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

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

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## Friday, February 13, 2015

### Geometry Problem 1082: Square, Circle, Center, Radius, Side, Isosceles Triangle, 90 Degrees, Area

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ReplyDelete1. In triangle BFD with median FO we have

BF^2+FD^2=2(BO^2+FO^2)

replace BF= a/sqrt(2), FD= a, BO= a/sqrt(2) in above

we get FO= a/2

Circles centered A ,B, C and D have symmetry axis Ox so OE=OF’

Circles centered B and D have symmetry axis BD so OF=OF’ => OE=OF= a/2

2. Due to symmetric property as shown above we have triangles OFC congruent to OF’A and OEB ( case ASA)

Triangle OEB is the image of OFC in the rotational transformation centered O , rotational angle = 90

so angle ( EOF)=90

3. Area S1= ½. a/2 . a/2= a^2/8= S/8