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

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

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## Friday, October 5, 2018

### Geometry Problem 1391: Two Triangles, Ratio of Areas, Circumcircle

Labels:
area,
circle,
circumcircle,
geometry problem,
ratio,
triangle

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According to the Law of Sines (https://en.wikipedia.org/wiki/Law_of_sines) a/sinA = 2R, so, S1 = 1/2*b*c*sinA = 1/2*a*b*c/2R = (a*b*c)/(4*R). Similarly S2 = (d*e*f)/(4*R) and S1/S2 = (a*b*c)/(d*e*f), Q.E.D.

ReplyDeleteLet h be the height of the perpendicular from A to BC.

ReplyDeleteIf R is the circumradius, from similar triangles,

(c/2)/R = h/b and S1 = ha/2

Eliminating h, S1 = abc/4R

Similarly S2 = def/4R

Dividing, S1/S2 = abc/def

Sumith Peiris

Moratuwa

Sri Lanka