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?

The answer to the question is;

The area of hexagon is 136.

Step-by-step 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

linear

step-by-step explanation:

fa tha first one you gonna add y to both sides

it look like this

x-y+y=6+y

x=y+6

therefore your answer is

x =y+6

fa tha second one

you gonna flip tha equation

it gonna look like x-4y=y

then add 4 to both sides

it gonna look like tha

x-4 +4 =y +4

so therefore tha answer is x=y+4

step-by-step explanation: