2007 AMC 10B Problem 7

All five sides are the same length; call it $s$. Two right angles next to each other suggest a square, so look for one.

1. Draw the segment from $C$ to $E$. It splits the pentagon into a four-sided piece $ABCE$ and a triangle $CDE$.
2. In $ABCE$, the angles at $A$ and $B$ are both $90^\circ$. So side $EA$ and side $BC$ both stand straight up from $AB$, on the same side.
3. Those two sides are both $s$ long. So $E$ and $C$ are the same height above $AB$, and $EC$ is parallel to $AB$ and also $s$ long. That makes $ABCE$ a square with side $s$.
4. So $CE = s$, and the square's corner at $E$ is $\angle AEC = 90^\circ$.
5. Now triangle $CDE$ has $CD = s$, $DE = s$ and $CE = s$. All three sides are equal, so it is equilateral and $\angle CED = 60^\circ$.
6. The pentagon's angle at $E$ is made of the two pieces:
$$\angle E = 90^\circ + 60^\circ = 150^\circ.$$

The answer is $\boxed{\textbf{(E)}\ 150}$.

Equal sides plus a pair of right angles is a strong hint that a square is hidden inside. One well-chosen diagonal shows it and leaves a triangle you also know completely. The angle sum of a pentagon alone would only tell you that the angles at $C$, $D$ and $E$ add to $360^\circ$, which is not enough.

Let every side have length $s$ and draw the diagonal $CE$.

Look at quadrilateral $ABCE$. It has $EA = AB = BC = s$ and right angles at $A$ and $B$. Two right angles at the ends of $AB$ force $EA \parallel BC$, and since those two sides are also equal, $ABCE$ is a parallelogram with a right angle and two adjacent sides equal, that is, a square. So $CE = s$ and $\angle AEC = 90^\circ$.

Now triangle $CDE$ has $CD = DE = CE = s$: it is equilateral, so $\angle CED = 60^\circ$.

The pentagon's angle at $E$ is made of these two pieces:
$$\angle E = \angle AEC + \angle CED = 90^\circ + 60^\circ = 150^\circ.$$

The answer is $\boxed{\textbf{(E)}\ 150}$.