AMC 10 · 2009 · #18
Grade 8 geometry-2dRectangle ABCD has AB=8 and BC=6. Point M is the midpoint of diagonal AC, and E is on AB with ME⊥AC. What is the area of △AME?
Pick an answer.
AMC 10 2009 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 sides are $AB=8$ and $BC=6$. Point $M$ is the middle of the diagonal $\overline{AC}$. Point $E$ sits on side $AB$ so that segment $\overline{ME}$ meets the diagonal at a right angle. Find the area of triangle $AME$.
Givens: $ABCD$ is a rectangle with $AB=8$ and $BC=6$; $M$ is the midpoint of diagonal $\overline{AC}$; $E$ lies on side $AB$; $\overline{ME}\perp\overline{AC}$; Answer choices: (A) $\frac{65}{8}$, (B) $\frac{25}{3}$, (C) $9$, (D) $\frac{75}{8}$, (E) $\frac{85}{8}$
Unknowns: The area of triangle $AME$
Understand
Restated: In rectangle $ABCD$ the sides are $AB=8$ and $BC=6$. Point $M$ is the middle of the diagonal $\overline{AC}$. Point $E$ sits on side $AB$ so that segment $\overline{ME}$ meets the diagonal at a right angle. Find the area of triangle $AME$.
Givens: $ABCD$ is a rectangle with $AB=8$ and $BC=6$; $M$ is the midpoint of diagonal $\overline{AC}$; $E$ lies on side $AB$; $\overline{ME}\perp\overline{AC}$; Answer choices: (A) $\frac{65}{8}$, (B) $\frac{25}{3}$, (C) $9$, (D) $\frac{75}{8}$, (E) $\frac{85}{8}$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #1 Draw a Diagram, #7 Identify Subproblems
The rectangle's corners are right angles, so tool #4 (Introduce a Variable) lets me drop the figure onto a coordinate grid, put corner $A$ at the origin, and name the unknown point $E=(e,0)$ on side $AB$. Tool #1 (Draw a Diagram) pins down where $M$ and $E$ sit so the right angle at $M$ is easy to see. Tool #7 (Identify Subproblems) splits the task into three clean pieces: first the length $AM$ from the diagonal, then the length $AE$ from the right angle, then the area from those two lengths.
Execute — Answer: D
6.G.A.3 Step 1 Place the rectangle on a grid
- Put $A=(0,0)$, $B=(8,0)$, $C=(8,6)$, $D=(0,6)$ so that $AB=8$ and $BC=6$ as required.
- The diagonal $\overline{AC}$ runs from $(0,0)$ to $(8,6)$, and its midpoint is the average of the endpoints, giving $M=(4,3)$.
- Since $E$ is on side $AB$, which lies along the $x$-axis, write $E=(e,0)$ for an unknown $e$.
💡 Coordinates turn lengths, midpoints, and right angles into exact numbers you can compute with.
8.G.B.7 Step 2 Get AM from the diagonal
- The diagonal $\overline{AC}$ is the hypotenuse of right triangle $ABC$, whose legs are $AB=8$ and $BC=6$.
- By the Pythagorean theorem $AC=\sqrt{8^2+6^2}=\sqrt{100}=10$.
- Because $M$ is the midpoint of $\overline{AC}$, the length $AM$ is half of that: $AM=5$.
💡 A rectangle's diagonal is just the hypotenuse of the 8-6-10 right triangle it cuts off.
8.G.A.5 Step 3 Get AE from the right angle
- Because $\overline{ME}\perp\overline{AC}$, triangle $AME$ has a right angle at $M$.
- It shares the corner angle at $A$ with the big right triangle $ABC$, which has its right angle at $B$.
- Two matching angles make the triangles similar ($AME\sim ABC$), so matching sides are in proportion: $\frac{AM}{AB}=\frac{AE}{AC}$.
- Solving, $AE=AC\cdot\frac{AM}{AB}=10\cdot\frac{5}{8}=\frac{25}{4}$, so $e=\frac{25}{4}$.
💡 Same corner angle plus a right angle in each triangle forces the same shape, so their sides scale together.
6.G.A.1 Step 4 Compute the area of triangle AME
- Take $AE$ as the base of triangle $AME$.
- It lies flat along the $x$-axis, so the height is just how high $M$ sits above that axis, namely its $y$-coordinate $3$.
- The area is $\tfrac12\cdot\text{base}\cdot\text{height}=\tfrac12\cdot\frac{25}{4}\cdot3=\frac{75}{8}$.
- This matches choice (D).
💡 With the base on the axis, the height is simply the vertex's distance up from it.
6.G.A.3 Put $A=(0,0)$, $B=(8,0)$, $C=(8,6)$, $D=(0,6)$ so that $AB=8$ and $BC=6$ as requ 8.G.B.7 The diagonal $\overline{AC}$ is the hypotenuse of right triangle $ABC$, whose le 8.G.A.5 Because $\overline{ME}\perp\overline{AC}$, triangle $AME$ has a right angle at $ 6.G.A.1 Take $AE$ as the base of triangle $AME$. It lies flat along the $x$-axis, so the Review
Reasonableness: The area $\frac{75}{8}=9.375$ is just above choice (C) $9$, which fits a triangle a bit larger than a $9$-unit reference. As a sanity check on the shape: triangle $AME$ is similar to $ABC$ with ratio $\frac{AM}{AB}=\frac{5}{8}$, so its area should be $\left(\frac{5}{8}\right)^2$ times $[\triangle ABC]=\tfrac12\cdot8\cdot6=24$. That gives $\frac{25}{64}\cdot24=\frac{75}{8}$, the same value, so the two methods agree.
Alternative: Skip coordinates and use the perpendicular directly. In right triangle $AME$ the angle at $A$ satisfies $\cos A=\frac{AB}{AC}=\frac{8}{10}=\frac{4}{5}$ (from triangle $ABC$). The side $AM=5$ is adjacent to this angle, so $AE=\frac{AM}{\cos A}=\frac{5}{4/5}=\frac{25}{4}$, and the other leg is $ME=AM\tan A=5\cdot\frac{6}{8}=\frac{15}{4}$. Then $[\triangle AME]=\tfrac12\cdot AM\cdot ME=\tfrac12\cdot5\cdot\frac{15}{4}=\frac{75}{8}$.
CCSS standards used (min grade 8)
6.G.A.3Draw polygons in the coordinate plane given coordinates for the vertices (Placing the rectangle's corners, the midpoint $M$, and the unknown point $E=(e,0)$ on a grid.)8.G.B.7Apply the Pythagorean Theorem to determine unknown side lengths in right triangles (Finding the diagonal $AC=10$ and hence $AM=5$.)8.G.A.5Use informal arguments to establish the angle-angle criterion for similarity of triangles (Showing $AME\sim ABC$ from the shared angle and the right angle, then setting up the proportion for $AE$.)6.G.A.1Find the area of triangles by composing or decomposing shapes (Computing $[\triangle AME]$ from base $AE$ and height $3$.)
⭐ Half the diagonal is $5$; the little right triangle is a shrunk copy of the big one, so scale $\frac{5}{8}$ gives $AE=\frac{25}{4}$ and area $\frac{75}{8}$.
⭐ Half the diagonal is $5$; the little right triangle is a shrunk copy of the big one, so scale $\frac{5}{8}$ gives $AE=\frac{25}{4}$ and area $\frac{75}{8}$.
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