Question

CD=3
DB=4
Given is the right triangle ABC with altitude CD. Apply the Pythagorean Theorem and similar triangles as needed to find the length of AD. Round the answer to two decimal places.

Answer 1

In the given right triangle \( ABC \) with \( C \) as the right angle, altitude \( CD \) is drawn from point \( C \) to side \( AB \), creating two smaller right triangles, \( ACD \) and \( BCD \).

Given:
- \( CD = 3 \)
- \( DB = 4 \)

Let:
- \( AD = x \) (the length we need to find)
- \( AB = AD + DB = x + 4 \)

Using the property of similar triangles, we can set up the relationships between the segments:

1. In triangle \( ACD \) and triangle \( BCD \):
- For triangle \( ACD \):
\[
AC^2 = AD \cdot AB = x(x + 4)
\]

- For triangle \( BCD \):
\[
BC^2 = DB \cdot AB = 4(x + 4)
\]

Since \( CD \) is the altitude from \( C \), we can also relate the area of triangles \( ACD \) and \( BCD \) with the lengths \( AD \), \( DB \), and \( CD \). Using the Pythagorean theorem in triangles \( ACD \) and \( BCD \), we have:

\[
CD^2 = AC^2 - AD^2 \quad \text{and} \quad CD^2 = BC^2 - DB^2
\]

Substituting \( CD = 3 \):

\[
9 = AC^2 - x^2 \quad (1)
\]
\[
9 = BC^2 - 16 \quad (2)
\]

From equation (2), we find \( BC^2 \):

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

Now substituting \( BC^2 \) back to \( 4(x + 4) \) gives us:

\[
25 = 4(x + 4)
\]

Expanding and solving for \( x \):

\[
25 = 4x + 16
\]
\[
25 - 16 = 4x
\]
\[
9 = 4x
\]
\[
x = \frac{9}{4} = 2.25
\]

Thus, rounding to two decimal places, the length of \( AD \) is:

\[
\boxed{2.25}
\]