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(B) In an isosceles triangle $ABC$ with $AB = AC = 15$ and $BC = 10$, $P$ is the midpoint of $BC$. Since $AP$ is the altitude to the base $BC$, we hav

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$\frac{5}{\sqrt{2}}$ unit

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(B) In an isosceles triangle $ABC$ with $AB = AC = 15$ and $BC = 10$, $P$ is the midpoint of $BC$. Since $AP$ is the altitude to the base $BC$, we have $BP = PC = 5$. Using the Pythagorean theorem in $\Delta ABP$: $AP = \sqrt{AB^2 - BP^2} = \sqrt{15^2 - 5^2} = \sqrt{225 - 25} = \sqrt{200} = 10 \sqrt{2}$. The area of the triangle $\Delta$ is given by $\frac{1}{2} 	imes BC 	imes AP = \frac{1}{2} 	imes 10 	imes 10 \sqrt{2} = 50 \sqrt{2}$. The semi-perimeter $s$ is $\frac{15 + 15 + 10}{2} = 20$. The inradius $r$ of the triangle is $r = \frac{\Delta}{s} = \frac{50 \sqrt{2}}{20} = \frac{5 \sqrt{2}}{2} = \frac{5}{\sqrt{2}}$. Since $O$ is the center of the inscribed circle and $P$ is the point of tangency on $BC$, the distance $OP$ is equal to the inradius $r$. Therefore, $OP = \frac{5}{\sqrt{2}}$ unit.

$ABC$ is an isosceles triangle with an inscribed circle with center $O$. Let $P$ be the midpoint of $BC$. If $AB = AC = 15$ and $BC = 10$, then $OP$ equals

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