+0

# Geometry

0
4
1
+729

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

Jun 27, 2024

#1
+14
+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.)

Jun 27, 2024
edited by AsianMathGuy  Jun 27, 2024
edited by AsianMathGuy  Jun 27, 2024
edited by AsianMathGuy  Jun 27, 2024

#1
+14
+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