Altitudes

For this question we know the shortest altitude would come from the longest side which in this case can be either BC or AC because they both equal 20. Now we need to find the area of triangle ABC. For this question we should use a formula known as Heron's formula. Which basically says the area of a triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\) s = the semiperimeter or (a+b+c)/2 and a,b,c are the three sides of a triangle. Using heron's formula we get the semiperimeter = 55/2. \(\sqrt{55/2(55/2-20)(55/2-20)(55/2-15)}\) this simplifies to \(\sqrt{55/2(15/2)(15/2)(25/2)} = \sqrt{(15^2*55*25)/2^4}\) the denominator when square rooted equals 4. 15^2 goes out to 15 and 25 goes out to 5 leaving 55. Giving us the area of ABC as \(75\sqrt{55}/4\). Assigning the altitude as x. We get (x*20)/2 = \(75\sqrt{55}/4\). Which gets us our final answer of:

\(x = 75\sqrt{55}/40\)

Nice work Yesoka,

I got the same answer via a different method.

Use pythagoras theorem twice to set up 2 equations

Sove simultaneously first to get x and then to get the height