Friday, March 11, 2016

Geometry Problem 1198: Right Triangle, Square, Rectangle, Quadrilateral, Altitude, Equal Areas

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

Click the figure below to view more details of problem 1198.

Geometry Problem 1198: Right Triangle, Square, Rectangle, Quadrilateral, Altitude, Equal Areas, Online College School

4 comments:

  1. Observe that triangle AEF congruent to tri. ABC ….( case ASA)
    So AF= AC= AG
    So S(AFDC)= S(ABDE)= AB^2
    We have S(AHKG)=AG.AH=AC.AH= AB^2.. ( relation in right triangle ABC)
    So S(AFDC)=S(AHKG)

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  2. S(AHKG)
    = 2×S(AHG)
    = 2×S(ABG)
    = 2×S(AEC)
    = 2×S(AEB)
    = S(AEDB)
    = S(AFDB) + S(AEF)
    = S(AFDB) + S(ABC)
    = S(AFDC)

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  3. Since <EAF = < BAC, triangles AEF and ABC are congruent ASA and so AC = AF

    Hence S(AFDC) = S(ABDE) = AB^2 = AH.AC since triangles ABH and ABC are similar

    Now AH.AC = AH.AG (since AC = AF = AG) = S(AHKG)

    So S(AFDC)= S(AHKG)

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  4. Angles FAE and CAB are equal, as angles with orthogonal hands. So, the triangles AEF and ABC are congruent; both have the same area. Thus, area of the square ABDE is equal to area ABDF plus area ABC, that is area(ACBDF) = area(ABDE) = c^2, where the length of side AB = c. On the other side, area(AGKH) = bp, where p = length of AH.
    However, in the right triangle ABC is valid c^2 = pb. It follows from the similar triangles ABC and AHB and c:p = b:c, that is c^2 = bp. Accordingly, area(AGKH) = area(ACDF).

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