In triangle $ABC,$ $AB = 10,$ $BC = 24,$ and $AC = 26.$ Find the length of the shortest altitude in this triangle.
+14915 In triangle $ABC,$ $AB = c = 10,$ $BC = a = 24,$ and $AC = b = 26.$ Find the length of the shortest altitude in this triangle.
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\(h_a:h_b:h_c=\frac{1}{a}:\frac{1}{b}:\frac{1}{c}=\frac{1}{24}:\frac{1}{26}:\frac{1}{10}\)
\(h_b \) is the shortest altitude.
\(h_b=\frac{2}{b}\cdot \sqrt{s(s-a)(s-b)(s-c)}\)
\(s=\frac{1}{2}(a+b+c)=\frac{1}{2}(24+26+10)\\ s=30\)
\(h_b=\frac{2}{26}\cdot \sqrt{30(30-24)(30-26)(30-10)}\)
\(h_b=9.231\)
\(h_b =9.231\) is the shortest altitude.
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