Area of rectangle of the following image

Question

With the right triangle EMT has 5,12, and 13 of length.

Any help would be appreciated

Answer 1

By Euclid's theorem, we have: $$\overline{OT}=\frac{\overline{TM}^2}{\overline{EM}}=\frac{25}{13}$$ Let $\alpha=\angle{TEM}$, we have: $$\sin(\alpha)=\frac{5}{13} \land \cos(\alpha)=\frac{12}{13}$$ Now, the triangle $OAT$ is a right triangle with hypotenuse $OT$, and so: $$\overline{AT}=\overline{OT}\cdot\sin(\alpha)=\frac{125}{169}$$ Observing $ST$, we can say: $$\overline{ST}=\overline{ET}-\overline{ES}=\overline{ET}-\overline{OT}\cdot\cos(\alpha)=13-\frac{300}{169}=\frac{1897}{169}$$ Finally, the area of $AYST$ is: $$\overline{ST}\cdot\overline{AT}=\frac{125}{169}\cdot\frac{1897}{169}=\frac{237125}{28561}$$