Relationship between medians

Let the position coordinates of $A, B, C$ be $(0,c),(0,0),(a, 0)$ respectively. Then we have two sets of equations :

(i) $\dfrac{a^2}{4}+c^2=25$

$a^2+\dfrac{c^2}{4}=49$

We have to find the value of $\dfrac{a^2+c^2}{4}$ , which is $\dfrac{74}{5}=14.8$ .

(ii) $\dfrac{a^2+c^2}{4}=25$

$a^2+\dfrac{c^2}{4}=49$

We have to find the value of $\dfrac{a^2}{4}+c^2$ , which is $76$ .

Hence the sum of these two values is $14.8+76=\boxed {90.8}$ .

Since $AC^2 = AB^2 + BC^2$ , $\triangle ABC$ is a right triangle right angled at $B$ by converse of pythagoras theorem.

Let $BC = a\,,\,AB = b\,,\,AC = c\,,\, m_a\,,\,m_b$ and $m_c$ are the lengths of medians on $BC\,,\,AB$ and $AC$ .

From pythagoras theorem,

$m_{a}^{2} = b^2 + \Large\frac{a^2}{4}$

$m_{b}^2 = a^2 + \Large\frac{b^2}{4}$ and

$m_{c}^2 = \Large\frac{c^2}{4} = \frac{a^2 + b^2}{4}\hspace{50pt}$ [Twice of median on hypotenuse of right triangle equals to hypotenuse]

Eliminating $a$ and $b$ from the above equations we get,

$m_{a}^2 + m_{b}^2 = 5m_{c}^2\hspace{40pt}\cdots\hspace{10pt}\textbf{Eq. 1}$

Also, when $m_a\,,\,m_b\,$ and $\,m_c$ are given, $a\,,\,b\,$ and $\,c$ can be find out using above $3$ equations as:

$c^2 = 4m_c^2$

$b^2 = \Large\frac{4(4m_a^2 - m_b^2)}{15}$ and

$a^2 = \Large\frac{4(4m_b^2 - m_a^2)}{15}$

Since $a\,,\,b\,$ and $c$ are greater zero. Constraints on $m_a\,,\,m_b\,$ and $m_c$ are as :

$2m_a > m_b\;$ and $\newline 2m_b > m_a$

Using these constraints in $\text{Eq. 1}$ we will get

$m_a > m_c\;$ and $\newline m_b > m_c$

Now coming to our question, based on these $4$ constraints there are $4$ possible cases:

1. $m_a = 5\,,\,m_b = 7\,,\,m_c$ to be determine.
2. $m_b = 5\,,\,m_a = 7\,,\,m_c$ to be determine.
3. $m_b = 7\,,\,m_c = 5\,,\,m_a$ to be determine.
4. $m_a = 7\,,\,m_c = 5\,,\,m_b$ to be determine.

Using $\text{Eq. 1}$ , we can get the length of third median. From the first two cases,

$m_c^2 = \Large\frac{m_a^2 + m_b^2}{5} = \frac{5^2 + 7^2}{5} = \frac{74}{5}$

From case 3,

$m_a^2 = 5m_c^2 - m_b^2 = 5\cdot5^2 - 49 = 76$

From case 4,

$m_b^2 = 5m_c^2 - m_a^2 = 5\cdot5^2 - 49 = 76$

So, square of possible lengths of third median are $\Large\frac{74}{5}$ and $76$ . Hence their sum is $\Large\frac{74}{5}$ $+ 76 =$ $\Large\frac{454}{5}$ $= \boxed{90.8}$