AMC 10 · 2020 · #20
Grade 8 geometry-2dProblem
Quadrilateral ABCD satisfies ∠ABC=∠ACD=90∘,AC=20, and CD=30. Diagonals AC and BD intersect at point E, and AE=5. What is the area of quadrilateral ABCD?
Pick an answer.
AMC 10 2020 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: Quadrilateral $ABCD$ has two right angles: $\angle ABC = 90^\circ$ and $\angle ACD = 90^\circ$. The diagonal $AC$ has length $20$, and $CD = 30$. The two diagonals $AC$ and $BD$ meet at point $E$, with $AE = 5$. Find the area of $ABCD$.
Givens: $\angle ABC = 90^\circ$ — so $\triangle ABC$ is right-angled at $B$ with hypotenuse $AC$; $\angle ACD = 90^\circ$ — so $\triangle ACD$ is right-angled at $C$ with legs $AC$ and $CD$; $AC = 20$, $CD = 30$; Diagonals $AC$ and $BD$ meet at $E$ with $AE = 5$ (so $EC = 15$); Choices: (A) $330$, (B) $340$, (C) $350$, (D) $360$, (E) $370$
Unknowns: Area of quadrilateral $ABCD$
Understand
Restated: Quadrilateral $ABCD$ has two right angles: $\angle ABC = 90^\circ$ and $\angle ACD = 90^\circ$. The diagonal $AC$ has length $20$, and $CD = 30$. The two diagonals $AC$ and $BD$ meet at point $E$, with $AE = 5$. Find the area of $ABCD$.
Givens: $\angle ABC = 90^\circ$ — so $\triangle ABC$ is right-angled at $B$ with hypotenuse $AC$; $\angle ACD = 90^\circ$ — so $\triangle ACD$ is right-angled at $C$ with legs $AC$ and $CD$; $AC = 20$, $CD = 30$; Diagonals $AC$ and $BD$ meet at $E$ with $AE = 5$ (so $EC = 15$); Choices: (A) $330$, (B) $340$, (C) $350$, (D) $360$, (E) $370$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #13 Convert to Algebra
Tool #7 (Subproblems): the diagonal $AC$ cuts $ABCD$ into two right triangles $\triangle ACD$ and $\triangle ABC$. Compute each area separately and add. $\triangle ACD$ is immediate since both legs are given. $\triangle ABC$ needs the altitude from $B$ to $AC$ — call it $BF$, with $F$ on $AC$. Tool #1 (Diagram): draw $AC$ horizontally with $A$ on the right, $C$ at the origin, $D$ above $C$, and mark $E$ on $AC$. Since $BF$ and $CD$ are both perpendicular to $AC$, they are parallel, so $\triangle EBF \sim \triangle EDC$ by AA. Tool #13 (Algebra): let $EF = x$, use the similarity to express $BF$ in terms of $x$, then apply the right-triangle altitude theorem $BF^2 = AF \cdot FC$ to get $x$.
Execute — Answer: D
6.G.A.1 Step 1 - Compute the easy half: $\triangle ACD$ has a right angle at $C$ with legs $AC = 20$ and $CD = 30$.
- Area $= \tfrac{1}{2} \cdot 20 \cdot 30 = 300$.
💡 Right triangle with both legs given — half the rectangle.
5.G.A.1 Step 2 - Drop the altitude from $B$ to $AC$ and call its foot $F$.
- The area of $\triangle ABC$ becomes $\tfrac{1}{2} \cdot AC \cdot BF = 10 \cdot BF$.
- So we need $BF$.
- Since both $BF$ and $CD$ are perpendicular to line $AC$, they are parallel.
💡 Use the hypotenuse as base; altitude tells the height.
8.G.A.4 Step 3 - Now exploit the similar triangles around $E$.
- With $BF \parallel CD$ and $E$ on segment $BD$, triangles $\triangle EBF$ and $\triangle EDC$ share vertical angles at $E$ and have $\angle EBF = \angle EDC$ as alternate interior angles, so $\triangle EBF \sim \triangle EDC$ by AA.
- Corresponding sides give $\dfrac{BF}{CD} = \dfrac{EF}{EC}$.
💡 Parallel lines $+$ shared $E$ $=$ similar triangles, so $BF$ is double $EF$.
8.G.B.7 Step 4 - Let $EF = x$, so $BF = 2x$.
- Now use the altitude-on-hypotenuse theorem for the right triangle $\triangle ABC$: the altitude to the hypotenuse satisfies $BF^2 = AF \cdot FC$.
- With $F$ between $A$ and $E$ on segment $AC$ (the configuration consistent with the figure), $AF = AE - EF = 5 - x$ and $FC = FE + EC = x + 15$.
💡 Right-triangle altitude rule: altitude squared $=$ product of hypotenuse pieces.
8.EE.C.7 Step 5 - Expand and solve: $4x^2 = 75 + 5x - 15x - x^2 = 75 - 10x - x^2$.
- So $5x^2 + 10x - 75 = 0$, divide by $5$: $x^2 + 2x - 15 = 0$.
- Factor: $(x + 5)(x - 3) = 0$.
- Since $x = EF \ge 0$, take $x = 3$, giving $BF = 2x = 6$.
💡 Quadratic factors cleanly — take the positive root.
6.G.A.1 Step 6 - Now compute $[\triangle ABC] = 10 \cdot BF = 10 \cdot 6 = 60$.
- Add the two triangle areas: $[ABCD] = 300 + 60 = 360$.
- That's choice (D).
💡 Add the two right-triangle areas split by the diagonal.
6.G.A.1 Compute the easy half: $\triangle ACD$ has a right angle at $C$ with legs $AC = 5.G.A.1 Drop the altitude from $B$ to $AC$ and call its foot $F$. The area of $\triangle 8.G.A.4 Now exploit the similar triangles around $E$. With $BF \parallel CD$ and $E$ on 8.G.B.7 Let $EF = x$, so $BF = 2x$. Now use the altitude-on-hypotenuse theorem for the r 8.EE.C.7 Expand and solve: $4x^2 = 75 + 5x - 15x - x^2 = 75 - 10x - x^2$. So $5x^2 + 10x 6.G.A.1 Now compute $[\triangle ABC] = 10 \cdot BF = 10 \cdot 6 = 60$. Add the two trian Review
Reasonableness: Sanity-check the geometry: place $C = (0, 0)$, $A = (20, 0)$, $D = (0, 30)$, $E = (15, 0)$. The line $BD$ goes through $D$ and $E$ with slope $-2$. Drop the foot of the altitude $F = (18, 0)$; then $B = (18, -6)$ satisfies $BF = 6$ and the right angle at $B$: $\vec{BA} = (2, 6),\, \vec{BC} = (-18, 6)$ with dot product $-36 + 36 = 0$ ✓. Also $|AB| = \sqrt{40},\, |BC| = \sqrt{360}$, and $\tfrac{1}{2}\sqrt{40 \cdot 360} = \tfrac{1}{2}\sqrt{14400} = 60$ ✓. The total $300 + 60 = 360$ matches choice (D), and the nearby distractors $330, 340, 350, 370$ correspond to wrong $BF$ values.
Alternative: Tool #1 (Diagram) with coordinates: set $C=(0,0)$, $A=(20,0)$, $D=(0,30)$. $B$ lies on the circle with diameter $AC$ (so the angle at $B$ is $90^\circ$) and on line $DE$ through $D=(0,30)$ and $E=(15,0)$. Solve the two equations to get $B=(18, -6)$, then compute the area with the shoelace formula: $[ABCD] = \tfrac{1}{2}|x_A(y_B - y_D) + x_B(y_C - y_A) + x_C(y_D - y_B) + x_D(y_A - y_C)| = 360$. Same answer, no similar-triangle setup needed.
CCSS standards used (min grade 8)
6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Splitting $ABCD$ by diagonal $AC$ into two right triangles and adding their areas.)5.G.A.1Use a pair of perpendicular number lines forming a coordinate system (Setting up the altitude $BF$ perpendicular to $AC$ and recognizing $BF \parallel CD$.)8.G.A.4Understand that a two-dimensional figure is similar to another using transformations (Using AA similarity of $\triangle EBF$ and $\triangle EDC$ to express $BF = 2 \cdot EF$.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Applying the right-triangle altitude-to-hypotenuse relation $BF^2 = AF \cdot FC$.)8.EE.C.7Solve linear equations in one variable (Solving the quadratic $x^2 + 2x - 15 = 0$ to find $EF = 3$.)
⭐ This AMC 10 problem only needs Grade 8 similar-triangle and Pythagorean reasoning you already know — slice the quadrilateral with diagonal $AC$, get $300$ from the easy right triangle, find altitude $BF = 6$ via similar triangles, then add $60$. The answer is $\textbf{(D)} \; 360$.
⭐ This AMC 10 problem only needs Grade 8 similar-triangle and Pythagorean reasoning you already know — slice the quadrilateral with diagonal $AC$, get $300$ from the easy right triangle, find altitude $BF = 6$ via similar triangles, then add $60$. The answer is $\textbf{(D)} \; 360$.
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