Find the ratio of AD and CD when AB = 3 cm BC = 4 cm `angle B = 90^(@)`. Also find area of `Delta ADB`

Answer

To solve the problem, we need to find the ratio of AD and CD in triangle ABC, where AB = 3 cm, BC = 4 cm, and angle B = 90 degrees. We also need to find the area of triangle ADB.

### Step 1: Identify the triangle and its sides
We have a right triangle ABC with:
- AB = 3 cm (one leg)
- BC = 4 cm (the other leg)
- Angle B = 90 degrees

### Step 2: Calculate the hypotenuse AC using the Pythagorean theorem
Using the Pythagorean theorem:
\[ AC = \sqrt{AB^2 + BC^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ cm} \]

### Step 3: Draw the perpendicular line BD
We draw a perpendicular line from B to AC, meeting at point D. This creates two segments AD and CD.

### Step 4: Calculate lengths AD and CD
Using the formulas for the segments in a right triangle:
\[ AD = \frac{AB^2}{AC} = \frac{3^2}{5} = \frac{9}{5} \text{ cm} \]
\[ CD = \frac{BC^2}{AC} = \frac{4^2}{5} = \frac{16}{5} \text{ cm} \]

### Step 5: Find the ratio of AD to CD
Now, we can find the ratio of AD to CD:
\[ \text{Ratio of } AD \text{ to } CD = \frac{AD}{CD} = \frac{\frac{9}{5}}{\frac{16}{5}} = \frac{9}{16} \]

### Step 6: Calculate the length of BD
Using the formula for BD:
\[ BD = \frac{AB \times BC}{AC} = \frac{3 \times 4}{5} = \frac{12}{5} \text{ cm} \]

### Step 7: Calculate the area of triangle ADB
The area of triangle ADB can be calculated using the formula for the area of a triangle:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base is BD and the height is AD:
\[ \text{Area of } \Delta ADB = \frac{1}{2} \times BD \times AD = \frac{1}{2} \times \frac{12}{5} \times \frac{9}{5} = \frac{1}{2} \times \frac{108}{25} = \frac{54}{25} \text{ cm}^2 \]

### Final Answers
- The ratio of AD to CD is \( \frac{9}{16} \).
- The area of triangle ADB is \( \frac{54}{25} \text{ cm}^2 \).