Triangle ABC is right angled at B. AD and CE are the two medians drawn from A and C respectively. If AC = 5cm, AD = 3√5/2 cm, find the length of CE.
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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)
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)
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