you will have similar triangles, and it can be shown quite easily that
x(20-x) = 82
-x^2 + 20x - 64 = 0
x^2 - 20x + 64 = 0
(x-16)(x-4) = 0
x = 16 or x = 4
if one part is 16, the other is 20-16 or 4
if one part is 4, the other is 20-4 = 16
so the two segments are 16 and 4
If the length of the altitude of the hypotenuse of a right triangle is 8, and the length of the hypotenuse is 20, what are the lengths of the segments of the hypotenuse? (Let x and 20-x be the lengths of the segments of the hypotenuse.)
2 answers
The altitude to the hypotenuse of a right triangle is the geometric mean between the segments of the hypotenuse created by the point where the altitude intersects the hypotenuse or h^2 = xy.
x and y are the two segments of the hypotenuse.
Therefore, h^2 = 64 = xy and x + y = 20.
64 = y(20 - y) or y^2 - 20y + 64 = 0.
Then, x = [20+/-sqrt(400 - 256)]/2 or y = 16 making x = 4.
x and y are the two segments of the hypotenuse.
Therefore, h^2 = 64 = xy and x + y = 20.
64 = y(20 - y) or y^2 - 20y + 64 = 0.
Then, x = [20+/-sqrt(400 - 256)]/2 or y = 16 making x = 4.