# Areas of triangles

• Feb 12th 2011, 06:50 AM
amey
Areas of triangles
In \$\displaystyle \triangle\$ABC, E and F are such that A-F-B and A-E-C. Segments BE and CF intersect at P. Area of \$\displaystyle \triangle\$PEC=4 and area of \$\displaystyle \triangle\$PFB=8 and area of \$\displaystyle \triangle\$PBC=10. Find the area of quadrilateral AFPE.
I tried using reallyyyy many many approaches, even basic-most properties, but i couldn't get to it. (Headbang)
• Feb 12th 2011, 07:20 AM
earboth
Quote:

Originally Posted by amey
In \$\displaystyle \triangle\$ABC, E and F are such that A-F-B and A-E-C. Segments BE and CF intersect at P. Area of \$\displaystyle \triangle\$PEC=4 and area of \$\displaystyle \triangle\$PFB=8 and area of \$\displaystyle \triangle\$PBC=10. Find the area of quadrilateral AFPE.
I tried using reallyyyy many many approaches, even basic-most properties, but i couldn't get to it. (Headbang)

I'm not sure that I understand your question correctly ...(?)

If F is the midpoint of AB and if E is the midpoint of AC

then the area of triangle BCE is as large as the area of triangle ABE.

Therefore
\$\displaystyle \Delta(BCE) = \Delta(ABE)\$

\$\displaystyle (10+4) = 8+area(AFPE)\$

Solve for \$\displaystyle area(AFPE)\$
• Feb 12th 2011, 07:38 AM
amey
Sorry, but A-F-B means points A, F and B are collinear. F is not midpoint of AB
• Feb 12th 2011, 08:27 AM
Plato
Quote:

Originally Posted by amey
Sorry, but A-F-E means points A, F and E are collinear. F is not midpoint of AB

Have you omitted some other bit of information?
• Feb 12th 2011, 06:44 PM
amey
No. Atleast the problem is that way only.
• Feb 20th 2011, 08:56 AM
Wilmer
Easy solution, once one realises this is a mini "crossing ladder" classic problem; go here:
Crossed Ladder Puzzle, Triangle Area Puzzle
• Feb 22nd 2011, 06:57 PM
amey
Thanks. I have got it now.