GRE triangle question

[dux3000]

Not sure if this forum is meant for solving GRE practice questions, but I see there are some experts here who can probably help. 

It's a geometry question, so I can't really transcribe the diagram - http://www.greenlighttestprep.com/module/gre-geometry/video/891

Is there are easier solution?

What's the level of difficulty? They say it's a medium question. 

 

[abohammed]

I don't think that there are easier solutions than this,

I gave it between medium (engineering students) to High (other disciplines)

 

BTW, is this website have practice tests or only explanation? 

[Brent@GreenlightGRE]

Agree, this question is on the cusp. We have 3 difficulty categories of questions. I'd say this is high medium or possibly low hard. 

To answer abohammed's question about practice tests, we have a recommended list of free practice tests here: http://www.greenlighttestprep.com/resources/practice-tests

Cheers,

Brent - Greenlight Test Prep 

[TakeruK]

Not sure if this forum is meant for solving GRE practice questions, but I see there are some experts here who can probably help. 

It's a geometry question, so I can't really transcribe the diagram - http://www.greenlighttestprep.com/module/gre-geometry/video/891

Is there are easier solution?

What's the level of difficulty? They say it's a medium question. 

 

I think the solution presented in the video is the "best" solution. I can't think of another solution that does not use the so-called "30-60-90 special triangle". I was experimenting with another method and got the answer (C) but I found that I made a mistake in the early logic, so it sounds like this question was designed to catch silly mistakes like the one I made! 

I think the 30-60-90 triangle approach is the best approach because it's one of the skills listed by ETS as essential for the GRE Q. Also, I think you are better off learning this approach rather than some method that would only work for this one question, because the 30-60-90 triangle will be used for a lot of other questions too.

However, I can offer an alternate explanation of the exact same method used in the video. Sometimes it might help to hear another way of explaining it? I don't think any particular way is superior, just different:

In the video, go to the 0:52 mark (everything up to this point would be the same). The video goes on to talk about an "enlargement factor". If this confuses you, here's another way to look at it: (note this is using the exact same mathematical method, just with different words)

1. Because both the ABC triangle (left) and the "base 30-60-90 triangle" (right) all have the same three angles, that is 30, 60 and 90, this means that the sides in the ABC triangle are proportional to each other. (You may recall that this is a AAA similar triangle, if that helps, but if that doesn't mean anything to you, then it's also not a problem).

2. You can now "match up" the corresponding sides between the two triangles like so:

h / SQRT(3) = 12 / 2  

What happened here was that I first picked the value I want to solve for, which is "h". Then I found the corresponding side in the "base triangle" (on the right). Since "h" is the side that is opposite to the 60 degree angle in the left triangle, the corresponding side in the "base triangle" (on the right) is SQRT(3), i.e. the one opposite to the 60 degree angle.

Then, I write this as a proportionality: h / SQRT(3)

Next, I matched up the other known side of the left triangle, which is 12, the side opposite the 90 degree angle. The corresponding side in the base triangle has length 2. So on the right hand side of the equality, I put 12 / 2.

3. Now, I can solve 

h / SQRT(3) = 12/2

for h by multiplying both sides by SQRT(3) and I get:

h = 12/2 * SQRT (3)

which is the same as

h = 6 * SQRT(3)

since 12/2 = 6.

Now the rest of the solution is the same. 

Again, this is the exact same method as the video! Just another way to say the same things. For questions that require comparing to the base 30-60-90 triangle, I prefer matching up the sides with fractions and solving for the unknown side as above.

[jiggyjig]

I thought to calculate any triangle, you need three known value, either two angles and a side or two sides and an angle. In this particular case, A=0.5 x sin 60 x 5 x 12=15sqrt(3)

[dux3000]

I think the solution presented in the video is the "best" solution. I can't think of another solution that does not use the so-called "30-60-90 special triangle". I was experimenting with another method and got the answer (C) but I found that I made a mistake in the early logic, so it sounds like this question was designed to catch silly mistakes like the one I made! 

I think the 30-60-90 triangle approach is the best approach because it's one of the skills listed by ETS as essential for the GRE Q. Also, I think you are better off learning this approach rather than some method that would only work for this one question, because the 30-60-90 triangle will be used for a lot of other questions too.

However, I can offer an alternate explanation of the exact same method used in the video. Sometimes it might help to hear another way of explaining it? I don't think any particular way is superior, just different:

In the video, go to the 0:52 mark (everything up to this point would be the same). The video goes on to talk about an "enlargement factor". If this confuses you, here's another way to look at it: (note this is using the exact same mathematical method, just with different words)

1. Because both the ABC triangle (left) and the "base 30-60-90 triangle" (right) all have the same three angles, that is 30, 60 and 90, this means that the sides in the ABC triangle are proportional to each other. (You may recall that this is a AAA similar triangle, if that helps, but if that doesn't mean anything to you, then it's also not a problem).

2. You can now "match up" the corresponding sides between the two triangles like so:

h / SQRT(3) = 12 / 2  

What happened here was that I first picked the value I want to solve for, which is "h". Then I found the corresponding side in the "base triangle" (on the right). Since "h" is the side that is opposite to the 60 degree angle in the left triangle, the corresponding side in the "base triangle" (on the right) is SQRT(3), i.e. the one opposite to the 60 degree angle.

Then, I write this as a proportionality: h / SQRT(3)

Next, I matched up the other known side of the left triangle, which is 12, the side opposite the 90 degree angle. The corresponding side in the base triangle has length 2. So on the right hand side of the equality, I put 12 / 2.

3. Now, I can solve 

h / SQRT(3) = 12/2

for h by multiplying both sides by SQRT(3) and I get:

h = 12/2 * SQRT (3)

which is the same as

h = 6 * SQRT(3)

since 12/2 = 6.

Now the rest of the solution is the same. 

Again, this is the exact same method as the video! Just another way to say the same things. For questions that require comparing to the base 30-60-90 triangle, I prefer matching up the sides with fractions and solving for the unknown side as above.

Thanks for your help TakerUK!

To be honest I have A LOT of trouble reading math solutions. When I see the solutions in the official GRE book, I go cross eyed! I think I'm more of a visual learner (which is why I prefer videos :-)

 

 

 

[Eigen]

I thought to calculate any triangle, you need three known value, either two angles and a side or two sides and an angle. In this particular case, A=0.5 x sin 60 x 5 x 12=15sqrt(3)

This is the "easier" way, but requires some general knowledge/memory of trig. 

The "longer" way can be worked out with just a general knowledge of geometry (i.e., all angles in a triangle add up to 180 °, area of a triangle, etc.

[AkashSky]

Well, if you know the 30, 60 90 rule (2x = hypo, root3 x = long side, x=short side), you can solve the question in 2 steps by computing the "height" of the triangle by pretending that 12 is the hypotenuse of a right triangle, with other side lenths 60 and 90.  This will lead 6 root 3 as the height, which you can then multiply by 1/2 base to get area which is 15 root 3

[Brent@GreenlightGRE]

I thought to calculate any triangle, you need three known value, either two angles and a side or two sides and an angle. In this particular case, A=0.5 x sin 60 x 5 x 12=15sqrt(3)

We know that we can create a 30-60-90 triangle, so we already have 3 angles. 

Once we know the length of any side, we can determine the remaining sides. 

Cheers,

Brent - Greenlight Test Prep 

[TakeruK]

Well, if you know the 30, 60 90 rule (2x = hypo, root3 x = long side, x=short side), you can solve the question in 2 steps by computing the "height" of the triangle by pretending that 12 is the hypotenuse of a right triangle, with other side lenths 60 and 90.  This will lead 6 root 3 as the height, which you can then multiply by 1/2 base to get area which is 15 root 3

Yes, this is the same solution that the GreenlightGRE video presented, and the same solution that I wrote up. The video and I went about explaining how to get there in different ways.

If I was solving the question myself, I would have done exactly what you said here, because the 30-60-90 rule is ingrained in my head and both of us are very familiar with all of the ways we can manipulate the special triangle. But the purpose of the instructional video is for test takers without this ingrained knowledge, so the video presented one way to think about it that is very intuitive and I presented a different way to think about the same concept!

[Eigen]

There's also the formula (derived from the explanation) from the side angle side method. 

For any triangle that you have the length of two adjacent sides a and b, and the included angle c, the area A = (ab sin c)/2. 

For a better explanation (with figures) see http://www.mathopenref.com/triangleareasas.html. It also has a quick flow chart at the bottom for shortcut methods to finding areas for triangles with different bits of information.

Basically, instead of "lengthening" the base you have, you bisect the largest angle to make two right triangles, then solve. You then know the largest side, b. And you know that  a sin c is the side opposite the angle (sin c = h/a). So then area is bh/2, substitute for h and you get the above formula. 

Edited September 10, 2015 by Eigen

[dux3000]

There's also the formula (derived from the explanation) from the side angle side method. 

For any triangle that you have the length of two adjacent sides a and b, and the included angle c, the area A = (ab sin c)/2. 

For a better explanation (with figures) see http://www.mathopenref.com/triangleareasas.html. It also has a quick flow chart at the bottom for shortcut methods to finding areas for triangles with different bits of information.

Basically, instead of "lengthening" the base you have, you bisect the largest angle to make two right triangles, then solve. You then know the largest side, b. And you know that  a sin c is the side opposite the angle (sin c = h/a). So then area is bh/2, substitute for h and you get the above formula. 

The link isn't working for me

Edited September 10, 2015 by Eigen

[Eigen]

Hmm, working fine for me. Is your browser catching the period after html in the link? If so, remove that.