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The sides of a triangle are three consecutive integers and its inradius is four units. Determine the circumradius.
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@Mertsj Help me sir
inradius: \[r = \frac{1}{2}\sqrt{\frac{(b + c - a)(a+c-b)(a+b-c)}{a + b + c}}\] given r = 4, let a = n, b = n+1, c = n+2: \[8=\sqrt{\frac{(n+3)(n+1)(n-1)}{3(n+1)}}\] \[8=\sqrt{\frac{n^2+2n-3}{3}}\] \[\sqrt{64}=\frac{\sqrt{n^2+2n-3}}{\sqrt{3}}\] \[\sqrt{192}=\sqrt{n^2+2n-3}\] Find n, then use: \[r=\frac{abc}{4sR}\] where s is the semiperimeter \(\frac{1}{2}(a+b+c)\) and R is the circumradius
Thank you Madam
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