The side lengths of the following right triangle 15,20 and 25, as shown below. The altitude from the right angle splits the hypotenuse into line segments of length x and y. Solve for h,x, and y.

Answer:
[tex]h = 12[/tex]
[tex]x = 9[/tex]
[tex]y= 16[/tex]
Step-by-step explanation:
Given
The attached triangle
Required
Find h, x and y
Let the base of the triangle be 15.
So, the area is:
[tex]A = \frac{1}{2} * 15 * 20[/tex]
[tex]i.e\ height = 20[/tex]
[tex]A = \frac{1}{2} * 300[/tex]
[tex]A = 150[/tex]
Let the base of the triangle be 25.
So, the area is:
[tex]A = \frac{1}{2} * 25 * h[/tex]
[tex]i.e\ height = h[/tex]
[tex]A = \frac{1}{2} * 25h[/tex]
Substitute [tex]A = 150[/tex]
[tex]\frac{1}{2}*25h = 150[/tex]
Solve for h
[tex]h = \frac{150 *2}{25}[/tex]
[tex]h = 6 *2[/tex]
[tex]h = 12[/tex]
Considering the smallest triangle
[tex]Hypotenuse = 15[/tex]
So:
[tex]15^2 = h^2 + x^2[/tex]
[tex]15^2 = 12^2 + x^2[/tex]
This gives:
[tex]15^2 - 12^2 = x^2[/tex]
[tex]81 = x^2[/tex]
Take square roots
[tex]9 =x[/tex]
[tex]x = 9[/tex]
Solving for y
[tex]x + y = 25[/tex]
[tex]y= 25 - x[/tex]
[tex]y= 25 - 9[/tex]
[tex]y= 16[/tex]