Let’s solve the two problems step by step!

Problem 11:

We are given a triangle with the following dimensions:

• One side is divided into segments of lengths 16 and 12.
• The total length of the side is $21$ (so it equals $x$ + 12 + 16).

This is a right triangle, and the property of similar triangles is used here.

Setup:

The triangles created by the altitude are similar, so we use the proportion: $\frac{\text{altitude}}{\text{segment 1}} = \frac{\text{segment 2}}{\text{altitude}}$

Let the altitude be $x$. Using the given values: $\frac{x}{16} = \frac{12}{x}$

Solve for $x$:

Cross multiply: $x^2 = 16 \cdot 12$

Simplify: $x^2 = 192$

Take the square root: $x = \sqrt{192} = 8\sqrt{3} \ (\text{exact value}) \approx 13.856$

So, the value of $x$ is approximately 13.86.

Problem 12:

We are given:

• A triangle with sides $6$, $10$, and parts of one side $x$ and $x - 4$.

This triangle uses the Angle Bisector Theorem, which states: $\frac{\text{side 1}}{\text{side 2}} = \frac{\text{segment 1}}{\text{segment 2}}$

Setup:

Using the theorem: $\frac{6}{10} = \frac{x - 4}{x}$

Solve for $x$:

Simplify the fraction: $\frac{3}{5} = \frac{x - 4}{x}$

Cross multiply: $3x = 5(x - 4)$

Distribute: $3x = 5x - 20$

Simplify: $20 = 5x - 3x$

$20 = 2x$

Divide: $x = 10$

So, the value of $x$ is 10.

Final Answers:

1. $x = 8\sqrt{3} \approx 13.86$ for Problem 11.
2. $x = 10$ for Problem 12.

Would you like detailed explanations for the steps, or do you have any further questions? 😊

5 Related Questions:

1. What are similar triangles, and how do we identify them?
2. How does the Angle Bisector Theorem apply to other triangle types?
3. How is the Pythagorean Theorem related to similar triangles?
4. What are other scenarios where proportions are used in geometry?
5. What is the relationship between the altitude and the hypotenuse in a right triangle?

Tip:

When solving geometry problems, carefully label all given dimensions and use diagrams to ensure proportions and relationships are clear.