Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Thursday, April 13, 2017
Geometry Problem 1331: Two Squares Side by Side, Parallel, Perpendicular, 90 Degrees, Measurement
Labels:
90,
measurement,
parallel,
perpendicular,
square
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HC=2, => BH=6 => AH²=100, HK²+AK²=100 => HK=5√2
ReplyDeleteApplying the knowledge from the previous problems, we have
ReplyDeleteAH=HM=GC=10 ----------(1)
and AHMG is cyclic with m(AHM) = 90-----------(2)
=> AM=10Sqrt(2) ---------(3)
=> K is the center of the circle on which A,H,M,G lie
=> HK=AM/2 = 5sQRT(2)
Problem 1331
ReplyDeleteSuppose that the BC intersects GM in point N then NM=FG=6, or MG=8-6=2.
AM^2=AG^2+MG^2=14^2+2^2=200, or AM=10√2 .So HK=AM/2=5√(2 ).
AB = 8, BH = 6 so AH = 10
ReplyDeleteTr. HAK is right isosceles
Hence HK = 10/sqrt2 = 5sqrt2
Sumith Peiris
Moratuwa
Sri Lanka