AMC 10 · 2007 · #7
Grade 8 geometry-2dAll sides of the convex pentagon ABCDE are of equal length, and ∠A=∠B=90∘. What is the degree measure of ∠E?
Pick an answer.
AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: In a convex pentagon $ABCDE$, all five sides have the same length, and the angles at $A$ and $B$ are both $90^\circ$. Find the degree measure of the angle at $E$.
Givens: $ABCDE$ is convex, with vertices in order; All five sides $AB$, $BC$, $CD$, $DE$, $EA$ have equal length; $\angle A = 90^\circ$ and $\angle B = 90^\circ$; Answer choices: (A) $90$, (B) $108$, (C) $120$, (D) $144$, (E) $150$
Unknowns: The degree measure of $\angle E$, the interior angle at vertex $E$
Understand
Restated: In a convex pentagon $ABCDE$, all five sides have the same length, and the angles at $A$ and $B$ are both $90^\circ$. Find the degree measure of the angle at $E$.
Givens: $ABCDE$ is convex, with vertices in order; All five sides $AB$, $BC$, $CD$, $DE$, $EA$ have equal length; $\angle A = 90^\circ$ and $\angle B = 90^\circ$; Answer choices: (A) $90$, (B) $108$, (C) $120$, (D) $144$, (E) $150$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems
Angles are hard to see in words but easy to see in a picture, so Tool #1 (Draw a Diagram) comes first: sketch $ABCDE$ and the two right angles at $A$ and $B$. That drawing exposes two familiar shapes hiding inside — a square and an equilateral triangle. Then Tool #7 (Identify Subproblems) finishes the job: the angle at $E$ is not one angle but two angles stacked together (a corner of the square plus a corner of the triangle), so find each piece separately and add them.
Execute — Answer: E
4.G.A.1 Step 1 Draw it and find a square
- Place side $AB$ flat, with side length $1$.
- At $A$ the angle is $90^\circ$, so side $AE$ turns straight up from $A$.
- At $B$ the angle is $90^\circ$, so side $BC$ turns straight up from $B$, the same way.
- Now $AE$ and $BC$ are both perpendicular to $AB$, both length $1$, and pointing the same direction.
- That makes $A$, $B$, $C$, $E$ the four corners of a unit square $ABCE$.
💡 Two equal sides both standing at right angles on the same base close up into a square.
4.G.A.2 Step 2 Spot the equilateral triangle
- The square $ABCE$ has a fourth side $CE$ joining the two top corners, and in a square every side is equal, so $CE = 1$ as well.
- The pentagon's remaining vertex is $D$, with sides $CD = DE = 1$.
- So triangle $CDE$ has all three sides equal to $1$: $CD = DE = EC = 1$.
- That is an equilateral triangle sitting on top of the square.
💡 Three sides all the same length can only make an equilateral triangle.
8.G.A.5 Step 3 Each triangle angle is 60
- In any triangle the three angles add up to $180^\circ$.
- An equilateral triangle has three equal angles, so each one is $180^\circ \div 3 = 60^\circ$.
- In particular the triangle's angle at $E$, namely $\angle DEC$, is $60^\circ$.
💡 Equal sides force equal angles, and three equal angles splitting $180^\circ$ are $60^\circ$ each.
4.MD.C.7 Step 4 Add the two pieces at E
- The pentagon's angle at $E$ is the angle $\angle DEA$ between sides $ED$ and $EA$.
- The diagonal $EC$ lies inside that angle and splits it into two parts: $\angle DEC$ from the equilateral triangle, and $\angle CEA$, which is a corner of the square and equals $90^\circ$.
- Add them: $\angle E = \angle DEC + \angle CEA = 60^\circ + 90^\circ = 150^\circ$.
- That is choice (E).
- Choice (A) $90$ forgets the triangle, (C) $120$ doubles the $60^\circ$, and (B) $108$ is the angle of a regular pentagon — a trap for equal sides but unequal angles.
💡 Two angles sharing a ray combine by simple addition, just like joining two slices of a pie.
4.G.A.1 Place side $AB$ flat, with side length $1$. At $A$ the angle is $90^\circ$, so s 4.G.A.2 The square $ABCE$ has a fourth side $CE$ joining the two top corners, and in a s 8.G.A.5 In any triangle the three angles add up to $180^\circ$. An equilateral triangle 4.MD.C.7 The pentagon's angle at $E$ is the angle $\angle DEA$ between sides $ED$ and $EA Review
Reasonableness: Check against the total: the interior angles of any pentagon add to $540^\circ$. Here $\angle A = \angle B = 90^\circ$ and, by the left-right symmetry of the figure, $\angle C = \angle E$ while $\angle D = \angle CDE = 60^\circ$ (the triangle's top). So $90 + 90 + 60 + 2\angle E = 540$, giving $2\angle E = 300$ and $\angle E = 150^\circ$, matching the answer. It is also convex ($150^\circ < 180^\circ$) as required, and larger than $90^\circ$, which fits the picture where side $ED$ leans well past straight up.
Alternative: Use coordinates instead of shape-spotting. Put $A=(0,0)$, $B=(1,0)$, then $E=(0,1)$ and $C=(1,1)$ from the right angles, and $D=(\tfrac12, 1+\tfrac{\sqrt3}{2})$ as the apex of the equilateral triangle on $CE$. The vector $EA=(0,-1)$ points at $-90^\circ$ and $ED=(\tfrac12,\tfrac{\sqrt3}{2})$ points at $60^\circ$; the angle between them is $60^\circ-(-90^\circ)=150^\circ$.
CCSS standards used (min grade 8)
4.G.A.1Draw points, lines, line segments, rays, angles, and identify in figures (Using the right angles at $A$ and $B$ to see that $AE$ and $BC$ are perpendicular to $AB$, forming the square $ABCE$.)4.G.A.2Classify two-dimensional figures based on presence of parallel or perpendicular lines (Recognizing $\triangle CDE$ as equilateral because its three sides $CD$, $DE$, $EC$ are all equal to the common side length $1$.)8.G.A.5Use informal arguments to establish facts about angle sum and exterior angles (Using the triangle angle sum $180^\circ$ to conclude each angle of equilateral $\triangle CDE$ is $60^\circ$.)4.MD.C.7Recognize angle measure as additive and solve addition and subtraction problems (Adding the square's corner $\angle CEA = 90^\circ$ and the triangle's $\angle DEC = 60^\circ$ to get $\angle E = 150^\circ$.)
⭐ Two right angles on equal sides build a square, the leftover equal sides build an equilateral triangle, and the corner angle at $E$ is just those two pieces added: $90^\circ + 60^\circ = 150^\circ$.
⭐ Two right angles on equal sides build a square, the leftover equal sides build an equilateral triangle, and the corner angle at $E$ is just those two pieces added: $90^\circ + 60^\circ = 150^\circ$.
More like this
Same archetype — closest grade level first.
