In △abc, p∈ ac so that ap: pc=3: 2. if abpc=14m2. find aabp and aabc.

Caution

I just finished doing this exact same problem. In nearly 4000 answers, I don't think that's ever happened. It's a great problem. Study it carefully but make sure you handle with care. Most area questions don't work this way.

Remark:

This problem will puzzle you greatly until you look at the diagram carefully.

Draw the diagram so that AC is on the horizontal and make it the longest line. This problem does not require that you have an accurate diagram.

Eyeball P so that it looks like it divides AC in such a way that PC is 2 units and AP is about 3 units. From B draw a line that is Perpendicular to AC. It shouldn't meet P otherwise you will get confused. Where the line from B hits AC label that point D. BD is perpendicular to AC.

What you need to know

BD is the height of both triangles. What does that mean? In practical terms it means that you can call the ratio of the areas 3/2

Ratio

Triangle ABP: Triangle BPC = x/14

3:2 = x:14         Change to a proportion

3/2 = x/14         Cross multiply

3*14 = 2x          Combine

42 = 2x             Divide by 2

42/2 = x            Divide

21 = x

Conclusion One

The area of ABP = 21 square units.

Conclusion Two

Area ABC = Area ABP + Area CBP

Area ABC = 21 + 14 = 35

Nice problem. Be careful how you treat it. Usually these kind of ratio problems don't work this way.