+868 In triangle $ABC,$ $AB = 15,$ $BC = 9,$ and $AC = 10.$ Find the length of the shortest altitude in this triangle.
+14 We can use Heron's formula to first find the area of the triangle, \(\sqrt{s(s-a)(s-b)(s-c)}\). Since the semiperimeter is \(17\), we have
\(\sqrt{17(17-15)(17-9)(17-10)} = \sqrt{1904} = 4\sqrt{119}\)
as the area of triangle \(ABC\). The shortest altitude of a triangle has the longest side as its base, so we have
\(\frac{15h}{2} = 4\sqrt{119}\),
where \(h\) is the altitude. Solving for \(h\) yields \(h = \boxed{\frac{8\sqrt{119}}{15}}\).
(I'm not sure if there's a way to solve this without using Heron's, but if there is, please let me know! Thanks.)
+14 We can use Heron's formula to first find the area of the triangle, \(\sqrt{s(s-a)(s-b)(s-c)}\). Since the semiperimeter is \(17\), we have
\(\sqrt{17(17-15)(17-9)(17-10)} = \sqrt{1904} = 4\sqrt{119}\)
as the area of triangle \(ABC\). The shortest altitude of a triangle has the longest side as its base, so we have
\(\frac{15h}{2} = 4\sqrt{119}\),
where \(h\) is the altitude. Solving for \(h\) yields \(h = \boxed{\frac{8\sqrt{119}}{15}}\).
(I'm not sure if there's a way to solve this without using Heron's, but if there is, please let me know! Thanks.)
