Thursday, April 13, 2017

Geometry Problem 1330: Two Squares Side by Side, Parallel, 45, 90 Degrees

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 1330: Two Squares Side by Side, Parallel, 45, 90 Degrees.

Posted by Antonio Gutierrez at 11:07 AM
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3 comments:

  1. APOSTOLIS MANOLOUDISApril 13, 2017 at 10:16 PM

    Problem 1330
    From problem 1329 the AH=HM and <AHM=90.Therefore <HMA=45.

    ReplyDelete
  2. c.t.e.oApril 14, 2017 at 12:12 AM

    From right isoceles tr AHM

    ReplyDelete
  3. Sumith PeirisOctober 23, 2025 at 3:37 AM

    Trigonometry Solution

    Let AB = a and DG = b
    Let < HAB = p and <MAG = q

    < CHG = < MGH = 45 so GHCM is an isosceles Trapezoid and hence
    MG = HC = CE = a - b

    Now Tan p = b/a
    and Tan q = MG/AG = (a - b) / (a + b) = (1 - b/a)/ (1+ b/a)
    Hence Tan q = (Tan 45 - Tan p)/(1 + Tan 45. Tan p) since Tan 45 = 1
    So Tan q = Tan (45 - p)
    Hence q = 45 - p and
    p + q = 45

    Therefore < MAH = 90 - 45 = 45

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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