Question

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

• A. $5a^2$ sq. units
• B. $\frac{5a^2}{2}$ sq. units
• C. $\frac{25a^2}{2}$ sq. units
• D. None of these

Answer 1

The correct option is B $\frac{5a^2}{2}$ sq. units

Let coordinates of the given vertices be $A(2a,0),B(0,a)$ and third vertex be $(2a,t)$. Now, $AC=BC$. Hence,
$t=\sqrt{4a^2+(a-t{)}^2}\Rightarrow t=\frac{5a}{2}$
So the coordinates of third vertex $C$ are $(2a,\frac{5a}{2})$.
Therefore, area of the triangle is using determinant formula i.e,
$=\pm \frac{1}{2}\left|\begin{array}{ccc}2a& \frac{5a}{2}& 1\\ 2a& 0& 1\\ 0& a& 1\end{array}\right|=\left|\begin{array}{ccc}a& \frac{5a}{2}& 1\\ 0& -\frac{5a}{2}& 0\\ 0& a& 1\end{array}\right|=\frac{5a^2}{2}$ sq. units