smc

Let us label the points on the diagram. By doing some angle chasing using the fact that $\angle ACE$ and $\angle CEG$ are right angles, we find that $\angle BAC\cong \angle DCE\cong \angle FEG$. Similarly, $\angle ACB\cong \angle CED\cong \angle EGF$. Therefore, $△ABC\sim △CDE\sim △EFG$. As we are given a rectangle and a square, $AB=4$ and $AC=5$. Therefore, $△ABC$ is a $3$-$4$-$5$ right triangle and $BC=3$. $CE$ is also $5$. So, using the similar triangles, $CD=4$ and $DE=3$. $EF=DF-DE=4-3=1$. Using the similar triangles again, $EF$ is $\frac{1}{4}$ of the corresponding $AB$. So, $$\begin{array}{rl}[△EFG]& ={\left(\frac{1}{4}\right)}^2\cdot [△ABC]\\ & =\frac{1}{16}\cdot 6\\ & =\frac{3}{8}.\end{array}$$ Finally, we have $$\begin{array}{rl}[ACEG]& =[ABDF]-[△ABC]-[△CDE]-[△EFG]\\ & =7\cdot 4-\frac{1}{2}\cdot 3\cdot 4-\frac{1}{2}\cdot 3\cdot 4-\frac{3}{8}\\ & =28-6-6-\frac{3}{8}\\ & =\boxed{15\frac{5}{8}}.\end{array}$$