AMC 10 · 2014 · #15
Grade 8 geometry-2dIn rectangle ABCD, DC=2⋅CB and points E and F lie on AB so that ED and FD trisect ∠ADC as shown. What is the ratio of the area of △DEF to the area of rectangle ABCD?
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 bottom side $DC$ is twice the vertical side $CB$. Two rays from corner $D$, namely $DE$ and $DF$, cut the right angle $\angle ADC$ into three equal pieces, and $E$ and $F$ land on the top side $AB$. Find how the area of triangle $DEF$ compares to the area of the whole rectangle.
Givens: $ABCD$ is a rectangle with $DC = 2\cdot CB$.; $E$ and $F$ lie on side $\overline{AB}$.; $\overline{ED}$ and $\overline{FD}$ trisect the right angle $\angle ADC$.
Unknowns: The ratio $\dfrac{[\triangle DEF]}{[ABCD]}$ of the triangle's area to the rectangle's area.
Understand
Restated: In rectangle $ABCD$ the bottom side $DC$ is twice the vertical side $CB$. Two rays from corner $D$, namely $DE$ and $DF$, cut the right angle $\angle ADC$ into three equal pieces, and $E$ and $F$ land on the top side $AB$. Find how the area of triangle $DEF$ compares to the area of the whole rectangle.
Givens: $ABCD$ is a rectangle with $DC = 2\cdot CB$.; $E$ and $F$ lie on side $\overline{AB}$.; $\overline{ED}$ and $\overline{FD}$ trisect the right angle $\angle ADC$.
Plan
Primary tool: #1 Draw a Diagram
Secondary: #4 Introduce a Variable, #7 Identify Subproblems, #3 Eliminate Possibilities
This is a positions-and-shapes problem, so tool #1 (Draw a Diagram) leads: put $D$ at the origin with $DC$ along the x-axis, which turns the trisected right angle into clean $30^\circ$ rays and makes $E$ and $F$ easy to pin down. Because only a ratio of areas is asked, tool #4 (Introduce a Variable) lets us fix the short side $CB=1$ once and for all. Then tool #7 (Identify Subproblems) splits the work into three bites: locate $E$ and $F$ on the top edge, find the area of $\triangle DEF$, and divide by the rectangle's area.
Execute — Answer: A
4.MD.C.7 Step 1 Set coordinates and split the right angle
- Put $D$ at the origin with $DC$ running right along the x-axis and $DA$ running up along the y-axis, so $\angle ADC = 90^\circ$.
- Since only a ratio of areas is wanted, choose the short side $CB = 1$; then $DC = 2$ and the top side $AB$ is the line at height $1$.
- The two rays trisect the $90^\circ$ corner into three equal $30^\circ$ angles, so measured up from $DA$ the ray $DE$ is at $30^\circ$ and $DF$ is at $60^\circ$.
💡 A trisected right angle is just three copies of $30^\circ$, so every ray's tilt is known exactly.
8.G.B.7 Step 2 Locate E and F on the top edge
- Drop from $E$ and $F$ straight down to the vertical side $DA$; the foot is $A$, since $A$ is the top-left corner at height $1$.
- Triangle $DAE$ has the right angle at $A$, angle $30^\circ$ at $D$, and $DA = 1$, making it a $30$-$60$-$90$ triangle whose leg ratios are $1 : \sqrt{3} : 2$.
- So the horizontal leg is $AE = \dfrac{DA}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}$.
- Triangle $DAF$ has angle $60^\circ$ at $D$, so its horizontal leg is the long one: $AF = DA\cdot\sqrt{3} = \sqrt{3}$.
💡 A $30$-$60$-$90$ triangle has fixed side ratios $1:\sqrt{3}:2$, so one known leg pins down the rest.
6.G.A.1 Step 3 Area of triangle DEF
- Both $E$ and $F$ sit on the top edge at height $1$, so segment $EF$ is horizontal and its length is the gap between the two feet: $EF = AF - AE = \sqrt{3} - \dfrac{\sqrt{3}}{3} = \dfrac{2\sqrt{3}}{3}$.
- Use $EF$ as the base of $\triangle DEF$; the height is the perpendicular distance from $D$ (at height $0$) up to that top edge, which is $1$.
- So the area is one-half base times height.
💡 With the base flat on the top edge, the triangle's height is simply the rectangle's height.
6.RP.A.1 Step 4 Compare to the rectangle and pick the choice
- The rectangle has area $DC \cdot CB = 2 \cdot 1 = 2$.
- Divide the triangle's area by the rectangle's area: $\dfrac{\sqrt{3}/3}{2} = \dfrac{\sqrt{3}}{6}$.
- Matching this against the five choices, it is exactly choice (A); the others do not equal $\dfrac{\sqrt{3}}{6}$.
- So the answer is (A).
💡 A ratio of two areas ignores the chosen size, so the single number $\sqrt{3}/6$ is the whole answer.
4.MD.C.7 Put $D$ at the origin with $DC$ running right along the x-axis and $DA$ running 8.G.B.7 Drop from $E$ and $F$ straight down to the vertical side $DA$; the foot is $A$, 6.G.A.1 Both $E$ and $F$ sit on the top edge at height $1$, so segment $EF$ is horizonta 6.RP.A.1 The rectangle has area $DC \cdot CB = 2 \cdot 1 = 2$. Divide the triangle's area Review
Reasonableness: The number $\dfrac{\sqrt{3}}{6}\approx 0.29$ is a believable fraction of the rectangle: the triangle spans from $E$ at $x=\tfrac{\sqrt3}{3}\approx0.58$ to $F$ at $x=\sqrt3\approx1.73$ across a rectangle that runs from $0$ to $2$, so it covers a bit more than a quarter — consistent with $0.29$. The answer also stays the same no matter what value we pick for $CB$, since choosing $CB=1$ was a free scaling choice for a ratio. Note this leans on the $30$-$60$-$90$ side ratios, which is early high-school geometry sitting at the top of the K-8 range.
Alternative: Skip coordinates and use trigonometric tangents directly: with $DA = 1$, $AE = \tan 30^\circ = \tfrac{\sqrt3}{3}$ and $AF = \tan 60^\circ = \sqrt3$, giving $EF = \tfrac{2\sqrt3}{3}$ and $[\triangle DEF] = \tfrac12\cdot\tfrac{2\sqrt3}{3}\cdot 1 = \tfrac{\sqrt3}{3}$. Dividing by the rectangle area $2$ again yields $\tfrac{\sqrt3}{6}$, confirming (A).
CCSS standards used (min grade 8)
4.MD.C.7Recognize angle measure as additive; solve addition problems to find unknown angles (Splitting the trisected right angle into three equal $30^\circ$ angles ($90 = 30+30+30$).)8.G.B.7Apply the Pythagorean Theorem to determine unknown side lengths in right triangles (Using the $30$-$60$-$90$ side ratios $1:\sqrt{3}:2$ to find the horizontal legs $AE$ and $AF$.)6.G.A.1Find the area of triangles and other polygons by composing or decomposing shapes (Computing $[\triangle DEF] = \tfrac12 \cdot EF \cdot \text{height}$ with base $EF$ on the top edge.)6.RP.A.1Understand the concept of a ratio and use ratio language (Forming the ratio of the triangle's area to the rectangle's area and reading off the answer.)
⭐ Anchor the corner at the origin, use the fixed $1:\sqrt{3}:2$ ratios of a $30$-$60$-$90$ triangle to place the points, then compare the triangle's area to the rectangle's.
⭐ Anchor the corner at the origin, use the fixed $1:\sqrt{3}:2$ ratios of a $30$-$60$-$90$ triangle to place the points, then compare the triangle's area to the rectangle's.
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