Let ABCD be a convex quadrilateral, and let P, Q, R, S, T, U, V, and W be points that trisect the sides of ABCD, as shown.
If the area of quadrilateral ABCD is 180 then find the area of hexagon APSCTW.
Triangle(BPS) is similar to Triangle(BAC) ---> Area(BPS) / Area(BAC) = BP2 / BA2 = 22 / 32 = 4/9
Consequently, Area(APSC) = 5/9 · Area(ABC)
Similarly, Area(DWT) / Area(DAC) = 4/9 ---> Area(WACT) = 5/9 · Area(DAC)
Therefore, Area(APSCTW) = 5/9 · Area(ABCD) ---> Area(APSCTW) = 5/9 · 180 = 20
Nothing like good manners and graciousness. ... That is sarcasm.
Note: I would not bother responding more to you.
I don't know if it is right or wrong but thanks for your efforts Geno.