If the extremities of the base of an isosceles triangle are the points (2a, 0) and (0, a) and equation of one of the sides is $x=2a,$ then the area of triangle is

1. $5a^2\text{sq. units}$
2. $\frac{5a^2}{2}\text{sq. units}$
3. None of these
4. $\frac{25a^2}{2}\text{sq. units}$

The correct option is B $\frac{5a^2}{2}\text{sq. units}$
Given: extremities of the base of an isosceles triangle are the points (2a, 0)
and (0, a)
Equation of one of the sides is $x=2a$
Let $A(2a,0),B(0,a)$
$\Rightarrow$Coordinates of the third side is $C(2a,t)$
As ABC is isosceles triangle
$AC=BC$
$\Rightarrow \sqrt{(2a-2a{)}^2+(t-0{)}^2}=\sqrt{(2a-0{)}^2+(t-a{)}^2}$
$\Rightarrow t^2=4a^2+t^2+a^2-2at$
$t=\frac{5a}{2}$
$\therefore \text{Coordinates of third vertex is}(2a,\frac{5a}{2})$
$\text{The area of}\Delta ABC$ is
$\pm \frac{1}{2}\left|\begin{array}{ccc}2a& \frac{5a}{2}& 1\\ 2a& 0& 1\\ 0& a& 1\end{array}\right|$
$=\pm \frac{1}{2}|-2a^2-5a^2+2a^2|=\frac{5a^2}{2}\text{sq. units}$