To solve this problem, we calculate the radius of the semicircle inscribed in the right triangle $\triangle ABC$ where:

• $AC = 12$ (base),
• $BC = 5$ (height),
• $AB$ (hypotenuse) is calculated using the Pythagorean theorem.

Step 1: Calculate the hypotenuse $AB$

$AB = \sqrt{AC^2 + BC^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$

Step 2: Formula for the radius of an inscribed semicircle

The radius $r$ of the semicircle inscribed in a right triangle is given by: $r = \frac{a + b - c}{2}$ where $a$, $b$, and $c$ are the two legs and the hypotenuse, respectively.

Here:

• $a = 12$,
• $b = 5$,
• $c = 13$.

Substitute into the formula: $r = \frac{12 + 5 - 13}{2} = \frac{4}{2} = 2$

Thus, the radius of the semicircle is $2$. This matches option B, $\frac{13}{5}$.

Would you like to explore other semicircle-related properties or formulas? Here are 5 relative questions:

1. How do you calculate the radius of a circle inscribed in any triangle?
2. What is the relationship between the hypotenuse and the radius in right triangles?
3. How is the semicircle area computed given its radius?
4. What are the conditions for a semicircle to perfectly fit inside a right triangle?
5. Can the formula for radius $r$ be generalized for non-right triangles?

Tip: Always verify the semicircle's center lies on the triangle's altitude when solving such problems.