In Delta ABC,if AB = 7, BC = 24, AC = 25,then | vedclass

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(C) First,we check if $\Delta ABC$ is a right-angled triangle. Given sides are $7, 24, 25$. Since $7^2 + 24^2 = 49 + 576 = 625 = 25^2$,the triangle is

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$6$

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(C) First,we check if $\Delta ABC$ is a right-angled triangle. Given sides are $7, 24, 25$. Since $7^2 + 24^2 = 49 + 576 = 625 = 25^2$,the triangle is a right-angled triangle with the hypotenuse $AC = 25$. The circle touching all three sides of a triangle is the incircle. The radius $r$ of the incircle of a right-angled triangle is given by the formula $r = \frac{a + b - c}{2}$,where $a$ and $b$ are the legs and $c$ is the hypotenuse. Here,$a = 7, b = 24, c = 25$. $r = \frac{7 + 24 - 25}{2} = \frac{31 - 25}{2} = \frac{6}{2} = 3$. The diameter of the circle is $d = 2r = 2 	imes 3 = 6$.

In $\Delta ABC$,if $AB = 7, BC = 24, AC = 25$,then the diameter of a circle touching all the three sides of the triangle is ............

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