In this problem, we have triangle $ABC$, where $AB \perp AD$. The given lengths are:

• $AB = 6 \, \text{cm}$
• $BD = 10 \, \text{cm}$
• $DC = 5 \, \text{cm}$

We are asked to find the area of triangle $ADC$.

Step 1: Find the total length of $BC$

Since $BC$ is made up of segments $BD$ and $DC$, the total length of $BC$ is:

$BC = BD + DC = 10 \, \text{cm} + 5 \, \text{cm} = 15 \, \text{cm}$

Step 2: Use area formula for right triangle

Since $AB \perp AD$, triangle $ABD$ is a right triangle. The area of triangle $ABD$ can be calculated using the area formula for a right triangle:

$\text{Area of } \triangle ABD = \frac{1}{2} \times AB \times BD = \frac{1}{2} \times 6 \, \text{cm} \times 10 \, \text{cm} = 30 \, \text{cm}^2$

Step 3: Use the similarity of triangles

Triangles $ABD$ and $ADC$ are similar because both share angle $A$, and they have a right angle. Since $D$ lies on line $BC$, the two triangles are similar by AA similarity.

The ratio of corresponding sides of the similar triangles $ABD$ and $ADC$ is:

$\frac{AD}{AB} = \frac{DC}{BD}$

Thus, the area ratio between triangles $ABD$ and $ADC$ is the square of the ratio of their corresponding sides:

$\frac{\text{Area of } \triangle ADC}{\text{Area of } \triangle ABD} = \left( \frac{DC}{BD} \right)^2 = \left( \frac{5}{10} \right)^2 = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$

Step 4: Calculate the area of triangle $ADC$

Since the area of $\triangle ABD$ is $30 \, \text{cm}^2$, the area of $\triangle ADC$ is:

$\text{Area of } \triangle ADC = \frac{1}{4} \times 30 \, \text{cm}^2 = 7.5 \, \text{cm}^2$

Step 5: Apply the full triangle $ABC$

By combining the final calculation above,