Given: $\square \mathrm{ABCD}$ and $\triangle \mathrm{AEB}$ An artist plans to paint an inside wall of a house as shown above. He measures the wall to have a width of 13 feet and that one slant the wall makes with the ceiling has a length of 12 feet. He also knows that EB is congruent to $B C$ and that the height of the wall is $16 \frac{8}{13}$ feet. What is the area of the wall he will be painting?

Step 1 ：Since $\triangle AEB$ is a right triangle and $EB = BC$, we can conclude that $\triangle AEB$ is a 5-12-13 right triangle. Therefore, $AE = 5$ feet.

Step 2 ：Let $F$ be the foot of the perpendicular drawn from $E$ to side $AB$. The distance from $E$ to side $AB$ is $AF$. By the similarity of triangles $AEF$ and $ABE$, we have $\frac{AF}{AE} = \frac{AE}{AB}$.

Step 3 ：Solving for $AF$, we find $AF = \frac{AE^2}{AB} = \frac{5^2}{13} = \frac{25}{13}$ feet.

Step 4 ：Since $AF$ is the height of $\triangle AEB$, the area of $\triangle AEB$ is $\frac{1}{2} \times AB \times AF = \frac{1}{2} \times 13 \times \frac{25}{13} = \frac{25}{2}$ square feet.

Step 5 ：The height of the wall is $16 \frac{8}{13}$ feet, so the area of the rectangle $ABCD$ is $AB \times AD = 13 \times 16 \frac{8}{13} = 216$ square feet.

Step 6 ：The area of the wall the artist will be painting is the area of the rectangle $ABCD$ minus the area of $\triangle AEB$, which is $216 - \frac{25}{2} = 203 \frac{1}{2}$ square feet.

Step 7 ：\(\boxed{203 \frac{1}{2}}\) square feet is the area of the wall the artist will be painting.