If the extremities of the base of an isosceles triangle are the points (2a,0) and (0,a) and the equation of one of its side is x=2a, then area of the triangle in sq. units is

Options

A. `5a^(2)`
B. `5/2a^(2)`
C. `25/2a^(2)`
D. None

Answer

To find the area of the isosceles triangle with given points and conditions, we can follow these steps:

### Step 1: Identify the Points
The extremities of the base of the isosceles triangle are given as:
- Point A (2a, 0)
- Point C (0, a)

The equation of one of its sides is given as \( x = 2a \). This means point B lies on this vertical line.

### Step 2: Determine the Coordinates of Point B
Since point B lies on the line \( x = 2a \), we can denote its coordinates as:
- Point B (2a, h), where h is the y-coordinate we need to determine.

### Step 3: Use the Distance Formula
Since triangle ABC is isosceles, we have:
- \( AB = BC \)

Using the distance formula, we can express the lengths of AB and BC:
- \( AB = \sqrt{(2a - 2a)^2 + (h - 0)^2} = \sqrt{h^2} = h \)
- \( BC = \sqrt{(2a - 0)^2 + (h - a)^2} = \sqrt{(2a)^2 + (h - a)^2} = \sqrt{4a^2 + (h - a)^2} \)

### Step 4: Set the Distances Equal
Since \( AB = BC \), we can set the equations equal to each other:
\[ h = \sqrt{4a^2 + (h - a)^2} \]

### Step 5: Square Both Sides
Squaring both sides to eliminate the square root gives:
\[ h^2 = 4a^2 + (h - a)^2 \]

### Step 6: Expand and Rearrange
Expanding the right side:
\[ h^2 = 4a^2 + (h^2 - 2ah + a^2) \]
This simplifies to:
\[ h^2 = 4a^2 + h^2 - 2ah + a^2 \]
Subtracting \( h^2 \) from both sides:
\[ 0 = 5a^2 - 2ah \]

### Step 7: Solve for h
Rearranging gives:
\[ 2ah = 5a^2 \]
Dividing both sides by \( 2a \) (assuming \( a \neq 0 \)):
\[ h = \frac{5a}{2} \]

### Step 8: Calculate the Area of Triangle ABC
The area \( A \) of triangle ABC can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base \( AC \) can be calculated as:
\[ AC = \sqrt{(2a - 0)^2 + (0 - a)^2} = \sqrt{4a^2 + a^2} = \sqrt{5a^2} = a\sqrt{5} \]
Thus, the area becomes:
\[ A = \frac{1}{2} \times a\sqrt{5} \times \frac{5a}{2} = \frac{5a^2\sqrt{5}}{4} \]

### Final Result
The area of the triangle in square units is:
\[ \frac{5a^2\sqrt{5}}{4} \]

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