The extremities of base of an isosceles triangle have coordinate (2a,0) and (0,a) if the equation of one of the equal sides be x=2a find the equation of other two sides and area of triangle.

Let the given isosceles triangle have its base extremities at \( A(2a, 0) \) and \( B(0, a) \), and one of its equal sides is given by the equation \( x = 2a \). This means that the third vertex \( C \) of the triangle lies on this line, so its x-coordinate is \( 2a \). Let the coordinates of \( C \) be \( (2a, y) \).

### Step 1: Find the Equation of Line \( AB \) (Base of the Triangle)
The equation of a line passing through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:

\[
y - y_1 = \frac{y_2 - y_1}{x_2 - x_1} (x - x_1)
\]

Substituting \( A(2a,0) \) and \( B(0,a) \):

\[
y - 0 = \frac{a - 0}{0 - 2a} (x - 2a)
\]

\[
y = -\frac{a}{2a} (x - 2a)
\]

\[
y = -\frac{1}{2} x + a
\]

Thus, the equation of base \( AB \) is:

\[
y = -\frac{1}{2} x + a
\]

### Step 2: Find the Third Vertex \( C(2a, y) \)
Since \( C \) lies on the line \( x = 2a \), we substitute \( x = 2a \) into the equation of base \( AB \):

\[
y = -\frac{1}{2} (2a) + a = -a + a = 0
\]

Thus, the third vertex is \( C(2a,0) \), which is the same as \( A(2a,0) \). This is incorrect because a triangle cannot have two coincident vertices. Therefore, the vertex \( C \) must be above the base.

To satisfy the isosceles condition, the y-coordinate of \( C \) must be symmetric to the midpoint of \( AB \). The midpoint of \( AB \) is:

\[
M = \left( \frac{2a + 0}{2}, \frac{0 + a}{2} \right) = (a, a/2)
\]

Since \( C \) is on \( x = 2a \), it should be symmetric about \( M \), meaning the y-coordinate of \( C \) must be:

\[
y = a + \left( a - \frac{a}{2} \right) = \frac{3a}{2}
\]

Thus, the correct coordinates of \( C \) are \( (2a, \frac{3a}{2}) \).

### Step 3: Find the Equation of Line \( BC \)
The line \( BC \) passes through \( B(0,a) \) and \( C(2a, \frac{3a}{2}) \). Its slope is:

\[
m = \frac{\frac{3a}{2} - a}{2a - 0} = \frac{\frac{a}{2}}{2a} = \frac{1}{4}
\]

Using the point-slope form:

\[
y - a = \frac{1}{4} (x - 0)
\]

\[
y = \frac{1}{4} x + a
\]

Thus, the equation of \( BC \) is:

\[
y = \frac{1}{4} x + a
\]

### Step 4: Find the Area of the Triangle
The formula for the area of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is:

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

Substituting \( A(2a, 0) \), \( B(0, a) \), and \( C(2a, \frac{3a}{2}) \):

\[
\text{Area} = \frac{1}{2} \left| 2a (a - \frac{3a}{2}) + 0 (\frac{3a}{2} - 0) + 2a(0 - a) \right|
\]

\[
= \frac{1}{2} \left| 2a (-\frac{a}{2}) + 0 + 2a (-a) \right|
\]

\[
= \frac{1}{2} \left| -a^2 - 2a^2 \right|
\]

\[
= \frac{1}{2} \left| -3a^2 \right|
\]

\[
= \frac{3a^2}{2}
\]

Thus, the area of the triangle is:

\[
\frac{3a^2}{2}
\]

### Final Answer
1. Equation of base \( AB \): \( y = -\frac{1}{2} x + a \)
2. Equation of side \( BC \): \( y = \frac{1}{4} x + a \)
3. Area of the triangle: \( \frac{3a^2}{2} \)