IN Triangle ABC, angle ACB=90 degrees. D and E lie on line AB such that line CD is perpendicular to line AB and line AE=line EB. F lies on line CB such that line EF is perpendicular to line CB. CE=8 and EF=10. In simplest radical form, CD=k times the Sq. root of w all over 9.
2007-01-25 14:22:36 · 2 answers · asked by Quagmire77 1 in Science & Mathematics ➔ Mathematics
sorry CE=18
2007-01-26 00:42:23 · update #1
By similar triangles, AC = 20.
By Pythagoras on triangle CEF, CF = sqr(18^2 - 10^2) = 4sqr(14).
This makes the coordinate of AE = 20-4sqr(14),10) if the origin were point A. So AE^2 = (20 - 4sqr(14))^2 + 10^2 = 724 - 160sqr(14).
So AB = 2*AE.
So CB^2 = AB^2 - 20^2 = 4( 724 - 160sqr(14)) - 400 = 2496 - 640sqr(14).
By similar triangles, CB/CD = AB/20 = 2*AE/20 = AE/10. So CD = 10*CB/AE
This is 10*sqr[ (2496 - 160sqr(14)]/(724 - 160sqr(14))], which simplifies to
20*sqr[ (156 - 40sqr(14)]/(181 - 40sqr(14))]. Now multiply by the conjugate of the denominator to simplify more. Etc.
2007-01-27 09:31:49 · answer #1 · answered by berkeleychocolate 5 · 0⤊ 0⤋
Are you sure the question is correct? CE=8 and EF=10 and CEF forms a right triangle but how can the hypotenuse be smaller than the other side?
2007-01-25 22:58:46 · answer #2 · answered by aznskillz 2 · 0⤊ 0⤋