Question Number 103389 by  M±th+et+s last updated on 14/Jul/20

Commented by M±th+et+s last updated on 14/Jul/20

find the hypotenuse of the triangle  as (x)

Answered by OlafThorendsen last updated on 14/Jul/20

(y/x) = (2/(x−3))  y = ((2x)/(x−3))  (√(x^2 +y^2 )) = (√(x^2 +(4/((x−3)^2 ))x^2 ))  (√(x^2 +y^2 )) = x(√(1+(4/((x−3)^2 ))))

Answered by bramlex last updated on 15/Jul/20

Commented bybramlex last updated on 15/Jul/20

((x−3)/3) = (2/(y−2)) ⇒((y−2)/2) = (3/(x−3))  y = (6/(x−3)) + 2 = ((6+2x−6)/(x−3))  y= ((2x)/(x−3)).   hypotenusa r = (√(x^2 +y^2 ))  r = (√(x^2 +(((2x)/(x−3)))^2 ))  r= (√((x^2 (x−3)^2 +4x^2 )/((x−3)^2 )))  r = (x/(x−3)).(√(4+x^2 −6x+9))  r = (x/(x−3)).(√(x^2 −6+13)) ■