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, in square units, is :

We are given an isosceles triangle with the base endpoints at (2a, 0) and (0, a). One side is given by the equation x = 2a. Since the point (2a, 0) lies on this line, and the other endpoint (0, a) does not, the side x = 2a must be one of the legs, not the base. Therefore, the third vertex, let's call it C, must lie on the line x = 2a. So, the coordinates of C are (2a, k) for some k.

In an isosceles triangle with base AB (where A is (2a, 0) and B is (0, a)), the two legs AC and BC must be equal. So, we set AC = BC.

First, calculate AC. The distance between A(2a, 0) and C(2a, k) is the absolute difference in y-coordinates since x-coordinates are the same: AC = |k - 0| = |k|.

Next, calculate BC. The distance between B(0, a) and C(2a, k) is found using the distance formula: BC = √[(2a - 0)² + (k - a)²] = √[4a² + (k - a)²].

Set AC equal to BC:

$$ |k| = \sqrt{4a^2 + (k - a)^2} $$

Since distances are positive, we can square both sides to eliminate the square root:

$$ k^2 = 4a^2 + (k - a)^2 $$

Expand the right side:

$$ k^2 = 4a^2 + k^2 - 2ak + a^2 $$

Simplify:

$$ k^2 = k^2 + 5a^2 - 2ak $$

Subtract k² from both sides:

$$ 0 = 5a^2 - 2ak $$

Rearrange:

$$ 2ak = 5a^2 $$

Assuming a ≠ 0 (as a = 0 would make the triangle degenerate), divide both sides by a:

$$ 2k = 5a $$

Solve for k:

$$ k = \frac{5a}{2} $$

So, the coordinates of C are (2a, 5a/2).

Now, we have the three vertices: A(2a, 0), B(0, a), and C(2a, 5a/2). To find the area of the triangle, use the formula for the area given three points (x₁, y₁), (x₂, y₂), (x₃, y₃):

$$ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| $$

Assign A(2a, 0) as (x₁, y₁), B(0, a) as (x₂, y₂), and C(2a, 5a/2) as (x₃, y₃). Substitute the values:

$$ \text{Area} = \frac{1}{2} \left| (2a)(a - \frac{5a}{2}) + (0)(\frac{5a}{2} - 0) + (2a)(0 - a) \right| $$

Simplify each term inside the absolute value:

First term: (2a)(a - 5a/2) = (2a)(-3a/2) = -3a²

Second term: (0)(5a/2) = 0

Third term: (2a)(-a) = -2a²

Sum the terms:

$$ -3a^2 + 0 - 2a^2 = -5a^2 $$

Take the absolute value:

$$ |-5a^2| = 5a^2 $$

Now multiply by 1/2:

$$ \text{Area} = \frac{1}{2} \times 5a^2 = \frac{5a^2}{2} $$

So, the area of the triangle is (5/2)a² square units.

Comparing with the options:

A. (5/4)a²

B. (5/2)a²

C. (25/4)a²

D. 5a²

The area matches option B. Hence, the correct answer is Option B.