finding median when all three sides are not given

Question

Let ABC be a triangle with AB=3cm, AC=5cm. If AD is a median drawn from the vertex A to the side BC, then which one of the following is correct?

a) AD is always greater than 4cm but less than 5cm

b) AD is always greater than 5cm

c) AD is always less than 4cm

d) none of the above

I assumed the triangle to be right angled at B and got BC=4. So, BD=2. I used Median theorem, AB^2 + AC^2 = 2(AD^2 + BD^2), and got AD=sqrt(13), which is less than 4. So, I am left with c) and d) options.

How to move forward?

And is there a way to solve this question instinctively without actually finding the value of AD?

Answer 1

Stewart's theorem gives that the length of the median $m_a$ between the two sides $b$ and $c$ satisfies: $$ 4m_a^2 = 2b^2+2c^2-a^2 \tag{1}$$ and since $a\geq |b-c|$ by the triangle inequality, we have: $$ 4m_a^2 \leq (b+c)^2 \tag{2}$$ hence the length of the median $m_a$ is always less or equal to the arithmetic mean of $b$ and $c$.

On the other hand, since $a\leq b+c$, $$ 4m_a^2 \geq (b-c)^2.\tag{3}$$ In you case, $(2)$ and $(3)$ give: $$1\leq m_a \leq 4 $$ hence the right answer is c).

Answer 2

Think about the two degenerate cases, one where $B$ lies between $A$ and $C$, and the other where $A$ lies between $B$ and $C$, and then about how the median point moves when $B$ does.

Answer 3

Since the maximum that side BC can be is 8 then using the Median theorem we get

$9 + 25 = 2AD^2 + 2(8/2)^2 ==> 9 + 25 = 2AD^2 + 32$

So, $2 = 2AD^2 ==> 1 = AD^2$ so AD can be as small as 1

Now if we look at the minimum that BC can be we get

$9 + 25 = 2AD^2$ + $2(2/1)^2$ ==> $9 + 25 = 2AD^2 + 2$

So, $32 = 2AD^2$ ==> $16 = AD^2$ and AD = 4

So the max AD can be is 4 and the min Ad can be is 1 so we have

$1 \leq AD \leq 4$.

So the answer would be C