Triples and Areas!

Since $(3j)^2 + (4j)^2 = (5j)^2$ , $\triangle ABC$ is a right triangle, and by Thales's Theorem right $\angle A$ will lie on the circle that has $BC$ as its diameter.

Since $M$ is the midpoint of $BC$ , $M$ is the center of that circle, and $MB = MA = MC = \cfrac{5}{2}j$ , so $\triangle ACM$ is an isosceles triangle.

As two base angles of an isosceles triangle, $\angle MCA = \angle MAC$ , which means $\triangle AMF \sim \triangle CAB$ by AA similarity.

Since $\triangle AMF \sim \triangle CAB$ , $MF = MA \cdot \cfrac{AB}{AC} = \cfrac{5}{2}j \cdot \cfrac{3j}{4j} = \cfrac{15}{8}j$ .

Then $A_{AEDF} = A_{MAED} - A_{\triangle AMF} = \bigg(\cfrac{5}{2}j\bigg)^2 - \cfrac{1}{2} \cdot \cfrac{15}{8} j \cdot \cfrac{5}{2} j = \cfrac{125}{32}j^2 = 250$ , which solves to $j = \boxed{8}$ .

$(3j,4j,5j)$ is a Pythagorean triple $\implies \triangle{ABC}$ is a right triangle with $m\angle{A} = 90^{\circ}$ and $\cos(\theta) = \dfrac{3}{5}$

Using the law of cosines on $\triangle{BAM}$ with included $\angle{ABC} = \theta \implies$

$\overline{AM}^2 = 9j^2 + \dfrac{25}{4}j^2 - 2(3j)(\dfrac{5j}{2})(\dfrac{3}{5}) =$ $\dfrac{25}{4}j^2 \implies \overline{AM} = \dfrac{5}{2}j \implies$

$\overline{MF} = \overline{AF}\cos(\theta) = \dfrac{3}{5}\overline{AF} \implies$

$\overline{AF}^2 = \dfrac{25}{4}j^2 + \dfrac{9}{25}\overline{AF}^2 \implies \dfrac{16}{25}\overline{AF}^2 = \dfrac{25}{4}j^2 \implies \overline{AF} = \dfrac{25}{8}j$

$\implies \overline{MF} = \dfrac{3}{5}(\dfrac{25}{8})j = \dfrac{15}{8}j \implies A_{\triangle{AMF}} =$ $\dfrac{1}{2}(\dfrac{5j}{2})(\dfrac{15j}{8}) = \dfrac{75}{32}j$

$\implies A_{R} = A_{AEDF} = \dfrac{25}{4}j^2 - \dfrac{75}{32}j^2 = \dfrac{125}{32}j^2 = 250 \implies j^2 = 64 \implies j = \boxed{8}$ .