3

In the triangle in the figure the area of the triangle $MBP$ is equal to $9$, the area of the triangle $NPC$ is equal to $8$ and the area of the triangle $BCP$ is equal to $24$. I need to figure out the area of the polygon $AMPN$.

enter image description here

Notice that $M$ and $N$ are NOT the midpoints of $AB$ and $AC$.

It seems to me we don't have enough elements to figure this out. Moreover, I've been trying to find similarities between the shown triangles but I couldn't find any.

Intelligenti pauca
  • 50,470
  • 4
  • 42
  • 77
Schiele
  • 325

1 Answers1

4

You do have enough data. Draw the segment $MN$, then considering the triangles $MBC,MNC$, which share the same split base $MPC$, we have $$\frac9{24}=\frac{A(\triangle MPN)}8$$ So $A(\triangle MPN)=3$. Now let $A(\triangle AMN)=x$, then by the same logic as before $$\frac x{9+3}=\frac{x+3+8}{9+24}$$ $$\frac x{12}=\frac{x+11}{33}$$ $$33x=12(x+11)=12x+132$$ $$21x=132\qquad x=\frac{44}7$$ Then $A(AMPN)=3+\frac{44}7=\frac{65}7$.

Parcly Taxel
  • 103,344