Competition · AMC preparation · step 4 of 4
AMC 8 · 2019 · #21
Grade 6 geometry-2dalgebraPick an answer.
AMC 8 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The problem hands us three lines but no picture, and the question is geometric — exactly the setup that begs for Tool #1 (Draw a Diagram). A quick sketch on the coordinate plane reveals that two of the three vertices lie on the horizontal line y = 5, which makes that segment a perfect horizontal base. Tool #7 (Identify Subproblems) then splits the work into three clean pieces: (a) find the three vertices, (b) read off the base length and the height from the picture, (c) apply the triangle area formula. Breaking the problem this way avoids any need for a distance formula or coordinate-geometry algebra beyond solving simple one-step equations.
Sketch the three lines
Sketch the three lines: y = 5 is horizontal, and y = 1 + x and y = 1 - x meet at (0, 1), fanning up to each side at 45°.
Plotting points and lines on a coordinate grid is the Grade 5 'use a pair of perpendicular number lines' skill.
5.G.A.1Draw A DiagramFind the top-right vertex
Where y = 5 meets y = 1 + x, substituting gives x = 4, so the top-right vertex is (4, 5).
Solving a one-step equation of the form px = q (here x = 4) is exactly the Grade 6 equation-solving standard.
6.EE.B.7Identify SubproblemsFind the top-left vertex
Where y = 5 meets y = 1 - x, the same substitution gives x = -4, so the top-left vertex is (-4, 5).
Locating (-4, 5) on the grid uses the Grade 6 understanding that negative coordinates name points to the left of the y-axis.
6.NS.C.6Identify SubproblemsFind the bottom vertex
Where the two slanted lines meet, 1 + x = 1 - x gives x = 0 and y = 1, so the bottom vertex is (0, 1).
Setting two expressions equal and solving for the variable is the Grade 6 'write and solve equations' move.
6.EE.B.7Identify SubproblemsRead the base and height
The top vertices share y = 5, giving a horizontal base 8, and the drop down to (0, 1) gives height 4.
Reading side lengths off a polygon drawn from its vertex coordinates is the Grade 6 'draw polygons in the coordinate plane' standard.
6.G.A.3Draw A DiagramApply the area formula
The area is half of base × height, so half of 8 × 4 = 16.
Finding the area of a triangle from its base and height is the Grade 6 triangle-area standard.
The triangle sitting on a base of 8 with a height of 4 covers an area of 16 square units.
▸ Why?
This triangle is exactly half of the 8-by-4 rectangle drawn snugly around it, and half of that rectangle's 32 square units is 16.
▸ Why?
The rectangle is 8 units wide and 4 units tall, so it is filled by 4 rows of 8 unit squares, which is 8 × 4 = 32 unit squares.
▸ Why?
The two corner pieces of the rectangle left outside the triangle can be turned to lie exactly on top of the two matching pieces inside it, so the triangle and the leftover parts have equal area — each is half of the rectangle.
This AMC 8 problem only needs Grade 6 coordinate-plane and triangle-area skills you already know!
- Sketch the three lines
- Find the top-right vertex
- Find the top-left vertex
- Find the bottom vertex
- Read the base and height
- Apply the area formula
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