Solving Right Triangle Problems with Altitude

This problem involves a right-angled triangle and an altitude drawn to the hypotenuse. Understanding the properties of similar triangles formed by the altitude is key to solving this.

Problem Description

We are given a triangle ABC, which is right-angled at B. An altitude BD is drawn from the vertex B to the hypotenuse AC, meeting AC at point D.

• Triangle ABC is right-angled at B ($\angle B = 90^\circ$).
• BD is the altitude from B to AC.
• The length of the hypotenuse AC is 9 cm.
• The length of the segment CD is 3 cm.

We need to find the length of the side AB.

Geometric Principles for Right Triangles with Altitude

When an altitude is drawn from the right angle to the hypotenuse of a right triangle, it creates two smaller triangles that are similar to the original triangle and also similar to each other.

In $\triangle ABC$ with altitude BD:

• $\triangle ADB \sim \triangle ABC$
• $\triangle BDC \sim \triangle ABC$
• $\triangle ADB \sim \triangle BDC$

From these similarity relationships, several important metric properties arise, often summarized by the Geometric Mean Theorem:

• The altitude is the geometric mean of the segments of the hypotenuse: $BD^2 = AD \times CD$.
• Each leg is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to that leg:
  • $AB^2 = AD \times AC$ (AB is adjacent to AD)
  • $BC^2 = CD \times AC$ (BC is adjacent to CD)

Step-by-Step Calculation of AB

We are given AC = 9 cm and CD = 3 cm. We need to find AB. The formula involving AB is $AB^2 = AD \times AC$. To use this, we first need to find the length of AD.

The hypotenuse AC is composed of the segments AD and CD.

$\text{AC} = \text{AD} + \text{CD}$

Substituting the given values:

$9 \text{ cm} = \text{AD} + 3 \text{ cm}$

Solving for AD:

$\text{AD} = 9 \text{ cm} - 3 \text{ cm} = 6 \text{ cm}$

Now we can use the Geometric Mean Theorem formula for the leg AB:

$AB^2 = AD \times AC$

Substitute the values AD = 6 cm and AC = 9 cm:

$AB^2 = 6 \times 9$

$AB^2 = 54$

To find AB, we take the square root of both sides:

$AB = \sqrt{54}$

Now, simplify the square root $\sqrt{54}$ by finding perfect square factors of 54. We know that $54 = 9 \times 6$, and 9 is a perfect square ($3^2$).

$AB = \sqrt{9 \times 6}$

$AB = \sqrt{9} \times \sqrt{6}$

$AB = 3\sqrt{6}$

So, the measure of AB is $3\sqrt{6}$ cm.

Summary of Calculation

The length of AB is $3\sqrt{6}$ cm.

Revision Table: Right Triangle Altitude Properties

Additional Information: Similar Triangles in Geometry

Understanding similar triangles is fundamental in geometry, especially for problems involving right triangles and altitudes. Two triangles are similar if their corresponding angles are equal. This implies that the ratio of their corresponding sides is constant.

In our case, the similarity $\triangle ADB \sim \triangle ABC$ means:

$\frac{AD}{AB} = \frac{AB}{AC} = \frac{BD}{BC}$

From $\frac{AD}{AB} = \frac{AB}{AC}$, cross-multiplying gives $AB^2 = AD \times AC$, which is exactly the formula we used. This shows how the Geometric Mean Theorem is derived directly from triangle similarity.

Similarly, $\triangle BDC \sim \triangle ABC$ gives:

$\frac{CD}{BC} = \frac{BC}{AC} = \frac{BD}{AB}$

From $\frac{CD}{BC} = \frac{BC}{AC}$, we get $BC^2 = CD \times AC$.

And $\triangle ADB \sim \triangle BDC$ gives:

$\frac{AD}{BD} = \frac{BD}{CD} = \frac{AB}{BC}$

From $\frac{AD}{BD} = \frac{BD}{CD}$, we get $BD^2 = AD \times CD$.

These relationships are powerful tools for solving problems involving right triangles with altitudes.