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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