2024 AMC 10B Problems/Problem 11

The following problem is from both the 2024 AMC 10B #11 and 2024 AMC 12B #7, so both problems redirect to this page.

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 $\triangle WMA$ ?

$[asy] pair X = (0, 0); pair W = (0, 4); pair Y = (8, 0); pair Z = (8, 4); label("$X$", X, dir(180)); label("$W$", W, dir(180)); label("$Y$", Y, dir(0)); label("$Z$", Z, dir(0)); draw(W--X--Y--Z--cycle); dot(X); dot(Y); dot(W); dot(Z); pair M = (2, 0); pair A = (8, 3); label("$A$", A, dir(0)); dot(M); dot(A); draw(W--M--A--cycle); markscalefactor = 0.05; draw(rightanglemark(W, M, A)); label("$M$", M, dir(-90)); [/asy]$

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

Solution

Solution 1

We know that $WX = 4$ , $WZ = 8$ , so $YZ = 4$ and $YX = 8$ . Since $\angle WMA = 90^\circ$ , triangles $WXM$ and $MYA$ are similar. Therefore, $\frac{WX}{MY} = \frac{XM}{YA}$ , which gives $\frac{4}{8 - XM} = \frac{XM}{4 - ZA}$ . We also know that the areas of triangles $WXM$ and $WAZ$ are equal, so $WX \cdot XM = WZ \cdot ZA$ , which implies $4 \cdot XM = 8 \cdot ZA$ . Substituting this into the previous equation, we get $\frac{4}{8 - 2ZA} = \frac{XM}{4 - ZA}$ , yielding $ZA = 1$ and $XM = 2$ . Thus,

$\triangle WMA = 4 \cdot 8 - \frac{4 \cdot 2}{2} - \frac{8 \cdot 1}{2} - \frac{6 \cdot 3}{2} = \boxed{\textbf{(C) }15}$

~Athmyx

Solution 2

Let $XM=b$ , $ZA = a$ , $4\cdot b= 8\cdot a$ , $b = 2a$ , $WM^2 + AM^2 = AW^2$ $(b^2+4^2) + (4-a)^2 + (8-b)^2 = (a^2 + 8^2)$ a=1, b=2 , $\triangle WMA = area(WXYZ) - \triangle WZA- \triangle WXM- \triangle MYA = 32 - 4-4-9 = \boxed{\textbf{(C) }15}$

~luckuso ~minor edits by EaZ_Shadow

Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)

https://youtu.be/YqKmvSR1Ckk?feature=shared

~ Pi Academy

2024 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2024 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2024 AMC 10B Problems/Problem 11

Contents

Problem

Solution 1

Solution 2

Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)

See also