In Delta ABC,angle B = 90^{circ}. The radius | vedclass

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(B) Let the sides of the right-angled triangle be $a = BC$,$c = AB$,and the hypotenuse $b = AC$.Let the radius of the incircle be $r$.The center of th

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$\frac{AB + BC - AC}{2}$

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(B) Let the sides of the right-angled triangle be $a = BC$,$c = AB$,and the hypotenuse $b = AC$.Let the radius of the incircle be $r$.The center of the incircle $(O)$ forms a square with the vertices at the point of tangency on sides $AB$ and $BC$ and the vertex $B$ because the radii are perpendicular to the sides.Thus,the distance from $B$ to the points of tangency on $AB$ and $BC$ is $r$.The remaining segments of the sides $AB$ and $BC$ are $(c - r)$ and $(a - r)$ respectively.By the property of tangents drawn from an external point to a circle,these segments are equal to the distances from vertices $A$ and $C$ to the points of tangency on the hypotenuse.Therefore,the hypotenuse $b = (c - r) + (a - r)$.$b = a + c - 2r$.$2r = a + c - b$.$r = \frac{a + c - b}{2}$.Substituting the side lengths: $r = \frac{BC + AB - AC}{2}$.

In $\Delta ABC$,$\angle B = 90^{\circ}$. The radius of the incircle touching all three sides of the triangle is $\ldots \ldots \ldots \ldots$

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