Ratio of Areas of Two Triangles

[Yuvraj Singh]

In Triangle ABC, P is a point on BC such that BP : PC = 1 : 2 and Q is the mid point of BP. Then, Area( Triangle ABQ ) : Area( Triangle ABC ) is :
Options Given Are :
1. 1 : 6
2. 1 : 5
3. 1 : 3
4. 1 : 4

I think the Concept of Similarity of Triangles will be used in this question. If we assume BP = 2x, then BQ = QP = x and PC = 4x.
Now we should prove that Triangle ABQ and ABC are similar and the sides BQ and BC are in a specific ratio, then Area( Triangle ABQ ) : Area( Triangle ABC ) =Square of (BQ : BC). But how to prove that both triangles are similar.

Or Is the Concept that Median divides the triangle in two equal areas will be used?

 

[ksdhart2]

You can't prove that the triangles are similar, because they're not. For two triangles to be similar, the lengths of their sides must be in the same proportion. In other words, it must be true that:

\(\displaystyle \frac{\overline{AB}}{\overline{AB}} = \frac{\overline{AC}}{\overline{AQ}} = \frac{\overline{BC}}{\overline{BQ}}\)

But this is definitely not the case, since the first fraction is equal to 1, but the others aren't. Intuitively, this makes sense as two triangles that share a side can only be similar if they are, in fact, the same triangle.

As for how to solve the problem at hand, I'm beginning to wonder if there's a typo here, as the answer seems to depend on the vertices of the triangles. As an example, let \(A = (0,0)\, \: B = (1,0), \: C = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\). In order for P to divide BC in the required ratio, it must be located at \(P = \left(\frac{2}{3}, \frac{1}{\sqrt{3}}\right)\), such that \(\overline{BP} = \frac{2}{3}\), and \(\overline{PC} = \frac{1}{3}\). From this, we can see that to make Q the midpoint of BP, it must be at \(Q = \left(\frac{5}{6}, \frac{1}{2\sqrt{3}}\right)\).

Then using the formula for area of a triangle given three vertices, we can see that the area of \(\triangle ABC\) is \(\frac{\sqrt{3}}{2}\), and the area of \(\triangle ABQ\) is \(\frac{1}{4\sqrt{3}}\), meaning the areas are in the ratio 1:6.

On the other hand, let \(A = (0,0)\, \: B = (2,0), \: C = (1, \sqrt{3})\). In order for P to divide BC in the required ratio, it must be located at \(P = \left(\frac{4}{3}, \frac{2}{\sqrt{3}}\right)\), such that \(\overline{BP} = \frac{4}{3}\), and \(\overline{PC} = \frac{2}{3}\). From this, we can see that to make Q the midpoint of BP, it must be at \(Q = \left(\frac{5}{3}, \frac{1}{\sqrt{3}}\right)\).

Then using the formula for area of a triangle given three vertices, we can see that the area of \(\triangle ABC\) is \(\sqrt{3}\), and the area of \(\triangle ABQ\) is \(\frac{1}{\sqrt{3}}\), meaning the areas are in the ratio 1:3.

By changing the location of the vertices, but keeping all of the required ratios in tact, we've changed the ratio of the areas of the triangles. Therefore, I conclude that something's off about the problem.

 

[Dr.Peterson]

ksdhart2 must have miscalculated something. If you think of BC as the base, then ABC and ABQ have the same altitude and bases in a known ratio. That determines the ratio of the areas.

 

[ksdhart2]

ksdhart2 must have miscalculated something...

That's entirely possible. I made sure that the points \(P\) and \(Q\) were chosen so they respected the given ratios... The only thing I can think of is that, by choosing my vertices like I did, I inadvertently assumed \(\triangle ABC\) is equilateral. When I have a bit more time later today, I'll try again with a scalene and/or isoceles triangle and see if that changes things.

 

[Dr.Peterson]

Looking closely at your work, I see a couple errors. First, you made the ratio BP : PC = 2 : 1 instead of 1 : 2; second, you missed a division by 2 in the area calculation for ABC, which should be sqrt(3)/4, not sqrt(3)/2. The ratio therefore is really 1:3. Your calculations are correct for the second triangle, making it also 1:3.

The second triangle has all dimensions doubled from the first, so all areas are 4 times the first, and the ratio is really the same, since the figures are similar.

It's not that anything doesn't work in the equilateral case. If you had put BC on the x-axis, everything would have been easier to calculate; in fact, you could have chosen A as an arbitrary point and shown the correct conclusion.

 

[ksdhart2]

D'oh! Well, still, it's good to know I was at least mostly on the right track about how to solve the problem, even if I did mess up along the way. Good thing someone else is here to set me straight so that students get good advice instead

 

[Yuvraj Singh]

If two triangles have a common vertex ( here it is vertex B ) and are based on a same straight line ( here it is BC ), then the area will be in the ratio of the length of their bases. ( Because both the triangles will have same altitude and area of triangle is (1/2) X base X height )
Answer will be 1 : 6 then.

Thanks for the help Dr. Peterson and ksdhart2

 

[Dr.Peterson]

If two triangles have a common vertex ( here it is vertex B ) and are based on a same straight line ( here it is BC ), then the area will be in the ratio of the length of their bases. ( Because both the triangles will have same altitude and area of triangle is (1/2) X base X height )
Answer will be 1 : 6 then.

Good reasoning. Often the solution is a matter of looking at a figure from the right perspective, such as making a wise choice of "base".

Good thing someone else is here to set me straight so that students get good advice instead

I often think the same thing. I'm also glad there are others who know the things I've forgotten or never learned.