Thursday, April 13, 2017

Geometry Problem 1333: Two Squares Side by Side, Three Triangles, Area, 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 1333: Two Squares Side by Side, Three Triangles, Area, Measurement.

6 comments:

  1. Quadrilateral AEGH is cyclic so
    ∠ (EAD)= ∠ (KHG)
    And triangle EKG similar to AKH ( case AA)
    Ratio of similarity= EK/AK= sqrt( area(EKG)/area(AKH))
    Triangle ADE congruent to HFE so AEH is isosceles right triangle
    Triangle AEC similar to AKH ( case AA)
    Ratio = AE/AK = sqrt( area(AEC)/area(AKH))= sqrt(42/56)= sqrt(3)/2 => AEK is 30-60-90 triangle
    EK/AK= ½ => area(EKG)/area(AKH)= ¼ => Area(EKG)= ¼ * 56= 14

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  2. Problem 1333
    Let (ACE) is area triangle ACE then (ACE)=CE.AD/2=GH.AD/2=(ADH)=42.
    (KDH)=(KAH)-(ADH)=56-42=14. But EH//CG (AE perpendicular CG) so GH=CE.
    Is EDHG trapezium so x=(KEG)=(KDH)=14.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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  3. Name: AD=a, DG=b, S(AED)=m, S(EDK)=n
    (a²)/2=42+m, (b²)/2=x+n, (AE²)/2=56+m+n,
    Using AE²=a²+b² => x=14

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  4. Let AD = a and DG = b
    AEGH is concyclic and so Tr. AEH is right isosceles.
    So Tr.s AED and EFH are congruent and
    GH = a-b.
    S(AGH)-S(HEG) = 56-x = 1/2((a+b)(a-b) - b(a-b))

    So 56-x = 1/2a(a-b) = S(AEC) = 42

    Hence x = 56-42 = 14

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  5. Three triangle are similar.
    by pytagorian theorem
    x=56-42=14

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  6. More generally: Area(ACG) + Area(EKG) = Area(AKH)

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