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

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

Details: Click on the figure below.

## Monday, March 4, 2019

### Geometry Problem 1420: Regular Hexagon, Inscribed Circle, Area, Tangent

Labels:
area,
circle,
geometry problem,
hexagon,
inscribed,
regular polygon,
tangent,
triangle

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From ∆CHE => R(G) = (3r√3)/2 r = r(Q)

ReplyDelete=> S = 27

Let the side of the regular hexagon be a. We note that B,H,O,G,E are collinear.

ReplyDeleteBE = 2a and BH = a/2 so EG = 3a/4 and r(Q)^2 = (9/16).a^2

S(G)/S(Q) = r(G)^2/r(Q)^2 = (a^2*(9/16)) / (a^2/12) = 6.75

So S(G) = 18

Sumith Peiris

Moratuwa

Sri Lanka

Got the geometry right and the arithmetic wrong!

Delete6.75 X 4 is indeed 27 not 18