ABC is a triangle with base AB. D is a point on AB such that AB = 5 and DB = 3. What is the ratio of the area of `Delta ADC` to the area of `Delta ABC` ?

Options

A. `3//2`
B. `2//3`
C. `3//5`
D. `2//5`

Answer

To find the ratio of the area of triangle ADC to the area of triangle ABC, we can follow these steps:

### Step 1: Identify the lengths of the segments
Given:
- The length of AB = 5
- The length of DB = 3

From this, we can find the length of AD:
- Since D is a point on AB, we can calculate AD as:
\[ AD = AB - DB = 5 - 3 = 2 \]

### Step 2: Understand the area of triangles
The area of a triangle can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
In our case, both triangles ADC and ABC share the same height (let's call it H), since they are both drawn from point C perpendicular to line AB.

### Step 3: Calculate the areas of triangles ADC and ABC
- **Area of triangle ABC**:
\[ \text{Area}_{ABC} = \frac{1}{2} \times AB \times H = \frac{1}{2} \times 5 \times H \]

- **Area of triangle ADC**:
\[ \text{Area}_{ADC} = \frac{1}{2} \times AD \times H = \frac{1}{2} \times 2 \times H \]

### Step 4: Set up the ratio of the areas
Now, we can find the ratio of the area of triangle ADC to the area of triangle ABC:
\[ \text{Ratio} = \frac{\text{Area}_{ADC}}{\text{Area}_{ABC}} = \frac{\frac{1}{2} \times 2 \times H}{\frac{1}{2} \times 5 \times H} \]

### Step 5: Simplify the ratio
The \(\frac{1}{2}\) and \(H\) cancel out:
\[ \text{Ratio} = \frac{2}{5} \]

### Conclusion
Thus, the ratio of the area of triangle ADC to the area of triangle ABC is:
\[ \frac{2}{5} \]