Tuesday, February 24, 2015

Geometry Problem 1089. Square, Semicircle, Tangent, Triangle, Area

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

Click the diagram below to enlarge it.

Online Math: Geometry Problem 1089. Square, Semicircle, Tangent, Triangle, Area.


  1. http://s21.postimg.org/byaoi9b2v/pro_1089.png
    Triangle OAE similar to CBE ( case ASA)
    So AE/BE=OA/CB= ½
    Triangle BEA similar to AGE and EFB ( case AA)
    Let AG= x , we have EG/AG=1/2 => EG= .5 x
    AG/EF=BF/EF= ½ => EF= 2x
    So EG/FG= .5x/(2.5x)= 1/5
    And area of AED/ Area of ABCD= ½ x 1/5= 1/10

  2. It's reasonably easy to show that CE will be represented by: y = 4x/3 - s/3 if A is (0,0) while C is (s,s) with the side of the square being=s. The semi-circle will be x² + (y - s/2)² = s²/4 and thus the point of tangency E will be (2s/5.s/5). Therefore, triangle AED=(1/2)*s*(s/5)=s²/10= S/10 since S = s².

  3. Similar reasoning, alternate presentation:

    Kites FAGE and CBFE are similar therefore: EG/EF=EF/EC=1/2 ⇒ EG/EC=1/4 ⇒ EG/CG=1/5
    Hence ΔAED height = CD/5 (Thales theorem applied to ΔGCD) ⇒ area(ΔAED) = AD.(CD/5)/2 = S/10

  4. Happy & Blessed 2019 to Antonio & to all Gogeometers, from Sri Lanka

    If S(AED) = S1, then S(BED) = 2S1 since Tr.s BED & AED are similar and BD = sqrt2. AD

    Also since Tr.s ABE & OED are similar S(ABE) = 2.S2 since AB = sqrt2. OD

    So S(ABD) = S/2 = (S1 + 2S1 + 2S1) and so
    S1 = S/10

    Sumith Peiris
    Sri Lanka

  5. Nice Pr