Isosceles right triangle and cosines law

Isosceles triangle ABC has a right angle at C . Point P is inside triangle ABC such that
PA = 11 , PB = 7 and PC = 6 . Find the area of triangle ABC .

Solution

$ \displaystyle \text{By}\ \text{the}\ \text{law}\ \text{of}\ \text{cosines}\ ,$ 

$ \displaystyle \cos \theta =\frac{{B{{C}^{2}}+P{{C}^{2}}-P{{B}^{2}}}}{{2BC.PC}}$

$ \displaystyle \ \ \ \ \ \ \ \ =\frac{{{{x}^{2}}+{{6}^{2}}-{{7}^{2}}}}{{2x\ .\ 6}}$ 

$ \displaystyle \ \ \ \ \ \ \ \ =\frac{{{{x}^{2}}-13}}{{12x}}$ 

$ \displaystyle \cos ({{90}^{\circ }}-\theta )=\frac{{A{{C}^{2}}+P{{C}^{2}}-P{{A}^{2}}}}{{2AC.PC}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \sin \theta =\frac{{{{x}^{2}}+{{6}^{2}}-{{{11}}^{2}}}}{{2x\ .\ 6}}\ \ \ \left( {\because \cos ({{{90}}^{\circ }}-\theta )=\sin \theta } \right)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{{{x}^{2}}-85}}{{12x}}$

$ \displaystyle {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ 

$\displaystyle {{\left( {\frac{{{{x}^{2}}-85}}{{12x}}} \right)}^{2}}+{{\left( {\frac{{{{x}^{2}}-13}}{{12x}}} \right)}^{2}}=1$ 

$\displaystyle \frac{{{{x}^{4}}-170{{x}^{2}}+7225}}{{144{{x}^{2}}}}\ +\ \frac{{{{x}^{4}}-26{{x}^{2}}+169}}{{144{{x}^{2}}}}=1$

$\displaystyle 2{{x}^{4}}-196{{x}^{2}}+7394=144{{x}^{2}}$

$\displaystyle 2{{x}^{4}}-340{{x}^{2}}+7394=0$ 

$\displaystyle \ \ {{x}^{4}}-170{{x}^{2}}+3697=0$ 

$\displaystyle \ {{x}^{4}}-170{{x}^{2}}+{{85}^{2}}={{85}^{2}}-3697$

$\displaystyle {{\left( {{{x}^{2}}-85} \right)}^{2}}=3528$

$\displaystyle \ \ \ \ {{x}^{2}}-85=42\sqrt{2}\ \ \ \ \left[ {\because \sin \theta >0} \right]$ 

$\displaystyle \ \ \ \ \ \ \ \ \ \ {{x}^{2}}=85+42\sqrt{2}$ 

$\displaystyle \therefore \ \alpha \ (\ \Delta \ ABC\ )=\frac{1}{2}{{x}^{2}}=\frac{{85+42\sqrt{2}}}{2}\ \ \text{sq}\ \text{units}$