Geometry Problem 1531. Post your solution in the comment box below.
Level: Mathematics Education, K-12 School, Honors Geometry, College.
Details: Click on the figure below.
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https://photos.app.goo.gl/2dcaP2AXaHV3zjWaA
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ReplyDeleteLe M is the midpoint of AB
DOM is 30-60-90 triangle => OM=1 and MD= sqrt(3)
MA*MA+OM*OM=OA*OA=3*3
MA=2.SQRT(2)
So AD= MA-MD=2.sqrt(2)-sqrt(3)
Let F be the midpoint of AB.
ReplyDeleteOF = 1 and DF = sqrt.3
AF^2 = 9^2 - 1^1 = 8
So AF = 2.sqrt.2
Hence AD = 2.sqrt.2 - sqrt.3
Sumith Peiris
Moratuwa
Sri Lanka
Hi!. I cannot get access in the solution of P 121?
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