In Delta ABC,m angle B = 90^{circ}. A circle | vedclass

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(D) Let the sides of the triangle be $a, b, c$ opposite to vertices $A, B, C$ respectively. Here,$a = BC = 24$,$b = AC$,and $c = AB$.Given $AB + AC =

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$3$

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(D) Let the sides of the triangle be $a, b, c$ opposite to vertices $A, B, C$ respectively. Here,$a = BC = 24$,$b = AC$,and $c = AB$.Given $AB + AC = 32$,so $c + b = 32$.In a right-angled triangle,by Pythagoras theorem,$a^2 + c^2 = b^2$.Substituting $a = 24$,we get $24^2 + c^2 = b^2$,which implies $b^2 - c^2 = 576$.$(b - c)(b + c) = 576$.Since $b + c = 32$,we have $(b - c)(32) = 576$,so $b - c = 18$.Adding the two equations: $(b + c) + (b - c) = 32 + 18 \implies 2b = 50 \implies b = 25$.Subtracting the two equations: $(b + c) - (b - c) = 32 - 18 \implies 2c = 14 \implies c = 7$.The radius of the incircle $r$ for a right-angled triangle is given by $r = \frac{a + c - b}{2}$.Substituting the values: $r = \frac{24 + 7 - 25}{2} = \frac{6}{2} = 3$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$. $A$ circle touches all the sides of $\Delta ABC$. If $AB + AC = 32$ and $BC = 24$,then find the radius of the incircle.

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