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

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

Details: Click on the figure below.

## Thursday, March 16, 2017

### Geometry Problem 1323: Square, Angle Trisector, Diagonal, Regular Dodecagon, Area, 30-60 degrees

Labels:
30 degrees,
angle,
area,
diagonal,
dodecagon,
regular polygon,
square,
trisection

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5)Alt ∆A1B2A4 is (a√3)/2. B4A5 = a -(a√3)/2 = (2a-a√3)/2 => B2B4= a√3-a

ReplyDeletearea =(B2B4)²/2 =(a√3-a)²/2 => a²(2-√3)

6)area B3B2D3= 1/8 of B1B2B3B4

7)Like 5) find C1C3

8)Find area of 4 trapezoids D1D10D11D12 bases D1D10 and D1D12 alt (B3B4-C1C2)/2

1) ang B3B2D3=120:8=15 => Ang B2= 15+60+15=90

ReplyDelete2) A3D4 bisector => A3D4 altitude of triangle A2A3B4

3) Red triangles D4D3C2 are congruent

4) Tr D4A2D5 congr tr A2D5D6 => D4D5=D5D6