AMC 10 · 2007 · #18
Grade 8 geometry-2dConsider the 12-sided polygon ABCDEFGHIJKL, as shown. Each of its sides has length 4, and each two consecutive sides form a right angle. Suppose that AG and CH meet at M. What is the area of quadrilateral ABCM?
Pick an answer.
AMC 10 2007 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: A staircase-shaped $12$-sided figure has every side of length $4$ and every corner a right angle. Two segments, $\overline{AG}$ and $\overline{CH}$, are drawn across it and cross at a point $M$. Find the area of the four-cornered region $ABCM$.
Givens: The polygon $ABCDEFGHIJKL$ has $12$ sides, each of length $4$; Every pair of consecutive sides meets at a right angle; Segments $\overline{AG}$ and $\overline{CH}$ cross at the point $M$; Answer choices: (A) $\frac{44}{3}$, (B) $16$, (C) $\frac{88}{5}$, (D) $20$, (E) $\frac{62}{3}$
Unknowns: The area of quadrilateral $ABCM$
Understand
Restated: A staircase-shaped $12$-sided figure has every side of length $4$ and every corner a right angle. Two segments, $\overline{AG}$ and $\overline{CH}$, are drawn across it and cross at a point $M$. Find the area of the four-cornered region $ABCM$.
Givens: The polygon $ABCDEFGHIJKL$ has $12$ sides, each of length $4$; Every pair of consecutive sides meets at a right angle; Segments $\overline{AG}$ and $\overline{CH}$ cross at the point $M$; Answer choices: (A) $\frac{44}{3}$, (B) $16$, (C) $\frac{88}{5}$, (D) $20$, (E) $\frac{62}{3}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #4 Introduce a Variable, #7 Identify Subproblems
Because every corner is a right angle, the whole figure fits perfectly on a grid, so Tool #1 (Draw a Diagram) turns each vertex into a pair of coordinates. Once the corners are numbers, Tool #4 (Introduce a Variable) lets me write each drawn segment as a line and solve for where they cross to pin down $M$. The region $ABCM$ has a slanted side, so Tool #7 (Identify Subproblems) rebuilds it as one big right triangle with a small right triangle cut off — two areas I can each find with base times height.
Execute — Answer: C
6.G.A.3 Step 1 Put the corners on a grid
- Since every angle is a right angle, each side runs straight across or straight up-and-down by $4$.
- Place the figure so that $H=(0,0)$.
- Reading the needed corners off the grid gives $A=(0,12)$, $B=(4,12)$, $C=(4,8)$, and $G=(4,0)$.
- Notice that $B$, $C$, and $G$ all sit on the same vertical line $x=4$, and $A$, $H$ sit on the line $x=0$.
💡 Right-angle corners mean every side is horizontal or vertical, so the figure drops straight onto grid points.
8.EE.B.6 Step 2 Turn the two segments into lines
- Write each drawn segment as a line using rise over run.
- Segment $\overline{AG}$ goes from $A=(0,12)$ down to $G=(4,0)$: it drops $12$ over a run of $4$, a slope of $-3$, so its line is $y = 12 - 3x$.
- Segment $\overline{CH}$ goes from $H=(0,0)$ up to $C=(4,8)$: it rises $8$ over a run of $4$, a slope of $2$, so its line is $y = 2x$.
💡 A straight segment keeps one steady slope, so rise over run between its ends is its whole rule.
8.EE.C.8 Step 3 Find where they cross
- The crossing point $M$ lies on both lines, so its height matches in both rules.
- Set them equal: $12 - 3x = 2x$.
- Adding $3x$ to each side gives $12 = 5x$, so $x = \frac{12}{5}$.
- Then $y = 2x = \frac{24}{5}$.
- So $M = \left(\frac{12}{5}, \frac{24}{5}\right)$.
💡 The one point on both lines is exactly where their two height rules agree.
6.G.A.1 Step 4 Rebuild ABCM as a big triangle minus a small one
- The corners $A$, $B$, $G$ form a right triangle whose slanted side $AG$ is one edge of our region.
- The line $CH$ slices through this triangle along $CM$, cutting off the little triangle $MCG$ near corner $G$ and leaving exactly the region $ABCM$.
- So $[ABCM] = [ABG] - [MCG]$.
- Both are right triangles resting against the vertical line $x=4$, so I can use base times height and halve.
💡 Cutting a clean triangle off a bigger triangle leaves the four-sided piece we want.
6.G.A.1 Step 5 Measure the two triangles
- Triangle $ABG$ has the vertical side $BG$ from $(4,12)$ to $(4,0)$ as its base, length $12$, and its far corner $A$ sits a horizontal distance $4$ away (from $x=0$ to $x=4$).
- So $[ABG] = \tfrac{1}{2}\cdot 12 \cdot 4 = 24$.
- Triangle $MCG$ has the vertical side $CG$ from $(4,8)$ to $(4,0)$ as its base, length $8$, and its far corner $M$ sits a horizontal distance $4 - \tfrac{12}{5} = \tfrac{8}{5}$ away.
- So $[MCG] = \tfrac{1}{2}\cdot 8 \cdot \tfrac{8}{5} = \tfrac{32}{5}$.
💡 For a triangle leaning on a vertical base, the height is just how far sideways the opposite corner reaches.
5.NF.A.1 Step 6 Subtract to get the area
- Take the big triangle and remove the small one: $[ABCM] = 24 - \tfrac{32}{5}$.
- Writing $24$ as $\tfrac{120}{5}$ gives $\tfrac{120}{5} - \tfrac{32}{5} = \tfrac{88}{5}$.
- So the area of quadrilateral $ABCM$ is $\tfrac{88}{5}$, which is choice (C).
💡 The leftover region is simply the whole triangle with the cut-off corner taken away.
6.G.A.3 Since every angle is a right angle, each side runs straight across or straight u 8.EE.B.6 Write each drawn segment as a line using rise over run. Segment $\overline{AG}$ 8.EE.C.8 The crossing point $M$ lies on both lines, so its height matches in both rules. 6.G.A.1 The corners $A$, $B$, $G$ form a right triangle whose slanted side $AG$ is one e 6.G.A.1 Triangle $ABG$ has the vertical side $BG$ from $(4,12)$ to $(4,0)$ as its base, 5.NF.A.1 Take the big triangle and remove the small one: $[ABCM] = 24 - \tfrac{32}{5}$. W Review
Reasonableness: The value $\frac{88}{5} = 17.6$ is one of the listed choices and it passes a size check: region $ABCM$ contains the full right triangle $ABC$ (legs $AB=4$ and $BC=4$, area $\tfrac{1}{2}\cdot4\cdot4 = 8$) plus the extra triangle $ACM$ below it, so the area must be comfortably more than $8$ but well under the $48$ of the whole rectangle $ABGH$ — and $17.6$ sits right in that band. A direct coordinate area check with the corners $A(0,12), B(4,12), C(4,8), M(\tfrac{12}{5},\tfrac{24}{5})$ using the shoelace formula also gives $\tfrac{88}{5}$.
Alternative: Split $ABCM$ the other way, along diagonal $AC$. Triangle $ABC$ is right-angled at $B$ with legs $4$ and $4$, giving area $8$. Triangle $ACM$ has $A=(0,12)$, $C=(4,8)$, $M=(\tfrac{12}{5},\tfrac{24}{5})$; the shoelace formula gives its area as $\tfrac{48}{5}$. Adding, $8 + \tfrac{48}{5} = \tfrac{40}{5} + \tfrac{48}{5} = \tfrac{88}{5}$, the same result.
CCSS standards used (min grade 8)
6.G.A.3Draw polygons in the coordinate plane given coordinates for the vertices (Placing the right-angled figure on a grid and reading off the corners $A, B, C, G, H$.)8.EE.B.6Use similar triangles to explain why the slope is the same between any two points (Writing each drawn segment $\overline{AG}$ and $\overline{CH}$ as a line using rise over run.)8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Solving $12 - 3x = 2x$ to locate the crossing point $M = (\tfrac{12}{5}, \tfrac{24}{5})$.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Rebuilding $ABCM$ as triangle $ABG$ minus triangle $MCG$ and finding each with base times height.)5.NF.A.1Add and subtract fractions with unlike denominators (Computing $24 - \tfrac{32}{5} = \tfrac{88}{5}$ for the final area.)
⭐ Drop a right-angled figure onto a grid, turn the crossing lines into slopes to find where they meet, then build the odd four-sided piece as one triangle with a smaller triangle cut off.
⭐ Drop a right-angled figure onto a grid, turn the crossing lines into slopes to find where they meet, then build the odd four-sided piece as one triangle with a smaller triangle cut off.
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