AMC 10 · 2024 · #11
Grade 8 geometry-2dProblem
In the figure below WXYZ is a rectangle with WX=4 and WZ=8. Point M lies XY, point A lies on YZ, and ∠WMA is a right angle. The areas of △WXM and △WAZ are equal. What is the area of △WMA?
Note: On certain tests that took place in China, the problem asked for the area of △MAY.
Pick an answer.
AMC 10 2024 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: 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
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) $17$
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
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.
💡 Labeling the picture turns three vague corner triangles into three formulas — the Grade 6 "area by composing/decomposing" idea.
6.EE.B.7 Step 2 - Use the equal-area condition $[\triangle WXM] = [\triangle WAZ]$.
- This gives the first equation linking $a$ and $b$.
💡 One sentence becomes one equation — Grade 6 "solve real-world problems by writing equations of the form $px = q$."
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 $YZ$).
💡 The right angle at $M$ is a Pythagorean trigger — turn the geometric condition into one numeric equation.
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 right angle plus an area condition was always going to collide into a quadratic — Grade 8 linear/quadratic solving handles it.
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 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."
6.G.A.1 Step 6 - Compute $[\triangle WMA]$ by subtracting the three corner triangles from the rectangle.
- $[\triangle WXM] = \tfrac{1}{2} \cdot 4 \cdot 2 = 4$, $[\triangle WAZ] = \tfrac{1}{2} \cdot 8 \cdot 1 = 4$, $[\triangle MAY] = \tfrac{1}{2} \cdot 6 \cdot 3 = 9$.
💡 "Big shape minus the corners" is the cleanest area decomposition — the diagram does all the work.
6.G.A.1 Label the unknowns on the picture: let $XM = b$ and $ZA = a$. Then $MY = 8 - b$ 6.EE.B.7 Use the equal-area condition $[\triangle WXM] = [\triangle WAZ]$. This gives the 8.G.B.7 Use the right angle at $M$ on triangle $WMA$. By the Pythagorean theorem on each 8.EE.C.7 Expand and substitute $b = 2a$ to reduce to a single-variable equation. Expandin 6.EE.B.5 Discard the extraneous root. If $a = 4$ then $ZA = 4 = YZ$, so $A$ coincides wit 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.1Find 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.7Solve 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.7Apply 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.7Solve linear equations in one variable (Reducing the Pythagorean equation under $b = 2a$ to $a^2 - 5a + 4 = 0$ and factoring it.)6.EE.B.5Understand 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!
⭐ This AMC 10 problem only needs Grade 8 Pythagorean theorem plus Grade 6 "rectangle minus the corners" area thinking that you already know!
More like this
Same archetype — closest grade level first.
