Question

If the sides of a triangle are 5, 6, 10, then the length of the median of the biggest side is:
A. $ \sqrt{\dfrac{13}{2}} $
B. $ \sqrt{\dfrac{11}{2}} $
C. $ \sqrt{\dfrac{17}{2}} $
D. $ \sqrt{\dfrac{19}{2}} $

Answer 1

Hint:The Apollonius theorem: states that "the sum of the squares of any two sides of any triangle equals twice the square on half the third side, together with twice the square on the median bisecting the third side". Specifically, in any triangle ABC, if AD is a median, then: $ A{{B}^{2}} $ + $ A{{C}^{2}} $ = $ 2\left(
A{{D}^{2}}+B{{D}^{2}} \right) $ .
If AD is the median to the longest side, what will be the values of AB, AC and BD in the above formula?

Complete step by step solution:
Let's say that the required median is AD, where D is the mid−point of the longest side BC opposite to
vertex A.
∴ BC = 10, AC = 6 and AB = 5.

Since D is the mid−point of BC, BD = DC = $ \dfrac{10}{2} $ = 5.

Using the Apollonius theorem, we have:
$ A{{B}^{2}} $ + $ A{{C}^{2}} $ = $ 2\left( A{{D}^{2}}+B{{D}^{2}} \right) $

Substituting the values, we get:
⇒ $ {{5}^{2}} $ + $ {{6}^{2}} $ = $ 2\left( A{{D}^{2}}+{{5}^{2}} \right) $
⇒ 25 + 36 = $ 2A{{D}^{2}} $ + 50

Subtracting 50 from both sides, we get:
⇒ $ 2A{{D}^{2}} $ = 11

Divide by 2 on both sides and take square root:
⇒ AD = $ \sqrt{\dfrac{11}{2}} $ .

Therefore, the correct answer is B. $ \sqrt{\dfrac{11}{2}} $ .

Note:Median is the line joining the mid−point of a side to the opposite vertex. All medians of a triangle are concurrent. The point of intersection of the medians is called the centroid and it divides each of the medians in the ratio 2 : 1, from the vertex. We can also drop a perpendicular on the longest side and solve using Pythagoras theorem, if we don't remember the formula well.