AMC 10 · 2024 · #11

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference

similar-trianglesarea-trianglespythagorean-theoremarea-rectangles identify-subproblemsconvert-to-algebraarea-difference ↑ Prerequisites: similar-trianglesarea-trianglespythagorean-theoremarea-rectangles

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

In the figure below $WXYZ$ is a rectangle with $WX=4$ and $WZ=8$ . Point $M$ lies $\overline{XY}$ , point $A$ lies on $\overline{YZ}$ , and $\angle WMA$ is a right angle. The areas of $\triangle WXM$ and $\triangle WAZ$ are equal. What is the area of $\triangle WMA$ ?

Note: On certain tests that took place in China, the problem asked for the area of $\triangle MAY$ .

$\textbf{(A) }13 \qquad \textbf{(B) }14 \qquad \textbf{(C) }15 \qquad \textbf{(D) }16 \qquad \textbf{(E) }17 \qquad$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: In rectangle $WXYZ$ with $WX = 4$ and $WZ = 8$, point $M$ is on side $\overline{XY}$ and point $A$ is on side $\overline{YZ}$. The angle at $M$ in triangle $WMA$ is a right angle, and the two corner triangles $\triangle WXM$ and $\triangle WAZ$ have equal areas. Find the area of $\triangle WMA$.

Givens: $WXYZ$ is a rectangle with $WX = 4$ (short side) and $WZ = 8$ (long side); $M$ lies on $\overline{XY}$ (the long side opposite $WZ$); $A$ lies on $\overline{YZ}$ (the short side opposite $WX$); $\angle WMA = 90^\circ$; $[\triangle WXM] = [\triangle WAZ]$; Answer choices: (A) $13$, (B) $14$, (C) $15$, (D) $16$, (E) $17$

Unknowns: The area of $\triangle WMA$

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #13 Convert to Algebra

The figure is the whole problem: a rectangle cut by the cevians $WM$ and $MA$ into four triangles whose areas must sum to $32$. Tool #1 (Draw a Diagram) makes that decomposition visible at a glance. Tool #7 (Identify Subproblems) turns it into two clean conditions — the equal-area condition links $XM$ and $ZA$, and the right angle at $M$ links them through the Pythagorean theorem. Tool #13 (Convert to Algebra) names $XM = b$ and $ZA = a$ so those two conditions become a $2 \times 2$ system that pins down $a$ and $b$. Once $a, b$ are known, the area falls straight out of $32 - (\text{three corner triangles})$.

Execute — Answer: C

#1 Draw a Diagram 6.G.A.1 Step 1

Label the unknowns on the picture: let $XM = b$ and $ZA = a$.
Then $MY = 8 - b$ and $AY = 4 - a$.
The three corner triangles each sit in a corner of the rectangle with two legs along the sides, so their areas are easy to read off.

$$[\triangle WXM] = \tfrac{1}{2} \cdot 4 \cdot b = 2b, \quad [\triangle WAZ] = \tfrac{1}{2} \cdot 8 \cdot a = 4a, \quad [\triangle MAY] = \tfrac{1}{2}(8-b)(4-a)$$

💡 Labeling the picture turns three vague corner triangles into three formulas — the Grade 6 "area by composing/decomposing" idea.

#13 Convert to Algebra 6.EE.B.7 Step 2

Use the equal-area condition $[\triangle WXM] = [\triangle WAZ]$.
This gives the first equation linking $a$ and $b$.

$$2b = 4a \;\Longrightarrow\; b = 2a$$

💡 One sentence becomes one equation — Grade 6 "solve real-world problems by writing equations of the form $px = q$."

#7 Identify Subproblems 8.G.B.7 Step 3

Use the right angle at $M$ on triangle $WMA$.
By the Pythagorean theorem on each of the three right triangles meeting at $M, A$, we have $WM^2 + MA^2 = WA^2$, where $WM^2 = 4^2 + b^2$, $MA^2 = (8-b)^2 + (4-a)^2$, and $WA^2 = 8^2 + a^2$ (drop a vertical from $W$ to the line $YZ$).

$$(16 + b^2) + \big((8-b)^2 + (4-a)^2\big) = 64 + a^2$$

💡 The right angle at $M$ is a Pythagorean trigger — turn the geometric condition into one numeric equation.

#13 Convert to Algebra 8.EE.C.7 Step 4

Expand and substitute $b = 2a$ to reduce to a single-variable equation.
Expanding: $16 + b^2 + 64 - 16b + b^2 + 16 - 8a + a^2 = 64 + a^2$, which simplifies to $2b^2 - 16b - 8a + 32 = 0$, i.e.
$b^2 - 8b - 4a + 16 = 0$.
Plugging $b = 2a$: $4a^2 - 16a - 4a + 16 = 0$, or $a^2 - 5a + 4 = 0$.

$$(a-1)(a-4) = 0 \;\Longrightarrow\; a = 1 \text{ or } a = 4$$

💡 A right angle plus an area condition was always going to collide into a quadratic — Grade 8 linear/quadratic solving handles it.

#7 Identify Subproblems 6.EE.B.5 Step 5

Discard the extraneous root.
If $a = 4$ then $ZA = 4 = YZ$, so $A$ coincides with $Y$; with $b = 2a = 8$, $M$ also coincides with $Y$, and $\triangle WMA$ collapses to a segment.
So $a = 1$, $b = 2$, giving $XM = 2$, $ZA = 1$, $MY = 6$, $AY = 3$.

$$a = 1, \;\; b = 2, \;\; MY = 6, \;\; AY = 3$$

💡 A root that makes a triangle degenerate is no triangle — kill it. Checking a solution against the picture is Grade 6 "is this value a real solution."

#1 Draw a Diagram 6.G.A.1 Step 6

Compute $[\triangle WMA]$ by subtracting the three corner triangles from the rectangle.
$[\triangle WXM] = \tfrac{1}{2} \cdot 4 \cdot 2 = 4$, $[\triangle WAZ] = \tfrac{1}{2} \cdot 8 \cdot 1 = 4$, $[\triangle MAY] = \tfrac{1}{2} \cdot 6 \cdot 3 = 9$.

$$[\triangle WMA] = 32 - 4 - 4 - 9 = 15 \;\Rightarrow\; \textbf{(C)}$$

💡 "Big shape minus the corners" is the cleanest area decomposition — the diagram does all the work.

[1] #1 6.G.A.1 Label the unknowns on the picture: let $XM = b$ and $ZA = a$. Then $MY = 8 - b$

[2] #13 6.EE.B.7 Use the equal-area condition $[\triangle WXM] = [\triangle WAZ]$. This gives the

[3] #7 8.G.B.7 Use the right angle at $M$ on triangle $WMA$. By the Pythagorean theorem on each

[4] #13 8.EE.C.7 Expand and substitute $b = 2a$ to reduce to a single-variable equation. Expandin

[5] #7 6.EE.B.5 Discard the extraneous root. If $a = 4$ then $ZA = 4 = YZ$, so $A$ coincides wit

[6] #1 6.G.A.1 Compute $[\triangle WMA]$ by subtracting the three corner triangles from the rec

Review

Reasonableness: Sanity-check with the right-triangle leg lengths: $WM = \sqrt{16 + 4} = \sqrt{20}$ and $MA = \sqrt{36 + 9} = \sqrt{45}$, so $\tfrac{1}{2} WM \cdot MA = \tfrac{1}{2}\sqrt{20 \cdot 45} = \tfrac{1}{2}\sqrt{900} = 15$. Same answer. Also the answer $15$ is just under half the rectangle area ($32$), which matches a glance at the picture where $\triangle WMA$ is the biggest of the four pieces but does not dominate.

Alternative: Tool #6 (Guess and Check) on the constraint $XM = 2 \cdot ZA$: try the smallest reasonable $ZA = 1$ (so $XM = 2$), check that $\angle WMA = 90^\circ$ by computing $WM^2 + MA^2 = 20 + 45 = 65 = 64 + 1 = WA^2$. Hits both conditions on the first try, then the area $32 - 4 - 4 - 9 = 15$ follows immediately. For an AMC multiple-choice setup this is faster than the full quadratic.

CCSS standards used (min grade 8)

6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing (Reading off the three corner-triangle areas $2b$, $4a$, $\tfrac{1}{2}(8-b)(4-a)$ from the labeled picture and subtracting them from the rectangle area $32$ to get $[\triangle WMA]$.)
6.EE.B.7 Solve real-world problems by writing and solving equations of the form px = q (Turning the equal-area condition into the equation $2b = 4a$, i.e. $b = 2a$.)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Writing $WM^2 = 16 + b^2$, $MA^2 = (8-b)^2 + (4-a)^2$, $WA^2 = 64 + a^2$, and combining via $WM^2 + MA^2 = WA^2$.)
8.EE.C.7 Solve linear equations in one variable (Reducing the Pythagorean equation under $b = 2a$ to $a^2 - 5a + 4 = 0$ and factoring it.)
6.EE.B.5 Understand solving an equation or inequality as a process of finding values (Checking $a = 4$ against the geometry and rejecting it as degenerate, keeping $a = 1$.)

⭐ This AMC 10 problem only needs Grade 8 Pythagorean theorem plus Grade 6 "rectangle minus the corners" area thinking that you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 8)

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