Question

Let ABCDEF be a convex hexagon, and denote by P, Q, R, S, T, U the midpoints of the sides AB, BC, CD, DE, EF, FA respectively. Suppose that the areas of the triangles ABR, BCS, CDT, DEU, EFP and FAQ are 12, 34, 56, 12, 34 and 56 respectively.Find the area of hexagon?

Answer 1

The answer to the question is;

The area of hexagon is 136.

Explanation:

We have area ABR = 0.5 (ABC+ABD) since R = midpoint of segment CD

Similarly we have area BCS = 0.5 ( BCD+BCE)

also area CDT = 0.5(CDE+CDF)

DEU = 0.5(DEF+DEA)

EFP = 0.5(EFA+EFB) and

FAQ = 0.5(FAB+FAC)

However 0.5 (ABC+ABD) + 0.5 ( BCD+BCE)+0.5(CDE+CDF)+0.5(DEF+DEA)+

0.5(EFA+EFB) + 0.5(FAB+FAC) = 0.5*6*0.5 * Area of a regular hexagon

= 3/2×area of the hexagon = 12+34+56+12+34+56 = 204

Area of the hexagon = 2/3*204= 136