+194 Two of the altitudes of an acute triangle divide the sides into segments of lengths 5,3,2 and x units, as shown. What is the value of x?
+128407 We can equate the area of the the triangle on both sides of an equation
(1/2)* (3 + 5) * sqrt [ (x + 2)^2 - 3^2 ] = (1/2) (x + 2) * sqrt [ ( 8^2 - 2^2 ]
8 * sqrt [ x^2 + 4x + 4 - 9] = (x +2 ) sqrt (60)
8 * sqrt [ x^2 + 4x - 5] = (x+ 2) sqrt (60) square both sides
64 (x^2 + 4x - 5 ] = (x^2 + 4x + 4 ) * 60
16(x^2 + 4x - 5) = 15(x^2 + 4x + 4)
16x^2 + 64x - 80 = 15x^2 + 60x + 60
x^2 + 4x - 140 = 0 factor
(x - 10) ( x + 14) = 0
Setting both factors to 0 and solving for x gives that x = -14 (reject) or
x = 10
