Medians and Area

Identity · 2007-11-29 18:37:22

Let ABC be any triangle. The two medians BN and CM intersect at the point P. Given that the area of the triangle BPC is 22 square units, how many square units are in the area of the triangle ABC ?

Thanks

JohnnyReinB · 2007-11-29 22:33:14

I'm guessing, but maybe it is 66 square units...

I only did this visually, and it looks as if triangle BPC is one third of the area of triangle ABC, so...
Voila!

Last edited by JohnnyReinB (2007-11-30 00:27:23)

JohnnyReinB · 2007-11-30 00:27:38

EDIT:
However, after much thought, I found proof, based on the formula of area of a triangle:

Since the third median would be the height of any triangle, it is split by the other two into 1/3 and 2/3 of the original median(which is the height). Since the 1/3 of the line is the height of triangle BPC, the ratio of their areas will then be 1:3. So, multiply the area of Triangle BPC by 3 and you get the area!

P.S. NEVER (mind)
P.S.S. Sorry, if the solution is only based on the area formula. Hey, before I read about medians after reading this post, I never even knew what it was. If anyone can find a more technical proof, please do.
P.S.S.S. So, DO never trust this guy from Pluto!(deleted from original postscript)

NullRoot · 2007-11-30 00:49:57

What if the angle of C in ABC was >60°? Would the median from A still be the height of the triangle?

Identity · 2007-11-30 01:59:59

Thanks, the answers said it was 66 so you're right Johnny, but I still don't understand how to work it out :

NullRoot · 2007-11-30 02:13:19

I'm still working on it myself. I agree with Johnny that it looks like it's 1/3rd, but I can't find an algebraic proof.

Last edited by NullRoot (2007-11-30 02:14:37)

JohnnyReinB · 2007-11-30 14:53:36

Or...

Hmm, If you use three medians, they will divide the whole triangle into six parts of equal area. Since triangle BPC has two of these areas, it is exactly one-third of the original triangle.

How about that?

JaneFairfax · 2007-11-30 23:58:29

Use this theorem:

The medians of a triangle intersect at a common point, called the centroid. The centroid divides each median in the ratio 2:1.

The centroid is P in this case. If you draw the third median AO, this will pass through P. We have |AP|:|PO| = 2:1, so |AO|:|PO| = 3:1. Therefore if you drop a perpendicular distance (i.e. altitude) from A to BC and one from P to BC, the altitudes will be in the ratio 3:1 (by similar triangles). Apply this to the ratio of the areas of the two triangles.

Last edited by JaneFairfax (2007-12-02 01:01:38)

Identity · 2007-12-01 00:21:20

Thanks Jane, I understand how to do it now

JohnnyReinB · 2007-12-01 16:12:54

Sorry, Jane, but i do think it divides the triangle into six equal (at least by area) parts.

JaneFairfax · 2007-12-02 01:20:08

JohnnyReinB wrote:

Sorry, Jane, but i do think it divides the triangle into six equal (at least by area) parts.

Yes, you are correct. I do apologize for not having thought clearly about it.

Last edited by JaneFairfax (2007-12-02 01:20:21)

Math Is Fun Forum

#1 2007-11-29 18:37:22

Medians and Area

#2 2007-11-29 22:33:14

Re: Medians and Area

#3 2007-11-30 00:27:38

Re: Medians and Area

#4 2007-11-30 00:49:57

Re: Medians and Area

#5 2007-11-30 01:59:59

Re: Medians and Area

#6 2007-11-30 02:13:19

Re: Medians and Area

#7 2007-11-30 14:53:36

Re: Medians and Area

#8 2007-11-30 23:58:29

Re: Medians and Area

#9 2007-12-01 00:21:20

Re: Medians and Area

#10 2007-12-01 16:12:54

Re: Medians and Area

#11 2007-12-02 01:20:08

Re: Medians and Area

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