AMC 10 · 2012 · #15
Grade 8 geometry-2dThree unit squares and two line segments connecting two pairs of vertices are shown. What is the area of △ABC?
Pick an answer.
AMC 10 2012 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: Three unit squares are stacked to make an L-shape. Two straight segments are drawn across it: one from the top-left corner $A$ down to the far bottom-right corner, and one from the top-middle point $B$ down to the bottom-left corner. These two segments cross at point $C$. Find the area of $\triangle ABC$.
Givens: The squares are unit squares, so every side has length $1$; $A$ is the top-left corner and $B$ is the point one unit to its right along the top edge; One segment runs from $A$ to the corner $2$ right and $1$ down from $A$; The other segment runs from $B$ to the corner $1$ left and $2$ down from $B$; $C$ is where the two segments cross
Unknowns: The area of $\triangle ABC$
Understand
Restated: Three unit squares are stacked to make an L-shape. Two straight segments are drawn across it: one from the top-left corner $A$ down to the far bottom-right corner, and one from the top-middle point $B$ down to the bottom-left corner. These two segments cross at point $C$. Find the area of $\triangle ABC$.
Givens: The squares are unit squares, so every side has length $1$; $A$ is the top-left corner and $B$ is the point one unit to its right along the top edge; One segment runs from $A$ to the corner $2$ right and $1$ down from $A$; The other segment runs from $B$ to the corner $1$ left and $2$ down from $B$; $C$ is where the two segments cross
Plan
Primary tool: #4 Introduce a Variable
Secondary: #1 Draw a Diagram, #7 Identify Subproblems
Tool #1 (Draw a Diagram): the picture is a grid, so I lay a coordinate system on it and read every corner off as a point. Tool #4 (Introduce a Variable): with coordinates in hand, each segment becomes a line equation $y=mx+b$, and the crossing point $C$ is found by treating its coordinates as unknowns. Tool #7 (Identify Subproblems): I split the work into two clean pieces — first locate $C$, then use $AB$ as a base to get the area.
Execute — Answer: B
6.NS.C.6 Step 1 Put the figure on a grid
- Place $A$ at the origin and let one unit square equal one grid box, with right as $+x$ and down as $-y$.
- Then $A=(0,0)$ and $B=(1,0)$.
- The first segment goes from $A$ to the corner two right and one down, the point $(2,-1)$.
- The second segment goes from $B$ to the corner one left and two down of $A$, the point $(0,-2)$.
💡 A grid of unit squares is already a coordinate plane, so every corner has a whole-number address.
8.EE.B.6 Step 2 Write each segment as a line
- For the segment from $A=(0,0)$ to $(2,-1)$: the slope is $\frac{-1-0}{2-0}=-\frac12$, and it passes through the origin, so its line is $y=-\frac12x$.
- For the segment from $B=(1,0)$ to $(0,-2)$: the slope is $\frac{-2-0}{0-1}=2$, and using point $B$ gives $y-0=2(x-1)$, that is $y=2x-2$.
💡 Two points fix a line's slope, and the slope plus one point gives its full equation.
8.EE.C.8 Step 3 Find where they cross
- At $C$ both equations hold, so set the two right sides equal: $-\frac12x=2x-2$.
- Multiply through by $2$ to clear the fraction: $-x=4x-4$.
- Then $4=5x$, so $x=\frac45$.
- Substitute back: $y=-\frac12\cdot\frac45=-\frac25$.
- So $C=\left(\frac45,-\frac25\right)$.
💡 The one point on both lines is exactly the pair $(x,y)$ that satisfies both equations.
6.G.A.1 Step 4 Turn the corner into area
- Use $AB$ as the base.
- Since $A=(0,0)$ and $B=(1,0)$ both sit on the line $y=0$, the base length is $1$ and the base lies flat.
- The height is then the straight-up-and-down distance from $C$ to that line, which is $\left|-\frac25\right|=\frac25$.
- Area of a triangle is $\frac12\cdot\text{base}\cdot\text{height}=\frac12\cdot1\cdot\frac25=\frac15$.
- So the area of $\triangle ABC$ is $\frac15$, which is choice $\textbf{(B)}$.
💡 When the base sits on a horizontal line, the height is just how far the third point is above or below it.
6.NS.C.6 Place $A$ at the origin and let one unit square equal one grid box, with right a 8.EE.B.6 For the segment from $A=(0,0)$ to $(2,-1)$: the slope is $\frac{-1-0}{2-0}=-\fra 8.EE.C.8 At $C$ both equations hold, so set the two right sides equal: $-\frac12x=2x-2$. 6.G.A.1 Use $AB$ as the base. Since $A=(0,0)$ and $B=(1,0)$ both sit on the line $y=0$, Review
Reasonableness: The whole L-shape covers $3$ square units, and $\triangle ABC$ is a thin sliver tucked near the top, so an area far below $1$ is expected — $\frac15$ fits. A quick shoelace check with $A=(0,0)$, $B=(1,0)$, $C=\left(\frac45,-\frac25\right)$ gives $\frac12\left|0(0-(-\tfrac25))+1((-\tfrac25)-0)+\tfrac45(0-0)\right|=\frac12\cdot\frac25=\frac15$, matching.
Alternative: Notice the two segments are perpendicular: their slopes $-\frac12$ and $2$ multiply to $-1$. So $\triangle ABC$ is right-angled at $C$, and its area is $\frac12\cdot AC\cdot BC$. Computing $AC=\sqrt{(\tfrac45)^2+(\tfrac25)^2}=\frac{2}{\sqrt5}$ and $BC=\sqrt{(\tfrac15)^2+(\tfrac25)^2}=\frac{1}{\sqrt5}$ gives area $\frac12\cdot\frac{2}{\sqrt5}\cdot\frac{1}{\sqrt5}=\frac15$, the same answer.
CCSS standards used (min grade 8)
6.NS.C.6Understand a rational number as a point on the number line (Laying a coordinate grid on the figure and reading each corner as a point.)8.EE.B.6Use similar triangles to explain why the slope is the same between any two points (Getting the slope and $y=mx+b$ equation for each of the two drawn segments.)8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Solving the two line equations together to locate the crossing point $C$.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Using base $AB$ and the height from $C$ to compute the triangle's area.)
⭐ Drop the picture onto a grid, turn each slanted segment into a line equation, solve the two lines together to find where they cross, then use the flat side $AB$ as a base — the area comes out to $\frac15$, choice $\textbf{(B)}$.
⭐ Drop the picture onto a grid, turn each slanted segment into a line equation, solve the two lines together to find where they cross, then use the flat side $AB$ as a base — the area comes out to $\frac15$, choice $\textbf{(B)}$.
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