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}
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