2007 AMC 10B Problem 7

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 $\boxed{\textbf{(E)}\ 150}$.

Equal sides plus a couple of right angles is a strong hint that a square is hiding inside the figure; a well-chosen diagonal exposes it and leaves an equilateral triangle. Decomposing an irregular polygon into pieces you fully understand beats attacking its angle sum directly (which alone only tells you $C + D + E = 360^\circ$).

• Using the regular-pentagon angle 108, or assuming the three unknown angles are equal and answering 120.
• Correctly finding the square but then treating $CDE$ as a right or isosceles-right triangle instead of equilateral.