`Delta ABC` is right angled at A, AB=3 units ,AC=4 units and AD is perpendicular to BC. What is the area of the `Delta ADB`

Answer

To find the area of triangle ADB, we will follow these steps:

### Step 1: Find the length of side BC using the Pythagorean theorem.
Given that triangle ABC is right-angled at A, with sides AB = 3 units and AC = 4 units, we can use the Pythagorean theorem:

\[ BC^2 = AB^2 + AC^2 \]

Calculating the squares:

\[ BC^2 = 3^2 + 4^2 = 9 + 16 = 25 \]

Taking the square root:

\[ BC = \sqrt{25} = 5 \text{ units} \]

### Step 2: Calculate the area of triangle ABC.
The area of triangle ABC can be calculated using the formula:

\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]

Here, we can take AB as the base and AC as the height:

\[ \text{Area}_{ABC} = \frac{1}{2} \times AB \times AC = \frac{1}{2} \times 3 \times 4 = 6 \text{ square units} \]

### Step 3: Relate the areas of triangles ABC and ADB.
Since AD is perpendicular to BC, we can express the area of triangle ABC in terms of triangle ADB:

\[ \text{Area}_{ABC} = \text{Area}_{ADB} + \text{Area}_{ADC} \]

### Step 4: Use the formula for the area of triangle ABC to find AD.
We know that:

\[ \text{Area}_{ABC} = \frac{1}{2} \times BC \times AD \]

Substituting the known values:

\[ 6 = \frac{1}{2} \times 5 \times AD \]

Multiplying both sides by 2:

\[ 12 = 5 \times AD \]

Solving for AD:

\[ AD = \frac{12}{5} \text{ units} \]

### Step 5: Find the length of BD using the relation in triangle ADB.
Using the relation from triangle ABC:

\[ AB^2 = BD \times BC \]

Substituting the known values:

\[ 3^2 = BD \times 5 \]

Calculating:

\[ 9 = 5 \times BD \]

Solving for BD:

\[ BD = \frac{9}{5} \text{ units} \]

### Step 6: Calculate the area of triangle ADB.
Now we can find the area of triangle ADB using the base BD and height AD:

\[ \text{Area}_{ADB} = \frac{1}{2} \times BD \times AD \]

Substituting the values:

\[ \text{Area}_{ADB} = \frac{1}{2} \times \frac{9}{5} \times \frac{12}{5} \]

Calculating:

\[ = \frac{1}{2} \times \frac{108}{25} = \frac{54}{25} \text{ square units} \]

### Final Answer:
The area of triangle ADB is \(\frac{54}{25}\) square units.

---