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

Step-by-step solution

Let y -coordinate of C =b aligned & AB = 4 a^2+a^2 = 5 a & Now, AC = BC b= 4 a^2+(b-a)^2 & b^2=4 a^2+b^2+a^2-2 a b & 2 a b=5 a^2 b= 5 a 2 & C = (2 a, 5 a 2 ) aligned Hence area of the triangle aligned & = 1 2 | array ccc 2 a & 0 & 1 0 & a & 1 2 a & 5 a 2 & 1 array |= 1 2 | array ccc 2 a & 0 & 1 0 & a & 1 0 & 5 a 2 & 0 array | & = 1 2 2 a (- 5 a 2 )=- 5 a^2 2 aligned Since area is always + ve , hence area = 5 a^2 2 sq. unit