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Difference between revisions of "2024 AMC 10B Problems/Problem 11"

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==Solution 1==
 
==Solution 1==
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We know that <math>WX = 4</math>, <math>WZ = 8</math>, so <math>YZ = 4</math> and <math>YX = 8</math>. Since <math>\angle WMA = 90^\circ</math>, triangles <math>WXM</math> and <math>MYA</math> are similar. Therefore, <math>\frac{WX}{MY} = \frac{XM}{YA}</math>, which gives <math>\frac{4}{8 - XM} = \frac{XM}{4 - ZA}</math>. We also know that the areas of triangles <math>WXM</math> and <math>WAZ</math> are equal, so <math>WX \cdot XM = WZ \cdot ZA</math>, which implies <math>4 \cdot XM = 8 \cdot ZA</math>. Substituting this into the previous equation, we get <math>\frac{4}{8 - 2ZA} = \frac{XM}{4 - ZA}</math>, yielding <math>ZA = 1</math> and <math>XM = 2</math>. Thus,
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<cmath>
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\triangle WMA = 4 \cdot 8 - \frac{4 \cdot 2}{2} - \frac{8 \cdot 1}{2} - \frac{6 \cdot 3}{2} = \boxed{\textbf{(B) }15}
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</cmath>
  
 
==See also==
 
==See also==

Revision as of 05:50, 14 November 2024

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,

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

See also

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

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