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Hi Sal, I drew a diagram and labeled some points. $P$ is the centre of the circle and $D, E$ and $F$ are points where the sides of the triangle are tangent to the circle. Thus angles $CAB, PDB$ and $BEP$ are right angles. Suppose the radius of the circle is $r$ and the side length of $CA$ is $s.$ By the symmetry of the diagram $E$ is the midpoint of $BC.$ Notice also that the length of $AD$ is $r.$ I used Pythagoras Theorem 3 times. Once for the triangle $ABC$ to find the length of $BC$ and hence $BE$ in terms of $s.$ A second for triangle $BEP$ to find the length of $BP,$ and a third time for triangle $BPD$ to write an equation involving $s$ and $r.$ Solve for $r.$ I hope this helps, | |||||||||||||||
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