well, if it is entirely within, it cannot touch any corner or side.
STart with the area along the sides the center of the coin cannot be in.
The center has to be greater than .5cm from any side. Draw the original triangle. Now mark a line parallel to each side, inside the triangle, .5 from the parallel side. These three lines make a triangle inside, in which the center may reside.
label the inner triangle a, b, c, with the a side parallel to 8, b parallel to 15, and c parallel to 17.
a/8=b/15=c/17 similar triangles.
but a= 8-.5-.5-.5(8/15) If you examine the 17 corner carefully, you can arrive at that with using smaller triangles which are similar.
solve for a: a= 6.86
which means b: b= a*15/8=12.86
area of inner triangle= 1/2 ab=44.1
area of original triangle= 1/2 8*15=60
Pr(coin inside)= areainner/areaouter
A coin of radius 1 cm is tossed onto a plane surface that has been tesselated by right
triangles whose sides are 8 cm, 15 cm, and 17 cm long. What is the probability that the
coin lands within one of the triangles?
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