What is the radius of the circle inscribed in triangle ABC if AB = AC=7 and BC=6? Express your answer in simplest radical form.
+2666 The inradius is \(\text{Area} \over \text{semiperimeter}\)
By Heron's formula, the area is \(\sqrt{s(s-a)(s-b)(s-c)} = \sqrt{10 \times 3 \times 3 \times \times 4} = \sqrt{360} = \sqrt{36} \times \sqrt {10} = 6 \sqrt{10}\)
The semiperimeter is \((7 + 7 + 6) \div 2 = 10\), so the inradius is \({6 \sqrt{10} \over 10} = \color{brown}\boxed{3 \sqrt {10} \over 5}\)
+2666 The inradius is \(\text{Area} \over \text{semiperimeter}\)
By Heron's formula, the area is \(\sqrt{s(s-a)(s-b)(s-c)} = \sqrt{10 \times 3 \times 3 \times \times 4} = \sqrt{360} = \sqrt{36} \times \sqrt {10} = 6 \sqrt{10}\)
The semiperimeter is \((7 + 7 + 6) \div 2 = 10\), so the inradius is \({6 \sqrt{10} \over 10} = \color{brown}\boxed{3 \sqrt {10} \over 5}\)
