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Mathematics 14 Online
OpenStudy (anonymous):

help! If AB = 12, CD = 12, and MD = 3, and if AM is the shorter segment, find AM

OpenStudy (anonymous):

OpenStudy (anonymous):

I remember some theorem ....AMxMB = CM x MD

OpenStudy (anonymous):

AM * MB = CM * 3 AM + MB = 12 CM + 3 = 12 CM = 9

OpenStudy (anonymous):

not sure though...... is m the center

OpenStudy (anonymous):

AM * MB = 27 AM = 27 / MB MB + 27/MB = 12 MB ^2 + 27 = 12MB MB = 9 AM = 3

OpenStudy (anonymous):

just checked the theorem is true....also works for secants,and tangents now we know CM = 9,DM = 3,AB = AM+MB and from the theorem AM = 27/MB solve the last 2 equations getting a quadratic .Which will give you the answer.

OpenStudy (anonymous):

AM = 9

OpenStudy (anonymous):

AM is the shortest segment so = 3 (MB =3 or 9 are the roots of the quadratric but AM is the shortest segment so its 12 - 9 = 3)

OpenStudy (anonymous):

yes, you're right ...I missed that..I meant AM could also be 9....

OpenStudy (anonymous):

ok - happy with that cherri?

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