`DeltaABC` and `DeltaDBC` are on the same base BC but on opposite sides of it. AD and BC intersect each other at O. If AO=a cm, DO=b cm and the area of `DeltaABC=x cm^(2)`, then what is the area (in `cm^(2)`) of `DeltaDBC`?

Options

A. `(bx)/(a)`
B. `(a)/(b)x`
C. `(ab)/(2)x`
D. `((a+b))/(2)x`

Answer

To find the area of triangle DBC given the area of triangle ABC, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We have two triangles, ABC and DBC, sharing the same base BC but positioned on opposite sides of it. The point O is where the lines AD and BC intersect.

2. **Given Information**:
- AO = a cm
- DO = b cm
- Area of triangle ABC = x cm²

3. **Use the Area Ratio Theorem**: The areas of triangles that share the same base and are between the same parallels are proportional to their heights. Here, the heights of triangles ABC and DBC from point A and D to line BC are AO and DO, respectively.

4. **Set Up the Ratio**:
\[ \frac{\text{Area of } \Delta ABC}{\text{Area of } \Delta DBC} = \frac{AO}{DO} \]
Substituting the known values:
\[ \frac{x}{\text{Area of } \Delta DBC} = \frac{a}{b} \]

5. **Cross-Multiply to Solve for Area of DBC**:
\[ x \cdot b = \text{Area of } \Delta DBC \cdot a \]

6. **Isolate the Area of DBC**:
\[ \text{Area of } \Delta DBC = \frac{b \cdot x}{a} \]

### Final Answer:
The area of triangle DBC is \(\frac{bx}{a}\) cm².