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

Given vertices are A2a,0 and B0,aLet the coordinates of the third vertex C are 2a, t. Now, AC=BC. Hence,t=4a2+a−t2 or t=5a2So, the coordinates of the third vertex C are 2a,  5a/2.Therefore, area of the triangle is 122a5a/212a010a1=5a22sq. units