Tuesday, 3 March 2009

Outline of solution, Ch.9 WS 1 Part Three!!

Finnally, I got the (standard) answer of no.4. 4. ABCD is a //gram, with E is a (free) point lying on AB, EC and BD intersect at F., find area of triangle DEF : area of triangle CBF. (HKCEE 2007 MC) Since EB//CD (by defination) we got 2 pairs of alt. angles, and 1 pair of vert. opp. angles, so that triangle EBF is similar to BCF (A.A.A.) EF/FC=BF/FD (2 sides prop.) Therefore, EF*FD=BF*FC You may not know what is it mean, but here's a formula about area of a triangle: area of a triangle = product of two sides * sin(intersect angle) * 1/2 since angle EFD = angle BFC their area is the same! I'm sorry that I can't think of other more solution (esspecially "primary way") P.S. area of triangle ABC(any triangle...)= AB*BC*sin angle ABC*0.5 = AB*AC*sin angle BAC*0.5=AC*CB*sin angle ABC*0.5

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