AMC 10 · 2019 · #7
Grade 6 geometry-2dProblem
Two lines with slopes 21 and 2 intersect at (2,2). What is the area of the triangle enclosed by these two lines and the line x+y=10  ?
Pick an answer.
AMC 10 2019 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: Two lines pass through the point $(2, 2)$ — one with slope $\dfrac{1}{2}$ and one with slope $2$. A third line is $x + y = 10$. These three lines bound a triangle; find its area.
Givens: Line 1: slope $\dfrac{1}{2}$, through $(2, 2)$; Line 2: slope $2$, through $(2, 2)$; Line 3: $x + y = 10$; Choices: (A) $4$, (B) $4\sqrt{2}$, (C) $6$, (D) $8$, (E) $6\sqrt{2}$
Unknowns: The area of the triangle bounded by the three lines
Understand
Restated: Two lines pass through the point $(2, 2)$ — one with slope $\dfrac{1}{2}$ and one with slope $2$. A third line is $x + y = 10$. These three lines bound a triangle; find its area.
Givens: Line 1: slope $\dfrac{1}{2}$, through $(2, 2)$; Line 2: slope $2$, through $(2, 2)$; Line 3: $x + y = 10$; Choices: (A) $4$, (B) $4\sqrt{2}$, (C) $6$, (D) $8$, (E) $6\sqrt{2}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
Tool #1: sketch the three lines and the triangle on a coordinate grid — this makes the three vertices visible. Tool #7 splits the job into three subproblems: (a) write the two line equations through $(2,2)$, (b) find where each meets $x+y=10$, (c) compute the area from the three vertices. Tool #3 confirms the answer matches a single choice.
Execute — Answer: C
5.G.A.1 Step 1 - Use point-slope form to write Line 1 and Line 2.
- Both pass through $(2, 2)$, so $y - 2 = \text{slope} \cdot (x - 2)$.
💡 Plug the known point and slope into $y = mx + b$ to find each line.
5.G.A.1 Step 2 - Find where Line 1 meets $x + y = 10$.
- Substitute $y = \tfrac{1}{2}x + 1$ into the third line: $x + \tfrac{1}{2}x + 1 = 10$, so $\tfrac{3}{2}x = 9$ and $x = 6$, giving $y = 4$.
- Vertex $B = (6, 4)$.
💡 Substitute one line's $y$ into the other line's equation to find the crossing.
5.G.A.1 Step 3 - Find where Line 2 meets $x + y = 10$.
- Substitute $y = 2x - 2$: $x + 2x - 2 = 10$, so $3x = 12$ and $x = 4$, giving $y = 6$.
- Vertex $C = (4, 6)$.
💡 Same substitution trick for the steeper line.
5.G.A.2 Step 4 - Draw the three vertices on a grid: $A = (2, 2)$, $B = (6, 4)$, $C = (4, 6)$.
- Plot and connect them.
💡 Three corners pinned down — now find the area.
4.MD.A.3 Step 5 - Enclose the triangle in the bounding square with corners $(2, 2), (6, 2), (6, 6), (2, 6)$ (side length $4$, area $16$).
- The three corners of the square that are NOT vertices of the triangle each lie at the tip of a right triangle whose legs are parts of the square's edges: $(6,2)$ corner has legs $4$ and $2$ (area $4$); $(6,6)$ corner has legs $2$ and $2$ (area $2$); $(2,6)$ corner has legs $2$ and $4$ (area $4$).
💡 Surround the tilted triangle with a square — then subtract the three right-triangle corners.
6.G.A.1 Step 6 Subtract the three corner triangles from the bounding square.
💡 Big square minus the three pointy pieces leaves the triangle's area.
4.NBT.A.2 Step 7 Area $= 6$ matches choice (C).
💡 Read off the matching choice.
5.G.A.1 Use point-slope form to write Line 1 and Line 2. Both pass through $(2, 2)$, so 5.G.A.1 Find where Line 1 meets $x + y = 10$. Substitute $y = \tfrac{1}{2}x + 1$ into th 5.G.A.1 Find where Line 2 meets $x + y = 10$. Substitute $y = 2x - 2$: $x + 2x - 2 = 10$ 5.G.A.2 Draw the three vertices on a grid: $A = (2, 2)$, $B = (6, 4)$, $C = (4, 6)$. Plo 4.MD.A.3 Enclose the triangle in the bounding square with corners $(2, 2), (6, 2), (6, 6) 6.G.A.1 Subtract the three corner triangles from the bounding square. 4.NBT.A.2 Area $= 6$ matches choice (C). Review
Reasonableness: Cross-check using base-and-height. The triangle is isosceles: $AB$ runs from $(2,2)$ to $(6,4)$, length $\sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$; $AC$ runs from $(2,2)$ to $(4,6)$, length $\sqrt{4 + 16} = 2\sqrt{5}$. Base $BC$ from $(6,4)$ to $(4,6)$, length $\sqrt{4 + 4} = 2\sqrt{2}$. The base $BC$ lies on $x + y = 10$. Distance from $A = (2,2)$ to the line $x + y = 10$ is $\dfrac{|2 + 2 - 10|}{\sqrt{2}} = \dfrac{6}{\sqrt{2}} = 3\sqrt{2}$. Area $= \tfrac{1}{2} \cdot 2\sqrt{2} \cdot 3\sqrt{2} = \tfrac{1}{2} \cdot 12 = 6$. ✓
Alternative: Tool #13 (Algebra) with the shoelace formula directly on the three vertices: Area $= \tfrac{1}{2}|x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)| = \tfrac{1}{2}|2(4 - 6) + 6(6 - 2) + 4(2 - 4)| = \tfrac{1}{2}|-4 + 24 - 8| = \tfrac{1}{2} \cdot 12 = 6$. Same answer.
CCSS standards used (min grade 6)
5.G.A.1Use a pair of perpendicular number lines forming a coordinate system (Writing each line's equation through $(2,2)$ and substituting to find intersection points.)5.G.A.2Represent real-world and mathematical problems by graphing points (Plotting the three vertices $A = (2,2)$, $B = (6,4)$, $C = (4,6)$ on the grid.)4.MD.A.3Apply area and perimeter formulas for rectangles in real-world problems (Computing the area of the $4 \times 4$ bounding square as $16$.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Decomposing the bounding square into the target triangle plus three right-triangle corners and subtracting to get the answer $6$.)4.NBT.A.2Read and write multi-digit whole numbers and compare using symbols (Matching the computed area to choice (C).)
⭐ This AMC 10 problem only needs Grade 6 area-by-pieces you already know: find the three triangle corners $(2,2)$, $(6,4)$, $(4,6)$, surround them with a $4 \times 4$ square (area $16$), then subtract the three right-triangle corners ($4 + 2 + 4 = 10$). The triangle's area is $16 - 10 = 6$.
⭐ This AMC 10 problem only needs Grade 6 area-by-pieces you already know: find the three triangle corners $(2,2)$, $(6,4)$, $(4,6)$, surround them with a $4 \times 4$ square (area $16$), then subtract the three right-triangle corners ($4 + 2 + 4 = 10$). The triangle's area is $16 - 10 = 6$.
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