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.
Contents
Problem
In the figure below 
is a rectangle with 
and 
. Point 
lies 
, point 
lies on 
, and 
is a right angle. The areas of 
and 
are equal. What is the area of 
?
![[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]](http://latex.artofproblemsolving.com/a/e/9/ae9f151d0a601f719d48015b0f36651f319802a3.png)
Solution 1
We know that 
, 
, so 
and 
. Since 
, triangles 
and 
are similar. Therefore, 
, which gives 
. We also know that the areas of triangles and 
are equal, so 
, which implies 
. Substituting this into the previous equation, we get 
, yielding 
and 
. Thus,
![\[\triangle WMA = 4 \cdot 8 - \frac{4 \cdot 2}{2} - \frac{8 \cdot 1}{2} - \frac{6 \cdot 3}{2} = \boxed{\textbf{(C) }15}\]](http://latex.artofproblemsolving.com/7/1/b/71bee82ee42923d2ff1c569ebce5b94fcdaf9f2d.png)
Solution 2
Let 
, 
, 
, 
,
![\[WM^2 + AM^2 = AW^2\]](http://latex.artofproblemsolving.com/3/6/d/36d751b67383d114bb859e9191df01043d8161dd.png)
![\[(b^2+4^2) + (4-a)^2 + (8-b)^2 = (a^2 + 8^2)\]](http://latex.artofproblemsolving.com/8/1/3/8134c67fc6cf055cc7298037d1a2de39fce9e346.png)
a=1, b=2 ,
![\[\triangle WMA = area(WXYZ) - \triangle WZA- \triangle WXM- \triangle MYA = 32 - 4-4-9 = \boxed{\textbf{(C) }15}\]](http://latex.artofproblemsolving.com/8/7/6/87617c92d320c908a7db950709273dfbcb267d31.png)
~luckuso ~minor edits by EaZ_Shadow
Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)
https://youtu.be/YqKmvSR1Ckk?feature=shared
~ Pi Academy
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
