#How should i continue
12 messages · Page 1 of 1 (latest)
Suppose S(DFGE) = S. Then by the similarity of triangles we have:
S(FCG) = 4S(DCE)
S(ACB) = 9S(DCE)
Moreover:
S(DFGE) = S(FCG) - S(DCE) = 3S(DCE)
Thus, S(DCE) = (1/3)S. So:
S(ABC) = 9S(DCE) = 3S
So, now just substitute S = 2100.
Not really sure for what purpose AB is given. Oh well.
yes
so for S(FCG) = 4S(DCE) how do we know the are of FCG is 4 times DCE
for triangle CDE
let the base = DE = b, and height = h
then S(CDE) = (1/2)bh
CDE ~ CFG
with lengths of CFG twice that of CDE
then S(CFG) = (1/2)(2b)(2h) = 4 * (1/2)bh = 4S(CDE)
what made u think that the lengths of CFG was twice of CDE
Well, since EG = CE, that means CG = 2CE.