Thursday, April 13, 2017

Geometry Problem 1331: Two Squares Side by Side, Parallel, Perpendicular, 90 Degrees, Measurement

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

Geometry Problem 1331: Two Squares Side by Side, Parallel, Perpendicular,  90 Degrees, Measurement.

4 comments:

  1. HC=2, => BH=6 => AH²=100, HK²+AK²=100 => HK=5√2

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  2. Applying the knowledge from the previous problems, we have
    AH=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)

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  3. Problem 1331
    Suppose 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 ).

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  4. AB = 8, BH = 6 so AH = 10

    Tr. HAK is right isosceles

    Hence HK = 10/sqrt2 = 5sqrt2

    Sumith Peiris
    Moratuwa
    Sri Lanka

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