AMC 8 · 2016 · #2

Easy mode Grade 6

📗 View original problem →

Problem

Picture rectangle $ABCD$ . The short side $AB$ is $6$ , and the long side $AD$ is $8$ .

Point $M$ sits exactly halfway along side $AD$ . Now draw the triangle with corners $A$ , $M$ , and $C$ .

What is the area of $\triangle AMC$ ?

$\textbf{(A) }12\qquad\textbf{(B) }15\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad \textbf{(E) }24$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

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 $\triangle AMC$.

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

#1 Draw a Diagram 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.

$$\text{Rectangle: } AB = CD = 6, \; AD = BC = 8$$

💡 Recognizing rectangle properties (opposite sides equal, adjacent sides perpendicular) from a sketched figure is a Grade 3 shape-attribute skill.

#7 Identify Subproblems 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 $8$.

$$AM = \tfrac{1}{2} \times AD = \tfrac{1}{2} \times 8 = 4$$

💡 Taking half of a whole number (the midpoint cuts a segment into two equal halves) is a Grade 4 fraction-of-a-whole operation.

#1 Draw a Diagram 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 $DC = AB = 6$.

$$h = DC = AB = 6$$

💡 Reading a perpendicular distance directly off a rectangle (right angle at $D$) uses Grade 4 understanding of perpendicular lines.

#7 Identify Subproblems 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 $h = 6$.

$$\text{Area}(\triangle AMC) = \tfrac{1}{2} \times 4 \times 6 = \tfrac{1}{2} \times 24 = 12 \;\Rightarrow\; \textbf{(A)}$$

💡 Using the triangle area formula with a known base and perpendicular height is the Grade 6 standard for finding triangle areas.

[1] #1 3.G.A.1 Draw rectangle $ABCD$ with $A$ at the bottom-left, $B$ bottom-right ($AB = 6$),

[2] #7 4.NF.B.4 Subproblem 1 — find the base. Choose $\overline{AM}$ as the base. Since $M$ is t

[3] #1 4.G.A.1 Subproblem 2 — find the height. The base $\overline{AM}$ lies along side $\overl

[4] #7 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.1 Understand 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.4 Apply 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.1 Draw 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.1 Find 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.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 6)

More like this