n 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.
h2=12−(2r−2)2=r2−(2−r)21−(4r2−8r+4)=r2−(4−4r+r2)1−4r2+8r−4=r2−4+4r−r24r2−4r−1=0r2−r−0.25=0
r=0.5±√0.25+0.25r∈{1.207,−0.207}r=0.5+√0.5=1.207h2=12−(2r−2)2h2=1−(1+√2−2)2h2=1−(√2−1)2h2=1−(2−2√2+1)h=√2√2−2h=¯CH=0.91
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