Two of the altitudes of an acute triangle divide the sides into segments of lengths 5,3,1 and x units, as shown. What is the value of x?
+129907 Length of altitude with an upward slope = sqrt (8^2 - 1^2) = sqrt (63)
Length of downward sloping altitude = sqrt [ ( x + 1)^2 - 3^2 ] = sqrt [ x^2 + 2x - 8 ]
Area of triangle = Area of Triangle
(1/2) (3 + 5)sqrt [ x^2 + 2x - 8 ] = (1/2) (1 + x) sqrt (63)
8 sqrt [ x^2 + 2x - 8 ] = ( 1 + x) sqrt (63)
Not hard to solve but tedious.......with a little help from WolframAlpha
x = 23
