If the lengths of the altitudes of a triangle are 4, 5, and 6, what is the area of the triangle?
Can you generalize?
From the
Dr. Math Forum , the area of a triangle equals
1/sqrt[(1/ha + 1/hb + 1/hc)*(-1/ha + 1/hb + 1/hc)*
(1/ha - 1/hb + 1/hc)*(1/ha + 1/hb - 1/hc) ],
where ha, hb and hc are the altitudes of the triangle. That web site derives this formula from Heron's formula.
Working this out for altitudes of 4, 5 and 6 results in
sqrt(2^8*3^4*5^4/(7*13*17*37))
= 3600 / sqrt(57239)
|
Posted by Charlie
on 2007-01-15 23:21:07 |