Lesson: Special Right Triangles

Comment on Special Right Triangles

great job .....i like you method sooooo much it is understandable
Abdul Hannan's picture

Hi Mr Hanneson,

Are Trigonometric identities (sin, cos, tan) etc and their principles part of GRE Syllabus.

greenlight-admin's picture

No, you don't need to know anything about trig identities.


what is a base triangle? and how are you assuming s particular base triangle for a particular question?
greenlight-admin's picture

At 0:40 and 1:22 in the above video, I explain what a base triangle is.

For 45-45-90 special triangles, the BASE triangle has measurements 1, 1 and √2
For 30-60-90 special triangles, the BASE triangle has measurements 1, √3 and 2

There are, of course, infinitely many 45-45-90 special triangles and 30-60-90 special triangles. However, if we know the measurements of the BASE triangles, we can use them to determine the measurements of all other sizes of special triangles.

Does that help?



I have a query regarding the triangle angles and the respective legs. If let's assume that 15 degrees equal to 1, then can we assume that 30 degrees would equal to 2?
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/in-the-figure-point-d-divides-side-bc-o...

Great idea! Unfortunately, the relationship between an angle and the length of its opposite side is not linear.


I think there's an error in the answer choices i got 18root3
greenlight-admin's picture

Good catch.
You're right; the correct answer is, indeed, 18√3

I've edited the question accordingly.


Are we going to be told when figures are drawn to scale or not
greenlight-admin's picture

No, you won't be told.
So, all geometric figures may or may not be drawn to scale.

I understand that is part / whole but why both quantities are equal ?

(4√3 - n)/2 = n/√3
Thank you

greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/topic14882.html

With any two SIMILAR triangles, the ratios of corresponding sides will be equal.
Notice that, for each fraction, I am comparing corresponding sides.

More here: https://www.greenlighttestprep.com/module/gre-geometry/video/872



I came across this GMAT question and wanted to know what is wrong with my idea. In order to find the shaded area I took the total area 144 pi - Non shaded areas. So looking at the circle we can see below the center O is a semi circle and we can find its area by doing pir^2/2 which will give us 72 pi since the radius is 12. Also we cnan use the area of the sector to find the area of triangle YZX which comes out to be 24 pi. Doing this we would take 144 pi -(72 pi + 24 pi) and get 48 pi as the shaded area. Why is this approach wrong?
greenlight-admin's picture

Question link: https://gmatclub.com/forum/the-area-of-the-circle-above-with-center-o-is...

Everything is looking good with your solution. The only problem is that you never completed it.

You're correct to say that 72pi = the area beneath the diameter XZ.
You're also correct to say that 24pi = the area of sector XOY

You still have one unshaded piece remaining: triangle YOZ.

Cheers, Brent


I wanted to know for this problem would it be possible to solve via 30-60-90 triangles. My first idea was to take (the area of the shaded)/( the area of the total figure)
I got 1/3 as the answer, but that isn't right.
greenlight-admin's picture

There are TONS of 30-60-90 triangles hiding in the diagram. So, that's a great approach.
It's possible you made a small error. Try again.

Another approach is to recognize that equilateral triangle PRS is comprised of 6 IDENTICAL 30-60-90 triangles.
Since two of those identical triangles are shaded, we know that 2/6 (aka 1/3) of triangle PRS is shaded.
Since equilateral triangles PQR and PRS are the same, the 2 shaded triangles represent 1/6 of the entire shape.

in question https://gre.myprepclub.com/forum/bcd-is-an-equilateral-triangle-and-ab-12750.html why CE is not root 3/2 since BD=1 and CE bisects the side? Hence, ED=x=1/2 and CE=root3x=root3/2.
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/bcd-is-an-equilateral-triangle-and-ab-1...

You're correct to say that CE bisects side BD. However, we aren't told that side BD has length 1. We're told that side AB has length 1.
So, we can't say that ED = 1/2

Does that help?

Thank you very much.

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