78}{15} & 1\end{array}\right|$

$=\frac{1}{2}\left[4-\frac{78}{15}+2\left(7+\frac{39}{15}\right)+7\left(\frac{78}{15}\right)+\frac{4 \times 39}{15}\right]$

$=\frac{1}{2}\left[18+18 \times \frac{13}{5}\right]$

$=9\left[\frac{18}{5}\right]=\frac{162}{5}=32.4$

$\Delta_{2}=\frac{1}{2}\left|\begin{array}{ccc}1 & 1 & 1 \\ -4 & 3 & 1 \\ -2 & -5 & 1\end{array}\right|$

Given $\frac{\Delta_{1}}{\Delta_{2}}=\frac{4}{7} \Rightarrow \frac{-2 x-5 y+7}{36}=\frac{4}{7}$ $\Rightarrow 14 x+35 y=-95 \ldots(1)$

Equation of $BC$ is $4 x + y =-13 \ldots(2)$

Solve equation $(1)$ and $(2)$

Point $P\left(\frac{-20}{7}, \frac{-11}{7}\right)$

Here point $Q\left(\frac{-1}{2}, 0\right) \& R\left(\frac{1}{2}, 0\right)$

So Area of triangle $AQR =\frac{1}{2} \times 1 \times 1=\frac{1}{2}$

Let orthocentre be $( h , k )$

Since it if an equilateral triangle hence orthocentre coincides with centroid.

$\therefore a + s + c =3 h , b + s - c =3 k$

$\therefore(3 h - a )^{2}+(3 k - b )^{2}=( s + c )^{2}+( s - c )^{2}=2\left( s ^{2}+ c ^{2}\right)=2$

$\therefore\left( h -\frac{ a }{3}\right)^{2}+\left( K -\frac{ b }{3}\right)^{2}=\frac{2}{9} \text {, }$

circle centre at $\left(\frac{ a }{3}, \frac{ b }{3}\right)$

Gives, $\frac{ a }{3}=1, \frac{ b }{3}=\frac{1}{3} \Rightarrow a =3, b =1$

$\therefore a^{2}-b^{2}=8$

since $AC$ is perpendicular to $AB$.

So, $\triangle ABC$ is right angled at $A$.

Circumcentre $=$ mid point of $BC .=\left(\frac{5 \alpha}{8}, 2\right)$

$\therefore \frac{5 \alpha}{8}=5 \& \frac{\alpha}{4}=2$

$\alpha=8$

Area $=\frac{1}{2}(6)(8)=24$

Perimeter $=24$

Circumradius $=5$

Inradius $=\frac{\Delta}{ s }=\frac{24}{12}=2$

$\Rightarrow x^{2}+y^{2}+x-3 y-2=0$

Put $x=0$

$\Rightarrow y^{2}-3 y-2=0$

$\Rightarrow y=\frac{3 \pm \sqrt{17}}{2}$

Put $y=0$

$\Rightarrow x^{2}+x-2=0$

$(x+2)(x-1)=0$

$\therefore$ A $(-2,0), B(1,0), C\left(0, \frac{3+\sqrt{17}}{2}\right), D\left(0, \frac{3-\sqrt{17}}{2}\right)$

Area $=\frac{1}{2} \cdot 3 \cdot \sqrt{17}=\frac{3 \sqrt{17}}{2}$

Solving with $2 x - y +2=0$ will give $(-2,2)$

$\& \sin 30^{\circ}=\frac{1}{2}=\frac{ r }{ R } \Rightarrow R =\frac{6}{\sqrt{2}}$

$\therefore r + R =\frac{9}{\sqrt{2}}$

$OA \cdot AB$

$= AP \cdot AQ$

$\Rightarrow 1.12=\lambda \cdot \lambda$

$\Rightarrow \lambda=2 \sqrt{3}$

Area $\Delta PQB =\frac{1}{2} \times 2 \lambda \times AB$

$\Delta=\frac{1}{2} \cdot 4 \sqrt{3} \times 12$

$=24 \sqrt{3}$

$50(20+5)$

$50(25)$

$=1250$

$\sin A =1$

$A =90^{\circ} \Rightarrow BC =13$

$BC =2 R =13$

$r =\frac{\Delta}{ S }=\frac{30}{15}=2$

$2 R + r =15$

$\Rightarrow\left|\begin{array}{ccc}\mathrm{a} & 0 & 1 \\ \mathrm{~b} & 2 \mathrm{~b}+1 & 1 \\ 0 & \mathrm{~b} & 1\end{array}\right|=\pm 2$

$\Rightarrow \mathrm{a}(2 \mathrm{~b}+1-\mathrm{b})-0+1\left(\mathrm{~b}^{2}-0\right)=\pm 2$

$\Rightarrow \mathrm{a}=\frac{\pm 2-\mathrm{b}^{2}}{\mathrm{~b}+1}$

$\therefore \mathrm{a}=\frac{2-\mathrm{b}^{2}}{\mathrm{~b}+1} \text { and } \mathrm{a}=\frac{-2-\mathrm{b}^{2}}{\mathrm{~b}+1}$

sum of possible values of $'a'$ is $=\frac{-2 \mathrm{~b}^{2}}{\mathrm{b}+1}$

$=\frac{1}{2}\left( OA ^{\prime}\right)^{2} \frac{\sin 30^{\circ}}{2}$

$=(3+1) \times \frac{1}{8}=\frac{1}{2}$

$\overrightarrow{ a }_{ New }=( p +1) \hat{ i }+\sqrt{10} \hat{ j }$

$\Rightarrow\left|\overrightarrow{ a }_{ Od }\right|=\left|\overrightarrow{ a }_{ New }\right|$

$\Rightarrow 9 p^{2}+1=p^{2}+2 p+1+10$

$8 p^{2}-2 p-10=0$

$4 p^{2}-p-5=0$

$(4 p-5)(p+1)=0 \rightarrow p=\frac{5}{4},-1$

Now, applying rotation theorem

$-\frac{1}{\sqrt{2}}+\frac{7}{\sqrt{2}} i=((b+2)+a i)\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)$

$\frac{-1}{\sqrt{2}}+\frac{7}{\sqrt{2}} i=\left(\frac{b+2}{\sqrt{2}}-\frac{a}{\sqrt{2}}\right)+i\left(\frac{b+2}{\sqrt{2}}+\frac{a}{\sqrt{2}}\right)$

$\Rightarrow b-a+2=-1....(1)$

$\text { and } b+2+a=7....(2)$

$\Rightarrow a=4 ; b=1$

$\Rightarrow 2 a+b=9$

$Q \equiv\left( x _{2}, mx _{2}\right)$

$A _{1}=\frac{1}{2}\left|\begin{array}{ccc}3 & 4 & 1 \\ 2 & 0 & 1 \\ -1 & 1 & 1\end{array}\right|=\frac{13}{2}$

$A _{2}=\frac{1}{2}\left|\begin{array}{ccc} x _{1} & mx _{1} & 1 \\ x _{2} & mx _{2} & 1 \\ 2 & 0 & 1\end{array}\right|$

$A _{2}=\frac{1}{2}\left|2\left( mx _{1}- mx _{2}\right)\right|= m \left| x _{1}- x _{2}\right|$

$A _{1}=3 A _{2} \Rightarrow \frac{13}{2}=3 m \left| x _{1}- x _{2}\right|$

$AC : x +3 y =2$

$BC : y =4 x -8$

$P : x +3 y =2$ and $y = mx \Rightarrow x _{1}=\frac{2}{1+3 m }$

$Q : y =4 x -8$ and $y = mx \Rightarrow x _{2}=\frac{8}{4- m }$

$\left| x _{1}- x _{2}\right|=\left|\frac{2}{1+3 m }-\frac{8}{4- m }\right|$

$=\left|\frac{-26 m }{(1+3 m )(4- m )}\right|=\frac{26 m }{(3 m +1)| m -4|}$

$=\frac{26 m }{(3 m +1)(4- m )}$

$\left| x _{1}- x _{2}\right|=\frac{13}{6 m }$

$\Rightarrow \frac{26 m }{(3 m +1)(4- m )}=\frac{13}{6 m }$

$\Rightarrow \quad 12 m ^{2}=-(3 m +1)( m -4)$

$\Rightarrow \quad 12 m ^{2}=-\left(3 m ^{2}-11 m -4\right)$

$\Rightarrow \quad 15 m ^{2}-11 m -4=0$

$\Rightarrow \quad 15 m ^{2}-15 m +4 m -4=0$

$\Rightarrow \quad(15 m +4)( m -1)=0$

$\Rightarrow m =1$

$\Rightarrow P \equiv\left(\frac{17}{6}, \frac{8}{3}\right)$

$\Rightarrow \mathrm{PQ}=5$

Radius $ = \sqrt {9 + 16 + 103}  = \sqrt {128}  = 8\sqrt 2 $

$\therefore \left( { - 5,4} \right)$ will be nearer to the $(0,0)$

$\therefore $ Ans. $\sqrt {{5^2} + {4^2}}  = \sqrt {25 + 16}  = \sqrt {41} $

$\frac{k}{h} =  - \frac{{4 - 0}}{{3 - 2}}\,\,\,\,\,\,\,\,\, - 4h = k$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,h = \frac{{ - 3}}{4}$

${x_2} =  - 3,{y_2} = 0$

$B\left( { - 3,0} \right)$

$\frac{{{x_3} + 1}}{2} = 2$ and $\frac{{{y_3} + 2}}{2} = 3$

${x_3} = 3,{y_3} = 4$

$C\left( {3,4} \right)$

Centroid $\left( {\frac{{{x_1} + {x_2} + {x_3}}}{3},\frac{{{y_1} + {y_2} + {y_3}}}{3}} \right)$

$\left( {\frac{{1 - 3 + 3}}{3},\frac{{2 + D + 4}}{3}} \right) = \left( {\frac{1}{3},2} \right)$

$\sqrt {{h^2} + {{\left( {k - 1} \right)}^2}}  + \sqrt {{h^2} + {k^2}}  + 1 = 4$

$\sqrt {{h^2} + {{\left( {k - 1} \right)}^2}}  + \sqrt {{h^2} + {k^2}}  = 3$

${h^2} + {\left( {k - 1} \right)^2} = 9 + {h^2} + {k^2} - 6\sqrt {{h^2} + {k^2}} $

$ - 2k - 8 =  - 6\sqrt {{h^2} + {k^2}} $

$k + 4 = 3\sqrt {{h^2} + {k^2}} $

${k^2} + 16 + 8k = 9\left( {{h^2} + {k^2}} \right)$

$9{h^2} + 8{k^2} - 8k - 16 = 0$

Locus of $P$ is $9{x^2} + 8{y^2} - 8y - 16 = 0$

$AB = \sqrt {40}  = 2\sqrt {10} $

Centroud divides orthocenter and cicumcentre of the triangle in ratio $2:1$

$\therefore $ $ AB:BC=2:1$

Now, $AB = \frac{2}{3}AC$

$AC = \frac{3}{2}AB = \frac{3}{2}\left( {2\sqrt {10} } \right)\,\, \Rightarrow AC = 3\sqrt {10} $

Radius of circle with $AC$ as diamentre

$ = \frac{{AC}}{2} = \frac{3}{2}\sqrt {10}  = 3\sqrt {\frac{5}{2}} $

$x = \sqrt 3 $

$y = 1$

$\frac{x}{{\cos \,{{120}^o}}}\frac{y}{{\sin \,{{120}^o}}} = 2$

$x =  - 1,y = \sqrt 3 $

$\frac{x}{{\cos \,{{75}^o}}} = \frac{y}{{\sin \,{{75}^o}}} = 2\sqrt 2 $

$x = \sqrt 3  - 1$

$y = \sqrt 3  + 1$

Sum $ = 2\sqrt 3  - 2$

$\frac{1}{2}\,\left\| \begin{array}{l}
\,k\,\,\, - 3k\,\,\,1\\
\,5\,\,\,\,\,\,\,\,k\,\,\,\,\,1\\
 - k\,\,\,\,\,2\,\,\,\,\,1
\end{array} \right\| = 28$

$ \Rightarrow 5{k^2} + 13k - 46 = 0$

or $5{k^2} + 13k - 66 = 0$

Now, $5{k^2} + 13k - 46 = 0$

$ \Rightarrow k = \frac{{ - 13 \pm \sqrt {1089} }}{{10}}$

$\therefore k = \frac{{ - 23}}{5};k = 2$

since $k$ is an integer, $\therefore k = 2$

Also $5{k^2} + 13k + 66 = 0$

$ \Rightarrow k = \frac{{ - 13 \pm \sqrt { - 1151} }}{{10}}$

So no real solution exist 

For cothocentre

$BH \bot AC$

$\therefore \left( {\frac{{\beta  - 2}}{{\alpha  - 5}}} \right)\left( {\frac{8}{{ - 4}}} \right) =  - 1$

$ \Rightarrow \,\,\alpha  - 2\beta  = 1\,\,\,\,\,\,\,\,\,\,.......\left( 1 \right)$

Also $CH \bot AB$

$\therefore \left( {\frac{{\beta  - 2}}{{\alpha  + 2}}} \right)\left( {\frac{8}{3}} \right) =  - 1$

$ \Rightarrow \,\,3\alpha  - 8\beta  = 1\,0\,\,\,\,\,\,\,\,\,.......\left( 2 \right)$

Solving $(1)$ and $(2)$ , we get

$\alpha  = 2,\beta  = \frac{1}{2}$

orthocentre is $\left( {2,\frac{1}{2}} \right)$

i.e. $\frac{{\lambda  - 5}}{2}\,\,,1,\frac{{\lambda  - 8}}{2}\,\,$

$\because$ This median is making equal angles with coordinate axes, therefore,

$\frac{{\lambda  - 5}}{2}\,\, = 1 = \frac{{\mu  - 8}}{2}\,$

$ \Rightarrow \lambda  = 7\,\,\& \,\,\mu  = 10$

$\therefore {\lambda ^3} + {\mu ^3} = 5 = 1348$

$T =\left\{ Y :( YQ )^2-( YP )^2=50\right\}$

for finding $S \equiv X(x, y, z)$ and for $T \equiv Y(x, y, z)$

$\left(( x -1)^2+( y -1)^2+( z -1)^2\right)-\left(( x -4)^2+( y -2)^2+( z -7)^2\right)=50$

$\Rightarrow \quad S =\{( x , y , z ): 6 x +8 z =105\}$

$\quad T =\{( x , y , z ): 6 x +8 z =5\}$

Since $S$ and $T$ both are plane ;

$(A)$ There exist a triangle in plane $S$ whose area $=1$ (always)

$(B)$ $L \& M$ lies on plane $T$, hence line segment joining $L \& M$ will lie on plane $T$.

$(C)$ Distance between $S \& T$

$d=\left|\frac{105-5}{10}\right|=10$

Hence for rectangle of perimeter $48$ can exist.

$ Q \equiv(\cos (\beta-\alpha), \sin \beta) \equiv\left(x_2, y_2\right) $

$ \text { and } R \equiv\left(x_2 \cos \theta+x_1 \sin \theta, y_2 \cos \theta+y_1 \sin \theta\right) $

$ \text { We see that } T \equiv\left(\frac{x_2 \cos \theta+x_1 \sin \theta}{\cos \theta+\sin \theta}, \frac{y_2 \cos \theta+y_1 \sin \theta}{\cos \theta+\sin \theta}\right)$

We see that $\mathrm{T} \equiv\left(\frac{\mathrm{x}_2 \cos \theta+\mathrm{x}_1 \sin \theta}{\cos \theta+\sin \theta}, \frac{\mathrm{y}_2 \cos \theta+\mathrm{y}_1 \sin \theta}{\cos \theta+\sin \theta}\right)$

and $\mathrm{P}, \mathrm{Q}, \mathrm{T}$ are collinear $\Rightarrow P, Q, R$ are non-collinear.

$ \frac{\frac{1}{\mathrm{PS}}+\frac{1}{\mathrm{ST}}}{2}>\sqrt{\frac{1}{\mathrm{PS}} \times \frac{1}{\mathrm{ST}}} $

$ \Rightarrow \frac{1}{\mathrm{PS}}+\frac{1}{\mathrm{ST}}>\frac{2}{\sqrt{\mathrm{QS} \times \mathrm{SR}}} $

$ \frac{\mathrm{QS}+\mathrm{SR}}{2}>\sqrt{\mathrm{QS} \times \mathrm{SR}} $

$ \frac{\mathrm{QR}}{2}>\sqrt{\mathrm{QS} \times \mathrm{SR}} \Rightarrow \frac{1}{\sqrt{\mathrm{QS} \times \mathrm{SR}}}>\frac{2}{\mathrm{QR}} $

$ \Rightarrow \frac{1}{\mathrm{PS}}+\frac{1}{\mathrm{ST}}>\frac{4}{\mathrm{QR}} .$

Equation of $BD$, $3x + 4y - 12 = 0;AE\, \bot \,BC$.....$(i)$

Slope of $AE = \frac{1}{4}$

Equation of $AE$, $x - 4y = 0$.....$(ii)$

From equation $(i)$ and $(ii)$, $x = 3,y = \frac{3}{4}$

orthocentre of the triangle is $\left( {3,\frac{3}{4}} \right)$

So, the incentre is the same as the centroid.

$\therefore $ Incentre $ = \,\left( {\frac{{1 + 0 + 2}}{3},\,\,\frac{{\sqrt 3 + 0 + 0}}{3}} \right) = \left( {1,\,\frac{1}{{\sqrt 3 }}} \right)$.

Hence $\frac{{a + 4}}{2} = \frac{{1 + 5}}{2}\,\, \Rightarrow \,\,a = 2$

$\frac{{b + 6}}{2} = \frac{{2 + 7}}{2}\,\, \Rightarrow \,\,b = 3$.

So, the coordinates of the centroid $\left( {\frac{{\Sigma {x_1}}}{3},\,\,\frac{{\Sigma {y_1}}}{3}} \right)$ will be rational. As $AB = c = \sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2},} \,\,c$ may or may not be rational and it may be an irrational number of the form $\sqrt p .$ Hence, the coordinates of the incentre $\left( {\frac{{\Sigma a{x_1}}}{{\Sigma a}},\,\,\frac{{\Sigma a{y_1}}}{{\Sigma a}}} \right)$ may or may not be rational.

If $(\alpha ,\,\,\beta )$ be the circumcentre or orthocentre, $\alpha$ and $\beta$ are found by solving two linear equations in $\alpha ,\,\,\beta $ with rational coefficients. So $\alpha ,\,\,\beta $must be rational numbers.

Therefore, the new coordinate of $(4, 5)$ with respect to new origin $(1, -2)$ are $(3, 7).$

${(x + 1)^2} + {y^2} + {(x - 2)^2} + {y^2} = 2\,[{(x - 1)^2} + {y^2}]$

$ \Rightarrow \,\,2x + 1 + 4 - 4x = - 4x + 2\,\, \Rightarrow \,\,x = - \frac{3}{2}$

Hence it is a straight line parallel to $y$-axis.

$AP\,\,:\,\,PB = 1\,\,:\,\,2,$ then $h = \frac{{1 \times 0 + 2 \times a}}{{1 + 2}} = \frac{{2a}}{3}$

or $a = \frac{{3h}}{2},$ similarly $b = 3k$

Now we have $O{A^2} + O{B^2} = A{B^2}$

$ \Rightarrow \,\,{\left( {\frac{{3h}}{2}} \right)^2} + {(3k)^2} = {l^2}$

Hence locus of $P(h,\,\,k)$ is given by $9{x^2} + 36{y^2} = 4{l^2}$.

for $(5, 0)$,$3 \times 5 + 0 > 0$ and

$( - 1,\,\,2)$ for $( - 1,\,\,2),\,\, - 3 + 4 > 0.$

Similarly other inequalities hold good.

Also perpendicular passes through vertices are $x + 4y = 13$ and $7x - 7y = - 19$.

Solving these lines, we get the orthocentre $\left( {\frac{3}{7},\frac{{22}}{7}} \right)$ i.e. in first quadrant.

Therefore, perpendicular from point third ${t_2}x + y = a[{t_1}{t_2}{t_3} + {t_1} + {t_3}]$ and

perpendicular from point first ${t_3}x + y = a[{t_1}{t_2}{t_3} + {t_1} + {t_2}]$

.$x = - a,y = a[{t_1}{t_2}{t_3} + {t_1} + {t_2} + {t_3}]$.

$ = \left| {\frac{{7x + 7y - 14}}{{49}}} \right| = \left| {\frac{{x + y - 2}}{7}} \right|$.

Hence the points are collinear.

Hence, area $ = \frac{1}{2}\,\left| {\,\begin{array}{*{20}{c}}{ - 2}&{ - 2}&1\\2&{ - 2}&1\\0&5&1\end{array}\,} \right| = 14\,sq.\,units.$

The area of the triangle $ = \frac{1}{2}bc\,\,\sin A$ $ = \frac{1}{2}.\,a\,.\,a\,\sin {60^o} = \frac{{\sqrt 3 }}{4}{a^2}$every angle is of ${60^o}\} $
Which is irrationals, because ${a^2}$ is positive integer.

But earlier we have calculated that the area of the triangle is rational number. Hence it is a contradiction. Therefore, if the vertices of a triangle are integers, the triangle cannot be equilateral.
Note : Students should remember this question as a fact.

but it passes through $(0,0)$ hence $k = 0$

thus equation of $AD = x - 4y = 0$.....$(i)$

Similarly the equation of $AC = 3x - y = 0$ and $BE$ is

$x + 3y + 1 = 0$.....$(ii)$

On solving $(i)$ and $(ii)$, the required orthocentre is $\left( {\frac{{ - 4}}{7},\frac{{ - 1}}{7}} \right)$.

Therefore check with the options. Obviously, point $\left( {\frac{{13}}{4},\,\,\frac{9}{4}} \right)$ is equidistant from the points $(2,\,\,1),\,\,(5,\,\,2)$ and $(3, 4)$.

Area of $\Delta POA = 2\,.\,$Area of $\Delta POB$

$ \Rightarrow \,\,\frac{1}{2} \times 4 \times x = \pm \,2 \times \frac{1}{2} \times 6 \times y$ or $x = \pm \,3y$

Hence the equation to both parts of the locus of $P$ is $(x - 3y)\,(x + 3y) = 0$.

(Applying ${C_2} \to {C_1} + {C_2})$

$ = \frac{{(a + b + c)}}{2}\,\left| {\,\begin{array}{*{20}{c}}a&1&1\\b&1&1\\c&1&1\end{array}\,} \right| = 0$.

$ \Rightarrow \,\,2(14x - 7) = \frac{{49}}{2}\,\, \Rightarrow \,\,x = \frac{{77}}{{56}} = \frac{{11}}{8}$.

$ \Rightarrow \,\, - 2x + 1 = - 6x + 9\,\, \Rightarrow \,\,x = 2$.

Hence the point is $(2, 2)$.

$ \Rightarrow \,\,{x^2} + {(a + b)^2} - 2x\,(a + b) + {y^2} + {(b - a)^2} - 2y(b - a)$

$ = {x^2} + {(a - b)^2} - 2x(a - b) + {y^2} + {(a + b)^2} - 2y(a + b)$

On simplification, we get $bx - ay = 0$

Trick : The locus will be right bisector of the line joining the given points, therefore the line must pass through the mid-points of the given point i.e. $(a, b)$. Obviously, the line given in option $(d)$ passes through $(a, b)$.

$ = a\sqrt {{{\sin }^2}\alpha + {{\cos }^2}\alpha + {{\cos }^2}\beta + {{\sin }^2}\beta - 2\cos \alpha \cos \beta - 2\sin \alpha \sin \beta } $

$ = a\sqrt {2\left\{ {1 - \cos \,(\alpha - \beta )} \right\}} $

$= 2a\,\sin \,\left( {\frac{{\alpha - \beta }}{2}} \right)$

Trick : Put $a = 1,\,\,\alpha = \frac{\pi }{2},\,\beta = \frac{\pi }{6},$ then the points will be $(0, 1)$ and $\left( {\frac{{\sqrt 3 }}{2},\,\,\frac{1}{2}} \right)$.

Obviously, the distance between these two points is 1 which is given by $(d)$.

$\left\{ {\,\,2a\,\sin \frac{{\alpha  - \beta }}{2} = 2 \times 1 \times \sin \frac{{(\pi /2) - (\pi /6)}}{2} = 2 \times \frac{1}{2} = 1} \right\}$

Trick : Check with option. Obviously, the point $(2, 2)$ is equidistant from the points $(2, 0)$ and $(0, 2)$.

$ \Rightarrow \,\,9 + {b^2} + 4 - 4b = 1 + {b^2} + 9 - 6b\,\, \Rightarrow \,\,b = - \frac{3}{2}$
Hence the point is $\left( {0,\, - \frac{3}{2}} \right)$.