The problem is as follows:
Figure 1 shows a triangle $\textrm{ABC}$ whose side $\textrm{BC = 4 inches}$. It is known that $\textrm{AM}$ is a median whose length is equal to $\textrm{1 inch}$. Find the sum of the squares from the other two sides of triangle $\textrm{ABC}$.
The existing alternatives in my book are:
- 12 inches
- 10 inches
- 14 inches
- 15 inches
- 8 inches
The figure is attached below:
In my attempt to solve this problem what I used was the median length formula which is shown below:
$$m = \frac{1}{2}\sqrt{2(a^{2}+b^{2})-c^{2}}$$
By plugin the values given:
$$1 = \frac{1}{2}\sqrt{2\left(a^{2}+b^{2}\right)-4^{2}}$$
$$2 = \sqrt{2\left(a^{2}+b^{2}\right)-4^{2}}$$
$$2^{2}= 2\left(a^{2}+b^{2}\right)-4^{2}$$
$$4 = 2\left(a^{2}+b^{2}\right)-4^{2}$$
$$16+4= 2\left(a^{2}+b^{2}\right)$$
$$10 = a^{2}+b^{2}$$
$$a^{2}+b^{2} = 10$$
This is also described in figure 2 from below which it proves how to relate the length of the median and the opposing sides of the triangle other than the base.
This is based on the parallelogram law.
It states as this:
the sum of the squares of the lengths of the four sides of a parallelogram equals the sum of the squares of the lengths of the two diagonals.
so,
$$2b^{2}+2a^{2}=4^{2} + 2^{2}$$
$$2\left ( b^{2} + a^{2} \right) = 2 \left ( 2^{3}+ 2 \right)$$
Dividing by $2$ on both sides:
$$b^{2} + a^{2} = 10$$
Therefore the answer would be $\textrm{10 inches}$ which does appears within the alternatives. But I wonder other than the method I used does it exist another ways to obtain what it is being asked?.
Not much left to add to the answers you provided in the post, except +1 for a nicely asked question.
For quick verification, and trusting that the problem has indeed a unique answer regardless of $\,A\,$, consider the degenerate case where $\,A\,$ lies on $\,BC\,$ one unit away from $\,B\,$ towards $\,C\,$. Then $\,AB=1, AC=3\,$ and $\,AB^2+AC^2=1+9=10\,$.
Or, in a complex plane where $\,B,C\,$ are at $\,\pm2\,$ and $\,A\,$ lies on the unit circle: $$\require{cancel} \begin{align} |a+2|^2 + |a-2|^2 &= (a+2)(\bar a+2) + (a-2)(\bar a-2) \\ &= a \bar a + \cancel{2a} + \bcancel{2 \bar a} + 4+ a \bar a - \cancel{2a} - \bcancel{2 \bar a}+ 4 \\ &= 2 |a|^2 + 8 \\ &= 10 \end{align} $$