# Help me understand this triangle problem

This looks on the face of it straightforward. And maybe it is. But I can’t get my head around it. I don’t understand the answer that’s given, can anyone help explain it to me in really simple terms?

It doesn’t look straightforward. I ain’t no professional mathematicalist but it looks like it assumes facts not in evidence. Perhaps you are supposed to assume that the triangle formed by regions 2 and 3 is Isosceles.

I believe 2 and 3 are measurements, not just labels.

Yes, 2 and 3 are the areas of the red and blue triangles, respectively.

The explanation is very, very poorly written. It refers to a “yellow triangle” which is not even in the diagram.

I’m getting the impression that English isn’t the first language of that site’s author.

Assume height of the rectangle is 3.
Red and green together have an area of 5 so the top edge of the red triangle (the base) is 10/3.
That means the height of red is 12/10 and the height of white is 18/10. From this ratio of their heights you can calculate the base of the white as 5.
Rectangle has an area of 15, each half has an area of 7.5. Green is 5.5, white is 4.5

Is this just an assumption, or is there something in the question that allows you to determine that the height is exactly 3?

You don’t need to assume a height for the rectangle, just call it “H”.

Call the base of the red triangle b1 and the height of the red triangle x1.

Areas of the red and blue triangles total 5, which has to be 0.5b1H, so b1 = (10/H).

The area of the red triangle is 2, which is 0.5x1b1. Substituting 10/H for b1 lets us get x1 = (4/10)*H

Because the red and white triangles are similar, they both have the same base:height ratio.

The height of the white triangle is (6/10)*H (because the heights of the two triangles must total H), and so the ratio of the heights of the two triangles is 6:4 or just 1.5:1.

The bases must also have the same ratio, so the base of the white triangle (the length of the whole rectangle) is 1.5*(10/H) or 15/H.

Almost done. The area of the entire rectangle is (15/H)*H = 15. Notice at this point that the unknowns have dropped out of the picture - the exact value of H doesn’t matter at all.

Half of 15 is 7.5, which is the area of the red+green segments. The area of the red segment is 2, so the green must be (7.5-2) = 5.5

Thanks Valgard! I completely lost sight of the fact that the sum of the similar triangles heights = the height of the rectangle. I was flummoxed on how to proceed next without that tidbit.

:smack: I had 5 equations and 6 variables, and thought I was stuck. I didn’t bother to see if any of the variables dropped out. Thanks for the walkthrough.

Now for extra credit, can you calculate the area of the yellow triangle?

I tried to solve this and I burned my fingers.

What the heck is wrong with these people?

Just J/K

Did you laugh?

0

Although I was always really good at geometry I was stuck on this one as well; once I read the step to calculate the heights of the red and white triangles the rest was pretty simple but until that I had a bunch of chicken scratches leading nowhere.

As Valgard pointed out you don’t have to have a value at all but using 3 (because the one triangle had an area of 3) made it easier for my thought process around work tasks.

My god I’m out of practice doing simple geometry. In high school, when I was doing math contests all the time, I would have whipped this out. But that was 25 years ago. As is, it took like four sheets of paper and any number of incorrect calculations, but I did eventually get the right answer.

I assumed that the whole rectangle was a 1 by 1 square, with the ratios of the two triangles being 3 to 2 (but the triangles not actually having area 3 or 2). Because both triangles have the same height, the length of the red triangle along the top of the square must be 2/3. That gives you the precise slope of the other line, and then I figured out the cartesian coordinates of all the points pictured, and figured out the area of the green bit by first extending it to be a full triangle, figuring out its full area, then subtracting off the triangle that’s outside the square. Finally to convert back to the original units, I had to scale it back up compared to the actual area of the blue triangle in my reduced square.

So, I got there, but… WOW is that not the best way to solve it.

I am so glad I wasn’t the only one having problems with this!

Thanks for the explanations.

Well, if I had realized that 2 and 3 represented areas then I’d still be getting nowhere. Great explanation by Valgard.

I thought the answer in the OP’s link was elegant. Cryptic and marred by changing colors in the middle, but elegant.

One thing that is interesting about this problem is that you never do find the height or width of the original rectangle. In fact, I have just verified the following: Take any a x b rectangle with ab=15. The draw a diagonal and choose the point on the diagonal that the line from the lower left vertex cuts off an area of 3 on the right. Then the cut off area on the left is always 2.

There is an interesting class of problems in which the answer is independent of some apparently crucial parameter. This one is independent of the height as seen. Often there is an easy solution if you assume that the answer does not depend on the parameter. Tim-n-va got the right answer by assuming the height was 3, but had he assumed 1 or 15, he’d have gotten it too.

Here is a simple example. A central core is removed from a spherical apple. The left over solid has a height of 3 inches. What is its volume?

If you assume it has a solution, you may as well assume the cure was 0 and figure the volume of a 3" sphere, which is 36*pi. This is easy, but it takes calculus to show that the answer does not depend on the height of the apple.

Math is like that.

Sorry, but this question just grabbed me and I must share. First let me say that the answer in my previous post was wrong; I got radius and diameter mixed up and the answer is (9/2) pi.

I got really carried away by the original question. The question really boils down to finding the area of the original rectangle, given the areas of the triangles. I decided to find out if there was anything special about the areas 2 and 3. There isn’t and there is a simple formula relating the areas of the two triangles to the area of the square. Here is the derivation.

Consider a rectangle ABCD, labeled clockwise with A at the lower right.
Draw the diagonal BD and choose a point E on the top edge CD. Let the
segment AE intersect the diagonal at F. Now let a be the length of the
right side DA and b be the length of the bottom edge AB. Let c be the
length of ED, d the length of the perpendicular from F to DA and e the
length of the perpendicular from F to ED. Let r,s,t denote the areas of
the rectangle, the triangle FDA, and the triangle FED, respectively.
Then the following equations are obvious:
r = ab; 2s = ad; 2t = ec; 2(s+t) = ac.
There is one more equation that follows easily from similarity of
triangles: b/a = d/e. Then we have:
e = 2t/c and c = 2(s+t)/a, so that:
e = 2t/[2(s+t)/a] = at/(s+t), which along with d = 2s/a gives:
d/e = (2s/a)/[at/(s+t)] = 2s(s+t)/(a^2t). But d/e = b/a so that:
b/a = 2s(s+t)/(a^2t) and finally, r = ab = 2s(s+t)/t.

When s = 3 and t = 2, this gives r = 6*5/2 = 15.