SOLUTION: Do not seem to be able to find the right substitutions to solve the problem. Problem: One of the legs of a right triangle has length 4cm. Express the length of the altitude per

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Question 1170331: Do not seem to be able to find the right substitutions to solve the problem.
Problem: One of the legs of a right triangle has length 4cm. Express the length of the altitude perpendicular to the hypotenuse as a function of the length of the hypotenuse? That is the question.
So I determined the following relationships from the triangle. I do not know how to draw pictures here so I will try to describe my picture. I have a right triangle with hypothenuse (h). Short side is 4cm, and the other side is side (b).
After drawing the altitude I have 2 new triangles within the original triangle. The one has side (a) the altitude and side (c) a portion of the original hypotenuse h and the new hypotenuse of the small triangle 4cm.
The other new triangle has side (h-c) other side (a), and hypotenuse (b)
So relationships I came up with from pythagorean are:
1. h^2=16+b^2
2. 16=a^2+c^2
3. b^2=(h-c)^2+a^2
My issue is no matter how I put this together I cannot get to the question "Express the length of the altitude perpendicular to the hypotenuse as a function of the length of the hypotenuse?"
Can you help me?

Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website!
.

Let the hypotenuse length be " H " cm Then the other leg of the triangle is = cm. The area of the triangle is half the product of the legs Area = . The area can be presented in other way as half the product of the hypotenuse and the perpendicular L drawn to the hypotenuse Area = . So, we have this equation = . From this equation, after canceling the factor in both sides, you get L = . It is the expression of the length of the altitude via the hypotenuse length. ANSWER. L = .

Solved.