Question

Abc is a right angled triangle at b ad and ce are two medians

Answer 1

We apply Pythagoras theorem.

ΔABC:  AC² = AB² + BC² 

             5² = (2 BE)² + BC²

             25 = 4 BE² + BC²    -- (1)

Also,  BE² + BC² = CE²      ----(2)

ΔABD,  AD² = AB² + BD²

                    = 4 BE² + BC²/4

=>   4 (3√5/2)² = 45 = 16 BE² + BC²  -- (3)

Solve the equations (1) and (2).

          55 = 3 BC²    => BC = √(55/3)

=>  BE² = (AC² - BC²)/4 = (5² - 55/3)/4 = 5/3

By (2), CE² = 5/3 + 55/3 = 20/3

CE = 2 √(5/3)