Question

raza cercului inscris in triunghiul ABC,stiind ca AB=6,AC=8 ,BC=10

Answer 1

r=S/p
p=(6+8+10)/2=12
$S= \sqrt{p(p-a)(p-b)(p-c)} = \sqrt{12*6*4*2} = \sqrt{576} =24 \\ \\ r= \frac{24}{12} =2$