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This right-angled triangle ABC has sides of lengths:
AC=12cm, BC=5cm and AB=13cm.
1. The diameter CD of the semicircle lies on the 12cm side.
2. The AB side is a tangent to the circle.

What is the radius R of the semi circle?

 See The Solution Submitted by Ady TZIDON No Rating

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 possible solution | Comment 1 of 2

Generalisation

Let r be the radius and {u,v} the generators of a Pythagorean triangle of side (v^2-u^2, 2uv, u^2+v^2).

If 2uv>(v^2-u^2), then r = u(v^2-u^2)/v.
If 2uv<(v^2-u^2), then r = (2uv(v-u))/(u+v)

In the given example 12>5, and the generators are {2,3}; 3^2-2^2=5, 2*3*2=12, 2^2+3^2=13. r is then 2(3^2-2^2)/3 =10/3.

Say the triangle had generators {1,4}, then 4^2-1^2=15, 2*1*4=8, 4^2+1^2=17, with 8<15. r is then (2*1*4(4-1))/(1+4) or 24/5.

Edited on March 7, 2018, 8:54 am
 Posted by broll on 2018-03-07 08:53:29

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