+400 In triangle ABC, angle ACB = 90 degrees. Let H be the foot of the altitude from C to side AB.
If BC = 1 and AH = 2, find CH.
+129830 A
2
H
x
C 1 B
Let HB = x
AC^2 = AB^2 - BC^2 = (2 +x)^2 - 1^2 = x^2 + 4x + 3
CH^2 = BC^2 - BH^2 = 1 - x^2 (1)
CH^2 = AC^2 - AH^2 = (x^2 + 4x + 3) - 2^2 = x^2 + 4x -1 (2)
Set (1) = (2)
1-x^2 = x^2 + 4x -1
2x^2 + 4x -2 = 0
x^2 + 2x - 1 = 0
x^2 + 2x = 1 complete the square on x
x^2 + 2x + 1 = 1 + 1
(x + 1)^2 = 2 take the positive root
x + 1 =sqrt (2)
x = sqrt (2) - 1
CH^2 = BC^2 - BH^2 = 1^2 - (sqrt 2 -1)^2 = 1 - (2 - 2sqrt 2 + 1) = 2sqrt (2) - 2
CH = sqrt (2sqrt (2) - 2) = sqrt (sqrt (8) - 2) ≈ .91
