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


  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)

  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)

  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
    Sri Lanka

  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).

  5. We see that triangle AEF is congruent with ABC therefore the area of ABDE is equal to the area of AFDC.

    Let the sides of the square ABDE equal a.

    Then the area of AFDC =a^2

    And let AC = c and AH =m.

    Therefore the area of AHKG

    Area (AHKG)=cm

    The triangles ABH and BHC are similar therefore :

    BH/m= (c-m)/BH

    Apply pythagoras on triangle ABH and we find :


    a^2 is the area of AFDC and cm is the area of AHKG