In triangle $ABC,$ $AB = 15,$ $BC = 9,$ and $AC = 10.$ Find the length of the shortest altitude in this triangle.

eramsby1O1O Jun 27, 2024

#1**+1 **

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.)}

AsianMathGuy Jun 27, 2024

edited by
AsianMathGuy
Jun 27, 2024

edited by AsianMathGuy Jun 27, 2024

edited by AsianMathGuy Jun 27, 2024

edited by AsianMathGuy Jun 27, 2024

edited by AsianMathGuy Jun 27, 2024

#1**+1 **

Best Answer

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.)}

AsianMathGuy Jun 27, 2024

edited by
AsianMathGuy
Jun 27, 2024

edited by AsianMathGuy Jun 27, 2024

edited by AsianMathGuy Jun 27, 2024

edited by AsianMathGuy Jun 27, 2024

edited by AsianMathGuy Jun 27, 2024