In `Delta ABC`, a line through a cuts the side BC at D such that BD : CD = 4: 5. If the area of `Delta ABD= 60 cm^(2)`, then the area of `DeltaADC` is

Options

A. `90 cm^(2)`
B. `50 cm^(2)`
C. `60 cm^(2)`
D. `75 cm^(2)`

Answer

To solve the problem step by step, we will use the information provided in the question regarding the triangle ABC and the ratio of segments BD and CD.

### Step-by-Step Solution:

1. **Understanding the Ratio**:
We are given that the ratio of BD to CD is 4:5. This means we can express BD and CD in terms of a common variable \( x \):
- Let \( BD = 4x \)
- Let \( CD = 5x \)

2. **Finding the Total Length of BC**:
The total length of side BC can be calculated as:
\[ BC = BD + CD = 4x + 5x = 9x \]

3. **Area of Triangle ABD**:
We are given that the area of triangle ABD is \( 60 \, \text{cm}^2 \).

4. **Using the Area Formula**:
The area of triangle ABD can be expressed using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} \]
Here, the base is \( BD = 4x \) and the height from A to line BD is the same for both triangles ABD and ADC. Let’s denote this height as \( h \). Therefore, we have:
\[ 60 = \frac{1}{2} \times 4x \times h \]
Simplifying this gives:
\[ 60 = 2xh \quad \Rightarrow \quad xh = 30 \quad \Rightarrow \quad h = \frac{30}{x} \]

5. **Finding the Area of Triangle ADC**:
Now, we will find the area of triangle ADC. The base of triangle ADC is \( CD = 5x \) and the height is still \( h \). Thus, the area of triangle ADC can be calculated as:
\[ \text{Area of } \triangle ADC = \frac{1}{2} \times CD \times h = \frac{1}{2} \times 5x \times h \]
Substituting \( h \) from the previous calculation:
\[ \text{Area of } \triangle ADC = \frac{1}{2} \times 5x \times \frac{30}{x} \]
Simplifying this gives:
\[ \text{Area of } \triangle ADC = \frac{1}{2} \times 5 \times 30 = \frac{150}{2} = 75 \, \text{cm}^2 \]

### Final Answer:
The area of triangle ADC is \( 75 \, \text{cm}^2 \).

---