AMC 10 · 2019 · #7

Grade 6 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Line Intersection Vertices then Area

coordinate-geometryslope-interceptlinear-equations-two-vararea-trianglesisosceles-triangle coordinate-geometryidentify-subproblems ↑ Prerequisites: coordinate-geometryslope-interceptarea-triangles

📏 Medium solution 💡 3 insights

Problem

Two lines pass through the point (2, 2) — one with slope

\frac{1}{2}

and one with slope 2. A third line is x + y = 10. These three lines bound a triangle; find its area.

Pick an answer.

(A)

(B)

$4\sqrt{2}$

(C)

(D)

(E)

$6\sqrt{2}$

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

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.

1STEP 1

Both lines pass through (2, 2), so point-slope gives Line 1: y = $\frac{1}{2}$ x + 1 and Line 2: y = 2x - 2.

Line 1: y =

\frac{1}{2}

x + 1, Line 2: y = 2x - 2

2STEP 2

Substitute Line 1 into x + y = 10: $\frac{3}{2}$ x = 9, so x = 6 and y = 4, giving vertex B = (6, 4).

x +

\frac{1}{2}

x + 1 = 10 →

\frac{3}{2}

x = 9 → x = 6, y = 4

3STEP 3

Substitute Line 2 into x + y = 10: 3x = 12, so x = 4 and y = 6, giving vertex C = (4, 6).

x + 2x - 2 = 10 → 3x = 12 → x = 4, y = 6

4STEP 4

Plot the three vertices A = (2, 2), B = (6, 4), C = (4, 6) on a grid and connect them.

A = (2, 2), B = (6, 4), C = (4, 6)

5STEP 5

Box the triangle in the 4-by-4 square from (2, 2) to (6, 6): area 16, leaving three right-triangle corners of areas 4, 2, and 4.

Bounding square area = 4 · 4 = 16

6STEP 6

Subtract the three corner triangles from 16: 16 - 4 - 2 - 4 = 6.

Area = 16 -

\frac{1}{2}

(4 · 2) -

\frac{1}{2}

(2 · 2) -

\frac{1}{2}

(2 · 4) = 16 - 4 - 2 - 4 = 6

7STEP 7

Area 6 is choice (C).

6 → (C)

Answer

Cross-check using base-and-height. The triangle is isosceles: AB runs from (2,2) to (6,4), length √(16 + 4) = √(20) = 2√(5); AC runs from (2,2) to (4,6), length √(4 + 16) = 2√(5). Base BC from (6,4) to (4,6), length √(4 + 4) = 2√(2). The base BC lies on x + y = 10. Distance from A = (2,2) to the line x + y = 10 is

\frac{|2 + 2 - 10|}{√(2)}

\frac{6}{√(2)}

= 3√(2). Area =

\frac{1}{2}

· 2√(2) · 3√(2) =

\frac{1}{2}

· 12 = 6. ✓

💡Key takeaway

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 × 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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