In Delta ABC,angle B is a right angle. If AB | vedclass

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(B) Let the radius of the incircle be $r$.Given,$AB = 8$,$BC = 15$,and $\angle B = 90^{\circ}$.Using the Pythagoras theorem in $\Delta ABC$:$AC^{2} =

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(B) Let the radius of the incircle be $r$.Given,$AB = 8$,$BC = 15$,and $\angle B = 90^{\circ}$.Using the Pythagoras theorem in $\Delta ABC$:$AC^{2} = AB^{2} + BC^{2}$$AC^{2} = 8^{2} + 15^{2}$$AC^{2} = 64 + 225 = 289$$AC = \sqrt{289} = 17$.The area of $\Delta ABC$ can be expressed as the sum of the areas of $\Delta AOB$,$\Delta BOC$,and $\Delta AOC$,where $O$ is the incenter:Area of $\Delta ABC = 	ext{Area}(\Delta AOB) + 	ext{Area}(\Delta BOC) + 	ext{Area}(\Delta AOC)$$\frac{1}{2} 	imes AB 	imes BC = \frac{1}{2} 	imes AB 	imes r + \frac{1}{2} 	imes BC 	imes r + \frac{1}{2} 	imes AC 	imes r$$\frac{1}{2} 	imes 8 	imes 15 = \frac{1}{2} 	imes r 	imes (AB + BC + AC)$$60 = \frac{1}{2} 	imes r 	imes (8 + 15 + 17)$$60 = \frac{1}{2} 	imes r 	imes 40$$60 = 20r$$r = \frac{60}{20} = 3$.Alternatively,for a right-angled triangle,the inradius $r = \frac{AB + BC - AC}{2} = \frac{8 + 15 - 17}{2} = \frac{6}{2} = 3$.

In $\Delta ABC$,$\angle B$ is a right angle. If $AB = 8$ and $BC = 15$,find the radius of the incircle of $\Delta ABC$.

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