Question

In a triangle ABC, AC = 36, BC = 48, and the medians BD and AE to sides AC and BC, respectively, are perpendicular. Find AB


I tryed to find the sides using Pythagorean theorem but it's not working

Answer 1

let the medians intersect at F, so we have 90° angles at F
Remember that the medians intersect at the centroid and the centroid divides each median in the ratio of 2 : 1, the longer side towards the vertex

So let FE = x, then AF = 2x
let FD = y, then FB = 2y

In triangle AFD: (2x)^2 + y^2 = 18^2
4x^2 + y^2 = 324 ---> y^2 = 324 - 4x^2
in triangle BEF: x^2 + (2y)^2 = 24^2
x^2 + 4y^2 = 576

subbing:
x^2 + 4(324-4x^2) = 576
x^2 + 1296 - 16x^2 = 576
-15x^2 = -720
x^2 = 48
x = √48 or 4√3
y^2 = 324 - 4(48) = 132

in triangle ABF
AB^2 = 4x^2 + 4y^2
= 4(48) + 4(132) = 720
AB = √720 or 12√5

check my arithmetic, I know my logic is valid.