AMC 8 · 2024 · #11

Grade 6 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Axis-Aligned Figure Area from Coordinates

area-trianglescoordinate-geometry coordinate-geometry ↑ Prerequisites: area-trianglesmulti-digit-arithmetic

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

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.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

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.

1STEP 1

Plot A(5,7), B(11,7), C(3,y). Since A and B both sit at y = 7, side AB is horizontal — the natural base.

A(5,7), B(11,7), C(3,y) with y ＞ 7

2STEP 2

Side AB is horizontal, so its length is just the x-gap: |11 - 5| = 6 units.

base = |11 - 5| = 6

3STEP 3

Height is C's vertical distance to line y = 7; since y ＞ 7, height = y - 7.

height = |y - 7| = y - 7 ( ∵ y ＞ 7)

4STEP 4

Area = $\frac{1}{2}$ × base × height, so 12 = $\frac{1}{2}$ × 6 × (y - 7), which simplifies to 12 = 3(y - 7).

12 =

\frac{1}{2}

× 6 × (y-7) = 3(y-7)

5STEP 5

Test the choices: y=8 gives 3, y=10 gives 9, y=11 gives 3·4 = 12 ✓, y=12 gives 15. Answer: (D).

y=11 → 3(11-7)=12 ✓

Answer

Quick sanity check from the picture: for a triangle with base 6 to have area 12, the height must be

\frac{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.

💡Key takeaway

This AMC 8 problem only needs the Grade 6 triangle-area formula and coordinate distance you already know!

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