AMC 8 · 2024 · #11
Grade 6 geometry-2dProblem
The coordinates of △ABC are A(5,7), B(11,7), and C(3,y), with y>7. The area of △ABC is 12. What is the value of y?
Pick an answer.
AMC 8 2024 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: Triangle $ABC$ has vertices $A(5,7)$, $B(11,7)$, and $C(3,y)$ with $y > 7$. Its area is $12$. Find the value of $y$.
Givens: Vertex coordinates: $A(5,7)$, $B(11,7)$, $C(3,y)$; Condition: $y > 7$ (so $C$ lies above the line through $A$ and $B$); Area of $\triangle ABC = 12$; Choices: (A) 8, (B) 9, (C) 10, (D) 11, (E) 12
Unknowns: The y-coordinate $y$ of vertex $C$
Understand
Restated: Triangle $ABC$ has vertices $A(5,7)$, $B(11,7)$, and $C(3,y)$ with $y > 7$. Its area is $12$. Find the value of $y$.
Givens: Vertex coordinates: $A(5,7)$, $B(11,7)$, $C(3,y)$; Condition: $y > 7$ (so $C$ lies above the line through $A$ and $B$); Area of $\triangle ABC = 12$; Choices: (A) 8, (B) 9, (C) 10, (D) 11, (E) 12
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #6 Guess and Check
Coordinates are given, so the very first move is #1 **Draw a Diagram** — plot the points and see what the triangle looks like. The picture instantly reveals that $A$ and $B$ share the same $y$-value, so side $AB$ is horizontal and makes a clean base. Next, #7 **Identify Subproblems** splits the work into two small questions: (i) base length, (ii) height. Finally, instead of reaching for algebra (#13), we use #6 **Guess and Check** — this is multiple choice, so we just plug each option into the area formula and see which one works.
Execute — Answer: D
5.G.A.2 Step 1 - Plot the three points on the coordinate plane.
- $A(5,7)$ and $B(11,7)$ both have $y = 7$, so they lie on the same horizontal line.
- $C(3,y)$ sits above that line (since $y > 7$) and a little to the left.
- The picture immediately suggests using $AB$ as the base of the triangle.
💡 Plotting points to visualize a figure is the very first thing students learn in the Grade 5 coordinate-plane standard.
6.NS.C.8 Step 2 - Now split the big question into smaller pieces.
- (i) How long is the base $AB$?
- Since $A$ and $B$ share the same $y$-coordinate, the segment is horizontal and its length is just the difference of the $x$-coordinates: $|11 - 5| = 6$ units.
💡 Finding the distance between two points that share a coordinate by subtracting the other coordinate is exactly the Grade 6 coordinate-distance standard.
6.NS.C.8 Step 3 - (ii) How tall is the triangle?
- The height is the perpendicular distance from $C$ down to the horizontal line $y = 7$ that contains $AB$.
- We don't need the $x$-coordinates here — only the difference in $y$: $|y - 7|$.
- The problem says $y > 7$, so we can drop the absolute value and get height $= y - 7$.
💡 The perpendicular distance from a point to a horizontal line is again a coordinate-difference, the same Grade 6 standard.
6.G.A.1 Step 4 - Apply the triangle-area formula: $\text{Area} = \tfrac{1}{2} \times \text{base} \times \text{height}$.
- Plug in base $= 6$, height $= y - 7$, and area $= 12$ to get $12 = \tfrac{1}{2} \cdot 6 \cdot (y-7) = 3(y-7)$.
- We've turned the geometry into a simple 'three times what equals twelve?' question.
💡 The triangle-area formula $\tfrac{1}{2}\cdot \text{base}\cdot \text{height}$ is a core Grade 6 geometry standard.
6.G.A.1 Step 5 - Instead of doing algebra, use #6 **Guess and Check** with the answer choices.
- (A) $y=8$: $3(8-7)=3$, too small.
- (C) $y=10$: $3(10-7)=9$, still small.
- (D) $y=11$: $3(11-7)=3\cdot 4 = 12$.
- **Exact match!** (E) $y=12$: $3(12-7)=15$, too big.
- So $y = 11$, choice **(D)**.
💡 Plugging each answer choice into the area formula is verification inside the same Grade 6 triangle-area standard.
5.G.A.2 Plot the three points on the coordinate plane. $A(5,7)$ and $B(11,7)$ both have 6.NS.C.8 Now split the big question into smaller pieces. (i) How long is the base $AB$? S 6.NS.C.8 (ii) How tall is the triangle? The height is the perpendicular distance from $C$ 6.G.A.1 Apply the triangle-area formula: $\text{Area} = \tfrac{1}{2} \times \text{base} 6.G.A.1 Instead of doing algebra, use #6 **Guess and Check** with the answer choices. (A Review
Reasonableness: Quick sanity check from the picture: for a triangle with base $6$ to have area $12$, the height must be $\tfrac{2 \times 12}{6} = 4$. So $C$ must sit exactly $4$ units above the line $y = 7$, giving $y = 7 + 4 = 11$. This also satisfies $y > 7$. The smallest choice (A) $y=8$ gives height $1$ and area $3$ (too small); the largest (E) $y=12$ gives height $5$ and area $15$ (too big). $y=11$ sits comfortably in between, which matches the size of the given area.
Alternative: We could also use #13 (Convert to Algebra): from $3(y-7)=12$, divide both sides by $3$ to get $y-7=4$, then add $7$ to get $y=11$. Same answer, but for an elementary student plugging the answer choices is more concrete and far less error-prone.
CCSS standards used (min grade 6)
5.G.A.2Represent real-world and mathematical problems by graphing points (Plotting the three vertices $A(5,7)$, $B(11,7)$, $C(3,y)$ on the coordinate plane to visualize the triangle.)6.NS.C.8Solve real-world problems by graphing points in all four quadrants (Computing the length of the horizontal base $|11-5|=6$ and the height $|y-7|=y-7$ as coordinate differences between points sharing a coordinate.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Applying $\text{Area} = \tfrac{1}{2} \times \text{base} \times \text{height}$ to set up $12 = 3(y-7)$ and verifying $y = 11$ by checking each answer choice.)
⭐ This AMC 8 problem only needs the Grade 6 triangle-area formula and coordinate distance you already know!
⭐ This AMC 8 problem only needs the Grade 6 triangle-area formula and coordinate distance you already know!
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