AMC 10 · 2014 · #22
Grade 8 geometry-2dIn rectangle ABCD, AB=20 and BC=10. Let E be a point on CD such that ∠CBE=15∘. What is AE?
Pick an answer.
AMC 10 2014 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 rectangle $ABCD$ the long sides $AB$ and $CD$ measure $20$ and the short sides $BC$ and $AD$ measure $10$. Point $E$ sits on the top side $\overline{CD}$ so that $\angle CBE=15^\circ$. Find the length of $\overline{AE}$.
Givens: $ABCD$ is a rectangle with $\overline{AB}=20$ and $\overline{BC}=10$; $E$ lies on side $\overline{CD}$; $\angle CBE=15^\circ$, measured at vertex $B$ between $\overline{BC}$ and $\overline{BE}$; Answer choices: (A) $\dfrac{20\sqrt3}{3}$, (B) $10\sqrt3$, (C) $18$, (D) $11\sqrt3$, (E) $20$
Unknowns: The distance $\overline{AE}$ from corner $A$ to the point $E$ on the far side
Understand
Restated: In rectangle $ABCD$ the long sides $AB$ and $CD$ measure $20$ and the short sides $BC$ and $AD$ measure $10$. Point $E$ sits on the top side $\overline{CD}$ so that $\angle CBE=15^\circ$. Find the length of $\overline{AE}$.
Givens: $ABCD$ is a rectangle with $\overline{AB}=20$ and $\overline{BC}=10$; $E$ lies on side $\overline{CD}$; $\angle CBE=15^\circ$, measured at vertex $B$ between $\overline{BC}$ and $\overline{BE}$; Answer choices: (A) $\dfrac{20\sqrt3}{3}$, (B) $10\sqrt3$, (C) $18$, (D) $11\sqrt3$, (E) $20$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #4 Introduce a Variable, #3 Eliminate Possibilities
The whole problem lives in one picture, so Tool #1 (Draw a Diagram) is the anchor: drawing the rectangle and dropping $E$ onto the top side reveals two right triangles hiding inside. Tool #7 (Identify Subproblems) splits the work cleanly — first the small right triangle $BCE$ that hides the $15^\circ$ angle, then the right triangle $ADE$ that actually contains $\overline{AE}$. Tool #4 (Introduce a Variable) names the unknown gap $CE$ so the $15^\circ$ angle can pin it down. Tool #3 (Eliminate Possibilities) closes it out by matching the clean final length to one of the five listed choices.
Execute — Answer: E
6.NS.C.8 Step 1 Set coordinates and spot two right triangles
- Lay the rectangle on the axes with $A=(0,0)$, $B=(20,0)$, $C=(20,10)$, and $D=(0,10)$.
- Then $E$ is somewhere on the top side, at $(x,10)$.
- Two right triangles appear: triangle $BCE$ has its right angle at $C$ (a corner of the rectangle), with vertical leg $\overline{BC}=10$ and horizontal leg $\overline{CE}$; triangle $ADE$ has its right angle at $D$, with legs $\overline{AD}=10$ and $\overline{DE}$.
- The segment $\overline{AE}$ we want is the hypotenuse of that second triangle.
💡 Pinning the rectangle to coordinate axes turns "which corner" into exact points you can measure between.
8.G.A.4 Step 2 Use the 15 degree angle to find CE and DE
- In right triangle $BCE$ the $15^\circ$ angle sits at $B$, with the leg $\overline{CE}$ opposite it and the known leg $\overline{BC}=10$ next to it.
- Every right triangle with a $15^\circ$ angle has the same shape, so its two legs are locked in a fixed ratio: the opposite leg is $\tan 15^\circ=2-\sqrt3$ times the adjacent leg.
- That gives $CE=10(2-\sqrt3)=20-10\sqrt3$.
- Since the whole top side $\overline{CD}=20$, the leftover piece is $DE=20-CE=20-(20-10\sqrt3)=10\sqrt3$.
💡 Two right triangles that share the same acute angle are similar, so the 15 degree angle alone fixes how far $E$ sits from $C$.
8.G.B.7 Step 3 Pythagorean theorem on triangle ADE
- Now work in right triangle $ADE$, whose right angle is at $D$.
- Its legs are $\overline{AD}=10$ and $\overline{DE}=10\sqrt3$, and $\overline{AE}$ is the hypotenuse.
- Square each leg and add: $AE^2=AD^2+DE^2=10^2+(10\sqrt3)^2=100+300=400$.
💡 The straight distance across a right corner is found by squaring the two legs and adding, straight from the Pythagorean theorem.
8.EE.A.2 Step 4 Take the square root and match a choice
- Take the square root of both sides.
- Because $400$ is a perfect square, the length comes out clean: $AE=\sqrt{400}=20$.
- Scanning the five options, that is exactly choice (E).
- None of the other choices equal $20$, so the answer is $\textbf{(E)}\ 20$.
💡 A square root just undoes the squaring, and $400$ being a perfect square makes the final length land on a whole number.
6.NS.C.8 Lay the rectangle on the axes with $A=(0,0)$, $B=(20,0)$, $C=(20,10)$, and $D=(0 8.G.A.4 In right triangle $BCE$ the $15^\circ$ angle sits at $B$, with the leg $\overlin 8.G.B.7 Now work in right triangle $ADE$, whose right angle is at $D$. Its legs are $\ov 8.EE.A.2 Take the square root of both sides. Because $400$ is a perfect square, the lengt Review
Reasonableness: The answer $20$ has to be believable as a slanted length inside the rectangle. It is longer than the short side $\overline{AD}=10$ and shorter than the full diagonal $\overline{AC}=\sqrt{20^2+10^2}=\sqrt{500}\approx22.4$, so $20$ fits the window. It also fits the picture: $CE=20-10\sqrt3\approx2.7$ is small, so $E$ sits close to $C$, which means $\overline{AE}$ should be nearly as long as the corner-to-corner diagonal $\overline{AC}\approx22.4$ — and $20$ is indeed just under it. Everything is consistent, and the answer is (E).
Alternative: Skip the square-root arithmetic. Once you have $DE=10\sqrt3$ and $AD=10$, triangle $ADE$ has legs in the ratio $1:\sqrt3$ — the signature of a $30\text{-}60\text{-}90$ triangle, whose hypotenuse is always twice the shorter leg. So $AE=2\cdot10=20$ directly. Either route relies on the single special fact $\tan15^\circ=2-\sqrt3$, which itself follows from the tangent subtraction formula applied to $15^\circ=45^\circ-30^\circ$.
CCSS standards used (min grade 8)
6.NS.C.8Solve real-world problems by graphing points in all four quadrants (Placing the rectangle's corners at exact coordinates so the two right triangles and their legs become measurable.)8.G.A.4Understand that a two-dimensional figure is similar to another using transformations (Recognizing that the $15^\circ$ angle fixes the shape of right triangle $BCE$, so its legs are in the constant ratio $\tan15^\circ=2-\sqrt3$, giving $CE=20-10\sqrt3$ and $DE=10\sqrt3$.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Combining the legs $AD=10$ and $DE=10\sqrt3$ of right triangle $ADE$ to get $AE^2=400$.)8.EE.A.2Use square root and cube root symbols to represent solutions (Taking $AE=\sqrt{400}=20$ to reach a whole-number length and match choice (E).)
⭐ The $15^\circ$ angle sets $CE=20-10\sqrt3$, so $DE=10\sqrt3$; then the right corner at $D$ makes $AE=\sqrt{10^2+(10\sqrt3)^2}=\sqrt{400}=20$.
⭐ The $15^\circ$ angle sets $CE=20-10\sqrt3$, so $DE=10\sqrt3$; then the right corner at $D$ makes $AE=\sqrt{10^2+(10\sqrt3)^2}=\sqrt{400}=20$.
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