ABCD is a rectangle with O as any point in its interior. If `ar(DeltaAOD)=3" cm"^(2)` and `ar(DeltaBOC)=6" cm"^(2)`, then area of rectangle ABCD is

Options

A. `9" cm"^(2)`
B. `12" cm"^(2)`
C. `15" cm"^(2)`
D. `18" cm"^(2)`

Answer

To find the area of rectangle ABCD given the areas of triangles AOD and BOC, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Areas**:
- Area of triangle AOD = 3 cm²
- Area of triangle BOC = 6 cm²

2. **Use the Property of Triangles and Parallelograms**:
- According to the theorem, the area of a parallelogram formed by extending the sides of the triangles is equal to twice the area of the triangle that shares the same base and height.
- For triangle AOD, we can form a parallelogram ALMD (where L and M are points on AD and BC respectively).
- Thus, the area of parallelogram ALMD = 2 × area of triangle AOD = 2 × 3 cm² = 6 cm².

3. **Calculate the Area of the Second Parallelogram**:
- Similarly, for triangle BOC, we can form another parallelogram LMCB.
- The area of parallelogram LMCB = 2 × area of triangle BOC = 2 × 6 cm² = 12 cm².

4. **Combine the Areas**:
- The total area of rectangle ABCD is the sum of the areas of the two parallelograms:
\[ \text{Area of rectangle ABCD} = \text{Area of parallelogram ALMD} + \text{Area of parallelogram LMCB} \]
\[ = 6 \text{ cm}² + 12 \text{ cm}² = 18 \text{ cm}² \]

5. **Final Answer**:
- Therefore, the area of rectangle ABCD is 18 cm².