The area of a parallelogram ABCD is equal to that right angled isosceles triangle whose hypotenuse is l cm. If O is a point inside the parallelogram ABCD then sum of areas of `DeltaAOBandDeltaCOD` is.

Options

A. `l^(2)/(2)"cm"^(2)`
B. `l^(2)/(4)"cm"^(2)`
C. `l^(2)/(8)"cm"^(2)`
D. `l^(2)/(16)"cm"^(2)`

Answer

To solve the problem, we need to find the sum of the areas of triangles AOB and COD inside the parallelogram ABCD, given that the area of the parallelogram is equal to the area of a right-angled isosceles triangle with a hypotenuse of \( l \) cm.

### Step-by-Step Solution:

1. **Understanding the Right-Angled Isosceles Triangle**:
- The hypotenuse of the triangle is given as \( l \) cm.
- In a right-angled isosceles triangle, if the legs are of length \( y \), then by the Pythagorean theorem:
\[ l^2 = y^2 + y^2 = 2y^2 \]
- Therefore, we can express \( y \) in terms of \( l \):
\[ y^2 = \frac{l^2}{2} \quad \Rightarrow \quad y = \frac{l}{\sqrt{2}} \]

2. **Calculating the Area of the Right-Angled Triangle**:
- The area \( A \) of the right-angled triangle is given by:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times y \times y = \frac{1}{2} \times \left(\frac{l}{\sqrt{2}}\right) \times \left(\frac{l}{\sqrt{2}}\right) \]
- Simplifying this gives:
\[ A = \frac{1}{2} \times \frac{l^2}{2} = \frac{l^2}{4} \]

3. **Area of the Parallelogram**:
- According to the problem, the area of the parallelogram ABCD is equal to the area of the right-angled triangle:
\[ \text{Area of parallelogram} = \frac{l^2}{4} \]
- The area of a parallelogram can also be expressed as:
\[ \text{Area} = \text{base} \times \text{height} = b \times h \]

4. **Finding the Areas of Triangles AOB and COD**:
- Since point O is inside the parallelogram, triangles AOB and COD will have equal areas because they share the same height from point O to line AD and line BC respectively.
- Let the height from point O to line AD be \( h_1 \) and to line BC be \( h_2 \). Since O divides the height of the parallelogram, we can say:
\[ h_1 + h_2 = h \]
- The area of triangle AOB is:
\[ \text{Area of } \triangle AOB = \frac{1}{2} \times b \times h_1 \]
- The area of triangle COD is:
\[ \text{Area of } \triangle COD = \frac{1}{2} \times b \times h_2 \]
- Thus, the sum of the areas of triangles AOB and COD is:
\[ \text{Area of } \triangle AOB + \text{Area of } \triangle COD = \frac{1}{2} \times b \times h_1 + \frac{1}{2} \times b \times h_2 = \frac{1}{2} \times b \times (h_1 + h_2) = \frac{1}{2} \times b \times h \]

5. **Substituting the Area of the Parallelogram**:
- Since we know that \( b \times h = \frac{l^2}{4} \):
\[ \text{Area of } \triangle AOB + \text{Area of } \triangle COD = \frac{1}{2} \times \frac{l^2}{4} = \frac{l^2}{8} \]

### Final Answer:
The sum of the areas of triangles AOB and COD is:
\[ \frac{l^2}{8} \text{ cm}^2 \]