In parallelogram ABCD, E is a point in AB and DE meets diagonal AC at point F. If DF:FE= `5:3` and area of `DeltaADF` is `60cm^(2)`, find :
(i) area of `DeltaADE`
(ii) if AE:EB= `4:5`, find the area of `DeltaADB`
(iii) also, find area of parallelogram ABCD.

Answer

To solve the problem step by step, we will break it down into three parts as per the question.

### Given Information:
- In parallelogram ABCD, E is a point on AB.
- DE meets diagonal AC at point F.
- DF:FE = 5:3
- Area of triangle ADF = 60 cm²

### (i) Find the area of triangle ADE

1. **Understanding the relationship between triangles ADF and AFE:**
- Triangles ADF and AFE share the same vertex A and have the same base DE.
- Therefore, the ratio of their areas is equal to the ratio of their corresponding segments on line DE.

2. **Setting up the ratio:**
- Given DF:FE = 5:3, we can denote the area of triangle AFE as \( x \).
- Thus, we have:
\[ \frac{\text{Area of } \triangle ADF}{\text{Area of } \triangle AFE} = \frac{DF}{FE} = \frac{5}{3} \]
- Substituting the known area of triangle ADF:
\[ \frac{60}{x} = \frac{5}{3} \]

3. **Cross-multiplying to find x:**
\[ 60 \cdot 3 = 5 \cdot x \implies 180 = 5x \implies x = \frac{180}{5} = 36 \text{ cm}^2 \]

4. **Finding the area of triangle ADE:**
- Area of triangle ADE = Area of triangle ADF + Area of triangle AFE
\[ \text{Area of } \triangle ADE = 60 + 36 = 96 \text{ cm}^2 \]

### (ii) If AE:EB = 4:5, find the area of triangle ADB

1. **Understanding the relationship between triangles ADE and EDB:**
- Triangles ADE and EDB share the same vertex D and lie on the same line AD.
- Therefore, the ratio of their areas is equal to the ratio of AE to EB.

2. **Setting up the ratio:**
- Let the area of triangle EDB be \( y \).
- Given AE:EB = 4:5, we have:
\[ \frac{\text{Area of } \triangle ADE}{\text{Area of } \triangle EDB} = \frac{AE}{EB} = \frac{4}{5} \]
- Substituting the known area of triangle ADE:
\[ \frac{96}{y} = \frac{4}{5} \]

3. **Cross-multiplying to find y:**
\[ 96 \cdot 5 = 4 \cdot y \implies 480 = 4y \implies y = \frac{480}{4} = 120 \text{ cm}^2 \]

4. **Finding the area of triangle ADB:**
- Area of triangle ADB = Area of triangle ADE + Area of triangle EDB
\[ \text{Area of } \triangle ADB = 96 + 120 = 216 \text{ cm}^2 \]

### (iii) Find the area of parallelogram ABCD

1. **Using the relationship between triangle ADB and parallelogram ABCD:**
- The area of triangle ADB is half of the area of parallelogram ABCD.
\[ \text{Area of } \triangle ADB = \frac{1}{2} \times \text{Area of parallelogram ABCD} \]
- Substituting the area of triangle ADB:
\[ 216 = \frac{1}{2} \times \text{Area of parallelogram ABCD} \]

2. **Solving for the area of parallelogram ABCD:**
\[ \text{Area of parallelogram ABCD} = 216 \times 2 = 432 \text{ cm}^2 \]

### Final Answers:
(i) Area of triangle ADE = 96 cm²
(ii) Area of triangle ADB = 216 cm²
(iii) Area of parallelogram ABCD = 432 cm²