AMC 10 · 2013 · #16
Grade 8 geometry-2dA triangle with vertices (6,5), (8,−3), and (9,1) is reflected about the line x=8 to create a second triangle. What is the area of the union of the two triangles?
Pick an answer.
AMC 10 2013 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 triangle has vertices $(6,5)$, $(8,-3)$, and $(9,1)$. Reflect it across the vertical line $x=8$ to get a second triangle. Find the area of the region covered by at least one of the two triangles (their union).
Givens: Triangle $T_1$ has vertices $A(6,5)$, $B(8,-3)$, $C(9,1)$; $T_2$ is the mirror image of $T_1$ across the vertical line $x=8$; Reflecting across $x=8$ sends a point $(x,y)$ to $(16-x,\,y)$; Answer choices: (A) $9$, (B) $\frac{28}{3}$, (C) $10$, (D) $\frac{31}{3}$, (E) $\frac{32}{3}$
Unknowns: The area of the union of the two triangles
Understand
Restated: A triangle has vertices $(6,5)$, $(8,-3)$, and $(9,1)$. Reflect it across the vertical line $x=8$ to get a second triangle. Find the area of the region covered by at least one of the two triangles (their union).
Givens: Triangle $T_1$ has vertices $A(6,5)$, $B(8,-3)$, $C(9,1)$; $T_2$ is the mirror image of $T_1$ across the vertical line $x=8$; Reflecting across $x=8$ sends a point $(x,y)$ to $(16-x,\,y)$; Answer choices: (A) $9$, (B) $\frac{28}{3}$, (C) $10$, (D) $\frac{31}{3}$, (E) $\frac{32}{3}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #16 Change Focus / Count the Complement, #4 Introduce a Variable, #7 Identify Subproblems
Everything lives on the coordinate plane, so Tool #1 (Draw a Diagram) is the anchor: plot the three points, reflect them across $x=8$, and the shape of the overlap becomes visible. Two triangles that overlap are hard to measure directly, so Tool #16 (Change Focus) reframes the union as "both triangles added together, minus the part counted twice": $\text{union} = \text{area}(T_1) + \text{area}(T_2) - \text{overlap}$. Each triangle's area is a small subproblem (Tool #7), and locating the one interior corner of the overlap is a Tool #4 job — name the two slanted top edges as lines and find where they cross. The reason this works is that a reflection copies area exactly, so the only real unknown left is the shared overlap.
Execute — Answer: E
8.G.A.3 Step 1 Reflect the three vertices
- Reflecting across the vertical line $x=8$ leaves the height $y$ alone and flips the horizontal distance to the line, so $(x,y)\to(16-x,\,y)$.
- Apply it to each vertex.
- $B$ sits on the line, so it does not move.
💡 Mirroring over a vertical line keeps how high a point is and only swaps how far left or right of the line it sits.
6.G.A.1 Step 2 Area of one triangle
- Use the coordinate area formula (the Shoelace formula) on $A(6,5)$, $B(8,-3)$, $C(9,1)$.
- A reflection never changes area, so the second triangle has this same area.
💡 The mirror image is congruent to the original, so both triangles cover exactly the same amount of space.
7.G.B.6 Step 3 Reframe as total minus overlap
- The union is everything covered by either triangle.
- Adding the two areas counts the shared middle twice, so subtract that overlap once.
- With both areas equal to $8$, the whole problem collapses to finding the overlap.
💡 When two shapes are glued together, the part they share gets double-counted, so you take it back out once.
6.G.A.3 Step 4 See the shared shape
- Check the slanted sides through $B$.
- Side $BA$ runs through $B(8,-3)$, $C'(7,1)$, $A(6,5)$ on one straight line, so the reflected corner $C'$ is exactly the midpoint of $BA$.
- Side $BC$ runs through $B(8,-3)$, $C(9,1)$, $A'(10,5)$, so $C$ is the midpoint of $BA'$.
- Both triangles open out from $B$ inside the same $\vee$ of two lines, so the overlap is the four-cornered region $B$, $C'$, $F$, $C$, where the two top edges cross at $F$.
💡 The reflected triangle sits in the very same corner at $B$, so the shared piece is just the lower part both tops agree on.
8.EE.C.8 Step 5 Find the top corner F
- The overlap's highest point $F$ is where the two top edges meet: edge $AC$ and its mirror edge $C'A'$.
- Write each as a line and set them equal.
- By symmetry they cross on the mirror line $x=8$.
💡 Where two lines meet is the single point that fits both equations at once.
7.G.B.6 Step 6 Overlap area, then the union
- The overlap $B,C',F,C$ is symmetric about $x=8$.
- Split it along the vertical segment $BF$, whose length is $\frac{7}{3}-(-3)=\frac{16}{3}$.
- Corner $C$ is $1$ unit right of the line and $C'$ is $1$ unit left, so each half-triangle has base $BF$ and width $1$.
- Add the two halves for the overlap, then subtract from $16$.
💡 A symmetric kite splits into two equal thin triangles, each easy to measure from the mirror line.
8.G.A.3 Reflecting across the vertical line $x=8$ leaves the height $y$ alone and flips 6.G.A.1 Use the coordinate area formula (the Shoelace formula) on $A(6,5)$, $B(8,-3)$, $ 7.G.B.6 The union is everything covered by either triangle. Adding the two areas counts 6.G.A.3 Check the slanted sides through $B$. Side $BA$ runs through $B(8,-3)$, $C'(7,1)$ 8.EE.C.8 The overlap's highest point $F$ is where the two top edges meet: edge $AC$ and i 7.G.B.6 The overlap $B,C',F,C$ is symmetric about $x=8$. Split it along the vertical seg Review
Reasonableness: Each triangle has area $8$, so the union must land between $8$ (if one sat entirely on the other) and $16$ (if they never overlapped). The overlap $\frac{16}{3}\approx 5.33$ is a real chunk but less than a whole triangle, giving union $\frac{32}{3}\approx 10.7$, comfortably inside that $8$-to-$16$ window. Cross-check the overlap with the Shoelace formula on $B(8,-3)$, $C(9,1)$, $F(8,\frac{7}{3})$, $C'(7,1)$: it also gives $\frac{16}{3}$, so union $=\frac{32}{3}$, choice (E).
Alternative: Tool #3 (Eliminate Possibilities): the union is strictly more than one triangle ($8$) and strictly less than two separate triangles ($16$), because the two copies clearly share the wedge at $B$ but do not coincide. Only choices $C=10$, $D=\frac{31}{3}\approx10.3$, and $E=\frac{32}{3}\approx10.7$ are even in the plausible upper range; a quick sketch shows the overlap is a fat kite (not tiny), pulling the union down toward the larger of these, which is $\frac{32}{3}$.
CCSS standards used (min grade 8)
8.G.A.3Describe the effect of dilations, translations, rotations, and reflections on coordinates (Reflecting each vertex across $x=8$ using the rule $(x,y)\to(16-x,y)$ to build the second triangle.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Computing one triangle's area as $8$ (and, by congruence, the reflected triangle's area) and the two thin overlap triangles.)7.G.B.6Solve real-world and mathematical problems involving area, surface area, and volume (Reframing the union as $16-\text{overlap}$ and combining the pieces into the final area.)6.G.A.3Draw polygons in the coordinate plane given coordinates for the vertices (Plotting the reflected points and identifying the overlap as the quadrilateral $B,C',F,C$.)8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Setting the two top-edge lines equal to locate their crossing point $F=(8,\tfrac{7}{3})$.)
⭐ To measure two overlapping copies, add both areas and subtract the shared middle once — a reflection keeps area unchanged, so all the work is just pinning down that shared piece.
⭐ To measure two overlapping copies, add both areas and subtract the shared middle once — a reflection keeps area unchanged, so all the work is just pinning down that shared piece.
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