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
Subscribe to:
Post Comments (Atom)
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