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If the extremities of the base of an isosceles triangle are the points $(2a,0)$ and $(0,a)$ and the equation of one of the sides is $x = 2a$, then the area of the triangle is
$5{a^2}sq$. units
$\frac{5}{2}{a^2}sq.$units
$\frac{{25{a^2}}}{2}sq.$units
None of these
Solution
(b) Let the co-ordinates of the third vertex be $(2a,\,\,t)$.
$AC = BC \Rightarrow t = \sqrt {4{a^2} + {{(a – t)}^2}} \Rightarrow $$t = \frac{{5a}}{2}$
So the coordinates of third vertex $C$ are $\left( {2a,\frac{{5a}}{2}} \right)$
Therefore area of the triangle
$ = \pm \frac{1}{2}\left| {\,\begin{array}{*{20}{c}}{2a}&{\frac{{5a}}{2}}&1\\{2a}&0&1\\0&a&1\end{array}\,} \right| = \left| {\,\begin{array}{*{20}{c}}a&{\frac{{5a}}{2}}&1\\0&{ – \frac{{5a}}{2}}&0\\0&a&1\end{array}\,} \right| = \frac{{5{a^2}}}{2}sq.$units.