In a right triangle ABC,the incircle touches | vedclass

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(D) Let the incircle have center $O$ and radius $r$. The incircle touches $AC$ at $D$,$BC$ at $E$,and $AB$ at $F$.Since $O$ is the center and $OE \per

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$2$

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(D) Let the incircle have center $O$ and radius $r$. The incircle touches $AC$ at $D$,$BC$ at $E$,and $AB$ at $F$.Since $O$ is the center and $OE \perp BC$,$OF \perp AB$,and $OE=OF=r$,the quadrilateral $O E B F$ is a square with side length $r$.Thus,$BE = BF = r$.By the property of tangents from an external point to a circle,we have:$AF = AD = 10$$CE = CD = 3$Therefore,the sides of the right-angled $	riangle ABC$ are:$AB = AF + BF = 10 + r$$BC = CE + BE = 3 + r$$AC = AD + DC = 10 + 3 = 13$Using the Pythagorean theorem,$AB^2 + BC^2 = AC^2$:$(10 + r)^2 + (3 + r)^2 = 13^2$$100 + 20r + r^2 + 9 + 6r + r^2 = 169$$2r^2 + 26r + 109 = 169$$2r^2 + 26r - 60 = 0$$r^2 + 13r - 30 = 0$$(r + 15)(r - 2) = 0$Since $r > 0$,we have $r = 2$.Thus,the inradius is $2$.

In a right $\triangle ABC$,the incircle touches the hypotenuse $AC$ at $D$. If $AD=10$ and $DC=3$,the inradius of $\triangle ABC$ is

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