Find the area of that triagle whose vertices are `(b+c,a),(b-c,a)and(a,-a).`

Answer

To find the area of the triangle with vertices at the points \((b+c, a)\), \((b-c, a)\), and \((a, -a)\), we can use the formula for the area of a triangle given its vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):

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

### Step 1: Assign the coordinates
Let:
- \( (x_1, y_1) = (b+c, a) \)
- \( (x_2, y_2) = (b-c, a) \)
- \( (x_3, y_3) = (a, -a) \)

### Step 2: Substitute the coordinates into the area formula
Substituting the coordinates into the area formula, we have:

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

### Step 3: Simplify the expression
Calculating each term:
- The first term:
\[ (b+c)(a + a) = (b+c)(2a) = 2a(b+c) \]
- The second term:
\[ (b-c)(-2a) = -2a(b-c) \]
- The third term:
\[ a(0) = 0 \]

Now substituting back into the area formula:
\[ \text{Area} = \frac{1}{2} \left| 2a(b+c) - 2a(b-c) + 0 \right| \]

### Step 4: Combine like terms
Now we can combine the terms:
\[ \text{Area} = \frac{1}{2} \left| 2a(b+c - (b-c)) \right| \]
\[ = \frac{1}{2} \left| 2a(b+c-b+c) \right| \]
\[ = \frac{1}{2} \left| 2a(2c) \right| \]
\[ = \frac{1}{2} \cdot 4ac = 2ac \]

### Final Answer
Thus, the area of the triangle is:
\[ \text{Area} = 2ac \]