+0

# Geometry

-3
2
1
+153

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.

Jun 12, 2024

#1
+1280
+1

First, we must set some variables to complete this problem. Let's set HB as x.

Note that $$AC^2 = AB^2 - BC^2 = (2 +x)^2 - 1^2 = x^2 + 4x + 3$$

We can write a system of equations to help us solve this question. We have

$$CH^2 = BC^2 - BH^2 = 1 - x^2 \\ CH^2 = AC^2 - AH^2 = (x^2 + 4x + 3) - 2^2 = x^2 + 4x -1$$

Since both equations are equal to CH^2, we set both of them to equal each other, and we get

$$1-x^2 = x^2 + 4x -1 \\ 2x^2 + 4x -2 = 0 \\ x^2 + 2x - 1 = 0 \\ x^2 + 2x = 1$$

$$x^2 + 2x + 1 = 1 + 1 \\ (x + 1)^2 = 2\\ x + 1 =\sqrt{2} \\ x =\sqrt{2} - 1$$

Now, we move on to the final step, which is to find CH. Plugging x back into the original formula, we get

$$CH^2 = BC^2 - BH^2 = 1^2 - (\sqrt{ 2} -1)^2 = 1 - (2 - 2\sqrt{ 2} + 1) = 2\sqrt {2} - 2 \\ CH = \sqrt {2\sqrt {2} - 2} = \sqrt{\sqrt {8} - 2} ≈ .91$$