## Tuesday, 10 February 2009

### Solution to Al's Right Triangle Problem.

A right triangle has a perimeter of 60 and the altitude to the right angle is 12. Find the area

When I first heard this problem I recognized the 60 and 12 as simply five times the values of the same measures for a 3-4-5 triangle and so I knew the answer, but also knew I could not answer the question in general. After spending several days seeming to go around in circles, I asked for a suggestion on a math problem solving site I belong to. With the help of the clue I managed to come up with the general solution. So Al, here it is:

If we let the legs of the right triangle be a and b, and let c represent the hypotenuse, then we can start with P=a+b+c... and squaring both sides gives P2 = a2+b2+c2 + 2ab + 2ac + 2bc

By the Pythagorean theorem we know that a2+b2=c2 so we can rewrite that as P2 = 2c2 + 2ab + 2ac + 2bc

Since a and b are the legs of a right triangle, the area, A = ab/2 or 2A= ab and so we replace that in the equation as well... P2 = 2c2 + 4 A + 2ac + 2bc

It was at this point that I seemed unable to make any furthur progress, but finally realized that by factoring I could rewrite 2ac + 2bc as 2c(a+b) which is also 2c(P-c) = 2PC - 2c2, so the equation could be written as P2 = 2c2 + 4 A + 2Pc-2c2 which simplifies to just P2 = 4 A + 2Pc.... So close, and yet it still avoided me... then I realized that since c was the hypotenuse, 2 A= cH (where H is the altitude to the right angle) and so c = 2A/H...

With this substitution I had P2 = 4 A + 4AP/H. The original question could be solved by plugging in values at this point and solving for area, but I wanted to write it out as a general solution, so factoring out the A...P2 = (4 + 4P/H ) A and then dividing by the 4+4P/H produced Next I want to talk about the Law of Sines and why it should NOT be written upside down. See ya' soon.