In triangle ABC, $AB = 12$ cm, $AC = 15$ cm and $BC = 18$ cm. Find the length of the median AD (in cm).
Answer and solution
Correct answer: A
Solution
D is midpoint of BC, $BD = 9$.
Apollonius: $AB^2 + AC^2 = 2(AD^2 + BD^2)$
$144 + 225 = 2(AD^2 + 81) \Rightarrow 369 = 2\,AD^2 + 162 \Rightarrow AD^2 = \tfrac{207}{2} \Rightarrow AD = \sqrt{103.5}$
Apollonius: $AB^2 + AC^2 = 2(AD^2 + BD^2)$
$144 + 225 = 2(AD^2 + 81) \Rightarrow 369 = 2\,AD^2 + 162 \Rightarrow AD^2 = \tfrac{207}{2} \Rightarrow AD = \sqrt{103.5}$