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Comment on Geometry Strategies - Part I
I chose D for the first GRE
Therefore
0<AB<16 so AB can be any number from 1-15 therefore D
Why was this approach wrong
The problem is that you're
The problem is that you're ignoring the fact that ∠ACB = 50°
When we combine that fact with the fact that the two sides are equal length, there is only ONE triangle that meets those conditions.
As such, there is no range of values for the length of side AB.
For example, if I tell you that all three angles in ∆ABC are 60°, and that side CB and side AC both have length 10, then the third side (side AB) is 100% fixed in place. That is, since all three angles are 60°, the triangle is an EQUILATERAL triangle, which means all three sides have length 10. So, it would be incorrect to say that 0 < AB < 20
The property you are referring to does not apply if we know the angles in a triangle.
Does that help?
Cheers,
Brent
https://gre.myprepclub.com/forum
For this problem I approached it from a geometric view point. Realizing the x : x : x√2, I found each side of the square to be (√10)/2.
Then using the given information of the midpoints I solved for QR, QP, ST, and TU which came out to be (√10)/4 finding the areas of the two triangles QRS and TUP, I found each to be 10/16.
I then found the total area of the square which was (rt(10)/2) ^2 - (2* (10/16)) in order to get the area of PQST. When i did all this I found QA to be 5/4 and QB to be 3/2. Thereby resulting in QB as the answer choice.
Please let me know where I went wrong. Hope you are having a great start to the new year Brent!
Hey Ravin,
Question link: https://gre.myprepclub.com/forum/q-and-t-are-the-midpoints-of-opposite-s...
Hey Ravin,
Unfortunately the error in your solution occurs at the very beginning.
Since we don't have any 45-45-90 special triangles, there are no triangles in the ratio x : x : x√2
Notice that ∆RQS has side measurements of 0.5x, x and √5 (where side QS is the hypotenuse).
We can use the Pythagorean theorem to write (0.5x)² + x² = (√5)²
Once we solve this equation for x, we'll have the length of all the sides needed to answer the question.
Happy New Year to you as well!!
https://gmatclub.com/forum/a
For this question why cant I approach it by doing the volume of the rectangle divided by the volume of the cubical block?
Question link: https:/
Question link: https://gmatclub.com/forum/a-box-measuring-11-inches-by-10-inches-by-7-i...
That approach would work perfectly IF it were the case that the cubes were able to fill the box completely (with no available airspace).