In right angled triangle ABC angle B = 900 where CD and AE are angle bisctors of angles C and A respectively. AE = 9 and CD = 8√2. What is the lenght of AC?
here let side of the triangle is 2x and 2y which are having <90 then AC^2=4(x^2+y^2) and also 81=4x^2+y^2 and 128=x^2+4y^2 which gives 4(x^2+y^2) = (81+128)x4/5 = 209x4/5 = 19x11x4/5 thus AC = 2(19x11/5)^1/2
Any one have solved it ???
what happen?
plz try.................
8√2 = √ [bh { (b + h)2 - a2 }] / (b + h)
and h = √ (a2 + b2 )
Three variables and three equation. . Any one has time to solve it for me. .
pls give the sol of this.....
here let side of the triangle is 2x and 2y which are having <90 then AC^2=4(x^2+y^2) and also 81=4x^2+y^2 and 128=x^2+4y^2 which gives 4(x^2+y^2) = (81+128)x4/5 = 209x4/5 = 19x11x4/5 thus AC = 2(19x11/5)^1/2
I realized my approach was wrong and I tried the problem again..
Meanwhile I come across a site where solution is available http://www.qbyte.org/puzzles/p023s.html
The solution is 6xsqrt(5)
Thanks for the link
but its more like a JEE problem than CAT
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