AMC 10 · 2025 · #15
Grade 8 geometry-2dIn the figure below, ABEF is a rectangle, AD⊥DE, AF=7, AB=1, and AD=5.
What is the area of △ABC?
Pick an answer.
AMC 10 2025 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 the figure, $ABEF$ is a rectangle with $AB=1$ and $AF=7$. A segment $\overline{AD}$ of length $5$ drops from $A$ down to a point $D$, with $\overline{AD}\perp\overline{DE}$. The slanted side $\overline{BE}$ of the rectangle crosses $\overline{AD}$ at a point $C$. Find the area of $\triangle ABC$.
Givens: $ABEF$ is a rectangle, so opposite sides are equal and every corner is a right angle; $AB=1$ and $AF=7$, hence the opposite side $BE=7$; $AD=5$ and $\overline{AD}\perp\overline{DE}$, so $\angle ADE=90^\circ$; $C$ is the point where side $\overline{BE}$ crosses $\overline{AD}$, so $A$, $C$, $D$ are collinear and $B$, $C$, $E$ are collinear; Answer choices: (A) $\tfrac{3}{8}$, (B) $\tfrac{4}{9}$, (C) $\tfrac{1}{8}\sqrt{13}$, (D) $\tfrac{7}{15}$, (E) $\tfrac{1}{8}\sqrt{15}$
Unknowns: The area of $\triangle ABC$
Understand
Restated: In the figure, $ABEF$ is a rectangle with $AB=1$ and $AF=7$. A segment $\overline{AD}$ of length $5$ drops from $A$ down to a point $D$, with $\overline{AD}\perp\overline{DE}$. The slanted side $\overline{BE}$ of the rectangle crosses $\overline{AD}$ at a point $C$. Find the area of $\triangle ABC$.
Givens: $ABEF$ is a rectangle, so opposite sides are equal and every corner is a right angle; $AB=1$ and $AF=7$, hence the opposite side $BE=7$; $AD=5$ and $\overline{AD}\perp\overline{DE}$, so $\angle ADE=90^\circ$; $C$ is the point where side $\overline{BE}$ crosses $\overline{AD}$, so $A$, $C$, $D$ are collinear and $B$, $C$, $E$ are collinear; Answer choices: (A) $\tfrac{3}{8}$, (B) $\tfrac{4}{9}$, (C) $\tfrac{1}{8}\sqrt{13}$, (D) $\tfrac{7}{15}$, (E) $\tfrac{1}{8}\sqrt{15}$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #7 Identify Subproblems, #1 Draw a Diagram
The area of $\triangle ABC$ needs the leg $BC$, and $BC$ is hidden inside the crossing point $C$. The crossing creates two triangles, $\triangle ABC$ and $\triangle EDC$, that share the vertical angle at $C$ and each have a right angle, so they are similar. Tool #4 (Introduce a Variable) is the engine: name the two unknown pieces $BC$ and $AC$, turn the similarity ratio and the two 'lengths add up' facts into a pair of linear equations, and solve. Tool #7 (Identify Subproblems) first splits off the smaller job of finding $DE$ with the Pythagorean theorem, which fixes the similarity ratio. Tool #1 (Draw a Diagram) keeps track of which segment lies on which line so the collinear sums are written correctly.
Execute — Answer: A
8.G.B.7 Step 1 Read the rectangle and its diagonal
- In rectangle $ABEF$ opposite sides are equal, so $BE=AF=7$.
- Draw the diagonal $\overline{AE}$.
- The corner at $B$ is a right angle, so $\triangle ABE$ is right-angled at $B$ with legs $AB=1$ and $BE=7$.
- The Pythagorean theorem gives the diagonal.
💡 A rectangle hands you a right triangle for free: two sides and the diagonal.
8.G.B.7 Step 2 Find DE
- Now look at $\triangle ADE$.
- It has a right angle at $D$ because $\overline{AD}\perp\overline{DE}$, with hypotenuse $\overline{AE}$ and one leg $AD=5$.
- The same diagonal $\overline{AE}=\sqrt{50}$ from Step 1 lets the Pythagorean theorem finish $DE$.
💡 Once you know a right triangle's hypotenuse and one leg, the other leg is forced.
8.G.A.5 Step 3 Spot the similar triangles
- At the crossing point $C$, compare $\triangle ABC$ and $\triangle EDC$.
- The angle at $B$ is a right angle (rectangle corner) and the angle at $D$ is a right angle (given $\overline{AD}\perp\overline{DE}$), so those match.
- The angles at $C$ are vertical angles, so they are equal too.
- Two equal angle pairs mean the triangles are similar by the angle-angle criterion.
- Matching the equal angles gives the correspondence $A\leftrightarrow E$, $B\leftrightarrow D$, $C\leftrightarrow C$, with ratio $AB:ED=1:5$.
💡 Two right angles plus a shared vertical angle is a guaranteed pair of similar triangles.
8.G.A.4 Step 4 Name the unknown pieces and build equations
- Let $BC=x$ and $AC=y$.
- In similar triangles matching sides are in the ratio $1:5$, so the sides of $\triangle EDC$ are $5$ times the matching sides of $\triangle ABC$: $DC=5x$ and $EC=5y$.
- Two collinearity facts finish the setup.
- Along $\overline{AD}$, the pieces $AC$ and $CD$ add to $AD=5$.
- Along $\overline{BE}$, the pieces $BC$ and $CE$ add to $BE=7$.
💡 The scale factor $5$ turns each unknown into its partner, so two segments become two equations.
8.EE.C.8 Step 5 Solve the system
- From the first equation $y=5-5x$.
- Substitute into the second equation and solve for $x$, which is the leg $BC$ we need.
💡 Substitution collapses two equations into one unknown you can just solve.
6.G.A.1 Step 6 Compute the area
- $\triangle ABC$ has its right angle at $B$, so its two legs $AB=1$ and $BC=\tfrac{3}{4}$ are the base and height.
- Area is half the product of the legs.
💡 For a right triangle the two legs are base and height, so area is just half their product.
8.G.B.7 In rectangle $ABEF$ opposite sides are equal, so $BE=AF=7$. Draw the diagonal $\ 8.G.B.7 Now look at $\triangle ADE$. It has a right angle at $D$ because $\overline{AD}\ 8.G.A.5 At the crossing point $C$, compare $\triangle ABC$ and $\triangle EDC$. The angl 8.G.A.4 Let $BC=x$ and $AC=y$. In similar triangles matching sides are in the ratio $1:5 8.EE.C.8 From the first equation $y=5-5x$. Substitute into the second equation and solve 6.G.A.1 $\triangle ABC$ has its right angle at $B$, so its two legs $AB=1$ and $BC=\tfra Review
Reasonableness: Back-substitute $x=\tfrac{3}{4}$: then $AC=y=5-5\cdot\tfrac{3}{4}=\tfrac{5}{4}$ and $EC=5y=\tfrac{25}{4}$. Check the $\overline{BE}$ length: $BC+EC=\tfrac{3}{4}+\tfrac{25}{4}=7=BE$, which matches. Also $\triangle ABC$ should satisfy the Pythagorean theorem: $AB^2+BC^2=1+\tfrac{9}{16}=\tfrac{25}{16}=\left(\tfrac{5}{4}\right)^2=AC^2$, consistent with $AC=\tfrac{5}{4}$. The area $\tfrac{3}{8}$ is a small, clean fraction, fitting a tiny triangle wedged in the corner near $B$, and it is choice (A).
Alternative: Coordinates avoid the similarity argument. Put $D=(0,0)$, $E=(5,0)$, $A=(0,5)$, so $\overline{AD}$ lies on the $y$-axis and $\overline{AD}\perp\overline{DE}$ automatically. The rectangle corner condition ($AB=1$, $BE=7$, right angle at $B$) places $B=(-\tfrac{3}{5},\tfrac{21}{5})$. Line $\overline{BE}$ then crosses the $y$-axis (that is, $\overline{AD}$) at $C=(0,\tfrac{15}{4})$. So $BC=\sqrt{\left(\tfrac{3}{5}\right)^2+\left(\tfrac{21}{5}-\tfrac{15}{4}\right)^2}=\tfrac{3}{4}$, matching the similarity answer, and the area is again $\tfrac{1}{2}\cdot 1\cdot\tfrac{3}{4}=\tfrac{3}{8}$.
CCSS standards used (min grade 8)
8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding the diagonal $AE=\sqrt{50}$ in right triangle $ABE$, then finding $DE=5$ in right triangle $ADE$.)8.G.A.5Use informal arguments to establish facts about angle sum and exterior angles (Using two right angles plus equal vertical angles at $C$ (angle-angle criterion) to conclude $\triangle ABC\sim\triangle EDC$.)8.G.A.4Understand that a two-dimensional figure is similar to another using transformations (Turning the $1:5$ similarity ratio into $DC=5\,BC$ and $EC=5\,AC$ so the unknown segments can be related.)8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Solving the system $y+5x=5$, $x+5y=7$ by substitution to get $BC=\tfrac{3}{4}$.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Computing the area of right triangle $ABC$ as half the product of its legs $AB$ and $BC$.)
⭐ When two lines cross, look for a pair of similar triangles from the equal vertical angles — name the mystery pieces, write two equations, and solve.
⭐ When two lines cross, look for a pair of similar triangles from the equal vertical angles — name the mystery pieces, write two equations, and solve.
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