Show that the inradius of a right-angled triangle is less than 1/4 of the length of the hypoteneuse and less than 1/2 the length of the shortest side.
Solution
The circle lies between BC and the parallel line through A, so its diameter is less than AC. It also lies between AB and the parallel line through C. But the distance between these two lines is less than CM (where M is the midpoint of AB) which is half AB.
© John Scholes
jscholes@kalva.demon.co.uk
1 Nov 2003
Last corrected/updated 1 Nov 03