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 side is `x=2a ,` then the area of the triangle is `5a^2s qdotu n i t s` (b) `(5a^2)/2s qdotu n i t s` `(25 a^2)/2s qdotu n i t s` (d) none of these

Options

A. `5a^(2) "sq. units"`
B. `5a^(2)//2 "sq. units"`
C. `25a^(2)//2 "sq. units"`
D. none of these

Answer

B

Explanation

Given vertices are A(2a,0) and B(0,a)
Let the coordinates of the third vertex C be (2a,t). Now, AC = BC. Hence,
`t = sqrt(4a^(2) + (a-t)^(2)) " or " t =(5a)/(2)`
So, the coordinates of the third vertex C are (2a,5a/2).
Therefore, area of the triangle is
`(1)/(2)||{:(2a, 5a//2, 1),(2a, 0, 1),(0, a, 1):}|| = (5a^(2))/(2)` sq. units