See complete Problem 123
Area of triangle, Similarity. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Monday, June 16, 2008
Elearn Geometry Problem 123
Subscribe to:
Post Comments (Atom)
See complete Problem 123
Area of triangle, Similarity. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
see P122
ReplyDeleteSAGEC = SAGFEC + SFGE
using the fact h is the same for AGF and GFE,
and AF = 6/10 AA", FE = 3/20 AA" ( 4/10 - 1/4 )
SFGE = 1/80
=>
SAGEC = 1/20 + 1/5 + 1/80 + 1/20
SAGEC = 5/16
for three sides of tr
3∙5/16 = 15/16
SGME = 16/16 - 15/16
SGME = 1/16
----------------------------------------
Apply Menelaus’s theorem to ABA’ and secant CMC”
ReplyDeleteCA’/CB x C”B/C”A x MA/MA’= 1
Replace value of CA’/CB and C”B/C”A in above we get MA/MA’= 3 so AM/AA’=3/4
Similarly apply Menelaus’s theorem to triangle CAA’ and secant BEB’
We get EA/EA”=3 => AE/AA’=>3/4
ME//BC and triangle MGE similar to triangle BAC with the factor ME/BC= ¼
So S1/S= (1/4)^2= 1/16
Solution to problem 123.
ReplyDeleteLet K be the center of gravity of triangle ABC. Let S be the area of ABC. As proved in problem 121, KE/KC = 1/4, and also KG/KA = 1/4. So triangles KGE and KAC are similar with
S(KGE) = (1/16).S(KAC). Similarly,
S(KEM) = (1/16).S(KCB) and S(KMG) = (1/16).S(KBA).
Adding these three equalities we get S1 = S/16.