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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,

In triangle ABC, given that AB is perpendicular to AD, AB = 6 cm, BD = 10 cm, and DC = 5 cm, what is the area of triangle ADC?

Learn how to calculate the area of triangle ADC in a right triangle setup where AB is perpendicular to AD. Given AB = 6 cm, BD = 10 cm, and DC = 5 cm, the problem explores triangle similarity and proportional areas. Suitable for students in Grades 9-12.

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