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.
First, we must set some variables to complete this problem. Let's set HB as x.
Note that AC2=AB2−BC2=(2+x)2−12=x2+4x+3
We can write a system of equations to help us solve this question. We have
CH2=BC2−BH2=1−x2CH2=AC2−AH2=(x2+4x+3)−22=x2+4x−1
Since both equations are equal to CH^2, we set both of them to equal each other, and we get
1−x2=x2+4x−12x2+4x−2=0x2+2x−1=0x2+2x=1
x2+2x+1=1+1(x+1)2=2x+1=√2x=√2−1
Now, we move on to the final step, which is to find CH. Plugging x back into the original formula, we get
CH2=BC2−BH2=12−(√2−1)2=1−(2−2√2+1)=2√2−2CH=√2√2−2=√√8−2≈.91
So our final answer is about .91.
Thanks! :)