AMC 8 · 2016 · #2
Grade 6 geometry-2dProblem
In rectangle ABCD, AB=6 and AD=8. Point M is the midpoint of AD. What is the area of △AMC?
Pick an answer.
AMC 8 2016 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: Rectangle $ABCD$ has $AB = 6$ and $AD = 8$. Point $M$ is the midpoint of side $\overline{AD}$. Find the area of $\triangle AMC$.
Givens: $ABCD$ is a rectangle with $AB = 6$ and $AD = 8$; $M$ is the midpoint of $\overline{AD}$, so $AM = MD = 4$; Opposite sides of a rectangle are equal, so $CD = AB = 6$ and $BC = AD = 8$; Adjacent sides of a rectangle are perpendicular; Answer choices: (A) $12$, (B) $15$, (C) $18$, (D) $20$, (E) $24$
Unknowns: The area of $\triangle AMC$
Understand
Restated: Rectangle $ABCD$ has $AB = 6$ and $AD = 8$. Point $M$ is the midpoint of side $\overline{AD}$. Find the area of $\triangle AMC$.
Givens: $ABCD$ is a rectangle with $AB = 6$ and $AD = 8$; $M$ is the midpoint of $\overline{AD}$, so $AM = MD = 4$; Opposite sides of a rectangle are equal, so $CD = AB = 6$ and $BC = AD = 8$; Adjacent sides of a rectangle are perpendicular; Answer choices: (A) $12$, (B) $15$, (C) $18$, (D) $20$, (E) $24$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems
This is a 2D geometry problem with named points, so Tool #1 (Draw a Diagram) is the natural first move: sketch the rectangle, mark $M$ as the midpoint of $\overline{AD}$, and draw $\triangle AMC$. The picture immediately reveals a key fact — side $\overline{AM}$ lies along side $\overline{AD}$ of the rectangle, which is perpendicular to side $\overline{DC}$. That means $\overline{AM}$ can serve as the base and $\overline{DC}$ gives the height for free. Tool #7 (Identify Subproblems) then splits the area calculation into two simple subproblems: (a) find the base $AM$, (b) find the perpendicular height, and (c) apply the triangle area formula.
Execute — Answer: A
3.G.A.1 Step 1 - Draw rectangle $ABCD$ with $A$ at the bottom-left, $B$ bottom-right ($AB = 6$), $C$ top-right, $D$ top-left ($AD = 8$).
- Mark $M$ on side $\overline{AD}$ halfway between $A$ and $D$.
- Draw segments $\overline{AM}$, $\overline{MC}$, and $\overline{AC}$.
- The picture shows $\triangle AMC$ sharing side $\overline{AM}$ with the left edge of the rectangle.
💡 Recognizing rectangle properties (opposite sides equal, adjacent sides perpendicular) from a sketched figure is a Grade 3 shape-attribute skill.
4.NF.B.4 Step 2 - Subproblem 1 — find the base.
- Choose $\overline{AM}$ as the base.
- Since $M$ is the midpoint of $\overline{AD}$ and $AD = 8$, the base $AM$ is half of $8$.
💡 Taking half of a whole number (the midpoint cuts a segment into two equal halves) is a Grade 4 fraction-of-a-whole operation.
4.G.A.1 Step 3 - Subproblem 2 — find the height.
- The base $\overline{AM}$ lies along side $\overline{AD}$.
- The perpendicular distance from $C$ to line $\overleftrightarrow{AD}$ is the length of side $\overline{DC}$, because $\overline{DC} \perp \overline{AD}$ in any rectangle.
- So the height is $DC = AB = 6$.
💡 Reading a perpendicular distance directly off a rectangle (right angle at $D$) uses Grade 4 understanding of perpendicular lines.
6.G.A.1 Step 4 Subproblem 3 — apply the triangle area formula $\text{Area} = \tfrac{1}{2} \times \text{base} \times \text{height}$ with $b = 4$ and $h = 6$.
💡 Using the triangle area formula with a known base and perpendicular height is the Grade 6 standard for finding triangle areas.
3.G.A.1 Draw rectangle $ABCD$ with $A$ at the bottom-left, $B$ bottom-right ($AB = 6$), 4.NF.B.4 Subproblem 1 — find the base. Choose $\overline{AM}$ as the base. Since $M$ is t 4.G.A.1 Subproblem 2 — find the height. The base $\overline{AM}$ lies along side $\overl 6.G.A.1 Subproblem 3 — apply the triangle area formula $\text{Area} = \tfrac{1}{2} \time Review
Reasonableness: Sanity check by comparing to the whole rectangle. The rectangle has area $6 \times 8 = 48$. The diagonal $\overline{AC}$ splits the rectangle into two equal triangles, so $\triangle ACD$ has area $24$. Triangle $AMC$ shares vertex $C$ with $\triangle ACD$ but uses only half of the base $\overline{AD}$ (since $AM = \tfrac{1}{2}AD$), so its area should be half of $24$, which is $12$. That matches choice (A).
Alternative: Tool #16 (Change Focus / Complement): instead of computing $\triangle AMC$ directly, compute the rest. $\triangle MDC$ has base $MD = 4$ and height $DC = 6$, so its area is $\tfrac{1}{2} \times 4 \times 6 = 12$. Then $\triangle AMC = \triangle ACD - \triangle MDC = 24 - 12 = 12$. Same answer.
CCSS standards used (min grade 6)
3.G.A.1Understand that shapes in different categories share attributes (Recognizing that $ABCD$ is a rectangle (opposite sides equal, all angles right), which gives $CD = AB = 6$ and $\overline{CD} \perp \overline{AD}$.)4.NF.B.4Apply understanding of multiplication to multiply a fraction by a whole number (Computing $AM = \tfrac{1}{2} \times 8 = 4$ from the midpoint condition on $\overline{AD}$.)4.G.A.1Draw and identify points, lines, line segments, rays, angles, and perpendicular and parallel lines (Identifying that $\overline{DC}$ is perpendicular to $\overline{AD}$ so its length serves as the triangle's height.)6.G.A.1Find the area of triangles and special quadrilaterals by composing or decomposing into known shapes (Computing the area of $\triangle AMC$ as $\tfrac{1}{2} \times 4 \times 6 = 12$ using base and perpendicular height.)
⭐ This AMC 8 problem only needs the Grade 6 triangle area formula — pick a smart base where the height is already drawn for you, and the answer falls out in one step.
⭐ This AMC 8 problem only needs the Grade 6 triangle area formula — pick a smart base where the height is already drawn for you, and the answer falls out in one step.
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