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(C) The sides of the triangle are $a = 18$,$b = 24$,and $c = 30$.First,calculate the semi-perimeter $s = \frac{a + b + c}{2} = \frac{18 + 24 + 30}{2}

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$6$

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(C) The sides of the triangle are $a = 18$,$b = 24$,and $c = 30$.First,calculate the semi-perimeter $s = \frac{a + b + c}{2} = \frac{18 + 24 + 30}{2} = \frac{72}{2} = 36 \, cm$.Since $18^2 + 24^2 = 324 + 576 = 900 = 30^2$,the triangle is a right-angled triangle.The area of the triangle $\Delta = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 18 	imes 24 = 216 \, cm^2$.The radius of the incircle $r$ is given by $r = \frac{\Delta}{s}$.$r = \frac{216}{36} = 6 \, cm$.

The radius of the incircle of a triangle with sides $18 \, cm$,$24 \, cm$,and $30 \, cm$ is ..... $cm$.

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