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From ∆CHE => R(G) = (3r√3)/2 r = r(Q)=> S = 27
Let the side of the regular hexagon be a. We note that B,H,O,G,E are collinear.BE = 2a and BH = a/2 so EG = 3a/4 and r(Q)^2 = (9/16).a^2S(G)/S(Q) = r(G)^2/r(Q)^2 = (a^2*(9/16)) / (a^2/12) = 6.75So S(G) = 18Sumith PeirisMoratuwaSri Lanka
Got the geometry right and the arithmetic wrong!6.75 X 4 is indeed 27 not 18
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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