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Comment on Special Right Triangles
great job .....i like you
Hi Mr Hanneson,
Are Trigonometric identities (sin, cos, tan) etc and their principles part of GRE Syllabus.
Thanks
No, you don't need to know
No, you don't need to know anything about trig identities.
Cheers,
Brent
what is a base triangle? and
At 0:40 and 1:22 in the above
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?
Cheers,
Brent
https://gre.myprepclub.com/forum
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?
Question link: https:/
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.
Cheers,
Brent
https://gre.myprepclub.com/forum
I think there's an error in the answer choices i got 18root3
Good catch.
Good catch.
You're right; the correct answer is, indeed, 18√3
I've edited the question accordingly.
Cheers,
Brent
Are we going to be told when
No, you won't be told.
No, you won't be told.
So, all geometric figures may or may not be drawn to scale.
https://gre.myprepclub.com/forum
I understand that is part / whole but why both quantities are equal ?
(4√3 - n)/2 = n/√3
Thank you
Question link: https:/
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
Cheers,
Brent
https://gmatclub.com/forum
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?
Question link: https:/
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
https://gmatclub.com/forum/if
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.
There are TONS of 30-60-90
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.
hello,
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.
Question link: https:/
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.