Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Tuesday, February 24, 2015

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

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http://s21.postimg.org/byaoi9b2v/pro_1089.png

ReplyDeleteTriangle 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

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.

ReplyDeleteSimilar reasoning, alternate presentation:

ReplyDeletehttp://bleaug.free.fr/gogeometry/1089.png

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