Question

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Answer 1

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Answer 2

$\large\text{\underline{Let's begin.}}$

Join O and the two tangent points of the triangle. This will form a square of the side length of the radius.

Let the inradius be $r\text{ cm}$.

$\text{\underline{Property of Tangents of a Circle}}$

The length of two segments drawn from the same point is equal.

Since $\triangle ABC$ is a right triangle, it satisfies the Pythagorean theorem.

This implies $\overline{AB}^{2}+\overline{BC}^{2}=\overline{CA}^{2}$.

$\implies 12^{2}+\overline{BC}^{2}=13^{2}$

$\implies \overline{BC}^{2}=13^{2}-12^{2}$

$\implies \overline{BC}^{2}=25$

So, the length of the remaining side is $5\text{ cm}$.

$\large\text{\underline{Solution A}}$

(Attachment is included for solution A.)

According to the property of tangents of a circle, we get the diagram.

$\implies r+8=12-r$

$\implies 2r=4$

$\implies r=2$

The value we need to find is half of the radius $r$.

$\implies \dfrac{r}{2}=1\text{ cm}$

$\large\text{\underline{Conclusion}}$

Half of the radius is 1 cm.

$\large\text{\underline{Solution B}}$

We know that the area of a triangle can be found by a perimeter and an inradius.

$\hookrightarrow\large\boxed{\text{(Area)}=\dfrac{1}{2}\times\text{(Perimeter)}\times\text{(Inradius)}}$

The area of $\triangle ABC$ is $30\text{ cm}^{2}$, and the perimeter is $30\text{ cm}$.

Then the inradius is $2\text{ cm}$.

The value we need to find is half of the radius $r$.

$\implies \dfrac{r}{2}=1\text{ cm}$

$\large\text{\underline{Conclusion}}$

Half of the radius is 1 cm.