AMC 10B/12B 2024 (Problem 11/7)In the figure below WXYZWXYZWXYZ is a rectangle with WX=4WX=4WX=4 and WZ=8WZ=8WZ=8. Point MMM lies on XY‾\overline{XY}XY, point AAA lies on YZ‾\overline{YZ}YZ, and ∠WMA\angle WMA∠WMA is a right angle. The areas of △WXM\triangle WXM△WXM and △WAZ\triangle WAZ△WAZ are equal. What is the area of △WMA\triangle WMA△WMA?(A) 13\text{(A)}\;13(A)13(B) 14\text{(B)}\;14(B)14(C) 15\text{(C)}\;15(C)15(D) 16\text{(D)}\;16(D)16(E) 17\text{(E)}\;17(E)17Related TopicsCoreSimilar TrianglesCheck AnswerYour answer:ABCDECheckHints (3)Hint 1[WXM]=[WAZ][WXM]=[WAZ][WXM]=[WAZ]4⋅XM2=8⋅ZA2\frac{4\cdot XM}{2}=\frac{8\cdot ZA}{2}24⋅XM=28⋅ZAXM=2ZA.XM=2ZA.XM=2ZA.Hint 2Let ZA=aZA=aZA=a. Then XM=2aXM=2aXM=2a.By AA similarity,△WXM∼△MYA.\triangle WXM\sim\triangle MYA.△WXM∼△MYA.WXMY=XMYA\frac{WX}{MY}=\frac{XM}{YA}MYWX=YAXM48−2a=2a4−a\frac{4}{8-2a}=\frac{2a}{4-a}8−2a4=4−a2a14−a=a4−a\frac{1}{4-a}=\frac{a}{4-a}4−a1=4−aaa=1.a=1.a=1.Hint 3[WMA]=8⋅4−8⋅12−3⋅62−4⋅22[WMA]=8\cdot4-\frac{8\cdot1}{2}-\frac{3\cdot6}{2}-\frac{4\cdot2}{2}[WMA]=8⋅4−28⋅1−23⋅6−24⋅2=32−4−9−4=15.=32-4-9-4=15.=32−4−9−4=15.Final Answer(C) 151515Related Problems (3)AMC 10A 2025 (Problem 15)2024 AMC 12A Problem 10AMC 10B 2024 (Problem 10)
