Let ABCD be a convex quadrilateral. Let P and Q be points on side \(\overline{AB}\) such that AP = PQ = QB. Similarly, BR = RS = SC, CT = TU = UD, and DV = VW = WA.

The area of quadrilateral ABCD is 180. Find the area of hexagon AWTCSP.

Guest May 13, 2020

#2**0 **

Draw line segment AC, dividing the figure into two triangles.

In the top triangle, triangle(BPS) ~ triangle(BAC) [side-angle-side for similarity]

Since BP = (2/3)^{rds} BA, the area of triangle(BPS) = (2/3 · 2/3)·triangle(BAC) = 4/9·triangle(BAC)

Similarly, the area of triangle(WDT) = 4/9·triangle(ADC).

Adding these two parts together, we get the fact that the unshaded area is 4/9^{ths} the quadrilateral.

However, we want the shaded area, so it will be 5/9^{ths} the quadrilateral.

(5/9)·180 = **100**

geno3141 May 14, 2020