# Two new gre problems need help with

1. Jul 19, 2011

### zaldar

expect I am going to have to skip ones like this during the test but thought I would get help with them anyway just for learning from the engineers here. Also putting both problems here in one post, hope that is ok.

Problem 1 15 in problem150002.pdf

1. The problem statement, all variables and given/known data
Quantitative comparison problems you have a square PRSU and a parallelogram drawn inside the square, PQ ST such that q and t are midpoints of opposite sides of the square. (file with picture attached). You know one side of the parallelogram (SQ) equals the square root of five. Question is how does the area of the region PQST compare with 3/2.

2. Relevant equations
Not sure honestly...

3. The attempt at a solution
Thought that you could use a 45 45 90 triangle as the lines to the midpoints would bisect the the 90 degree angle, quickly became obvious though that this was nonsensical as then tu would equal pu. Tried putting in a line QT making a rectangle on the bottom of the square but that didn't seem to help much either.

Problem 2 29 in problem 290003.pdf
Mostly looking for hints on how to do problems like this more quickly and without a calculator. I could do this one but it would take me ten minutes or so which on the GRE is not a good idea.
1. The problem statement, all variables and given/known data
You have an octagon with alternating sides of square root of 2 and 1 and you know that the octagon is equiangular. They want you to find the area of the polygon.

2. Relevant equations
You need to find the perimeter (which is easy for this one) and the apothem. You can find the angle measure using (180n-360)/n

3. The attempt at a solution
You can draw a right triangle from the corners and then use the properties of triangles to find the length of the apothem but you are not given any kind of radius (the geometry book I have gives that on the problems I found to practice) which makes the problem even more time consuming. Any tricks here appreciated!

Lastly a general question, is it always true that the height of a parallelogram is less than the length of the longest side or is that only true in certain circumstances?

Ok thanks, now to go see if there is a chemistry question I can help someone with or an algebra one.

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2. Jul 19, 2011

### Staff: Mentor

On the second one, I think you can solve it visually fairly easily.

Draw two horizontal lines connecting the top and bottom of the left and right sides "1". Draw two vertical lines connecting the left and right of the top and bottom "1" sides.

Now look at the triangles in the four corners, formed with the four "SQRT(2)" sides. Do you see something?

3. Jul 19, 2011

### Staff: Mentor

On question #1, what do the other 2 sides of the upper triangle need to be to have a hypoteneuse of SQRT(5)? (or at least what do they look like they are for this triangle?)

The area is the area of the square (which is what now?) minus the area of the bottom triangle (whose dimensions look to be similar to the top triangle...)...

4. Jul 20, 2011

### zaldar

ok on the first 1, the longest side of the triangle if you make it x then the shorter side would have to be 1/2 x since it is the midpoint. you can then you Pythagorean Therom to get x and both sides of the square (which are 2) so then the square has an area of 4. You can find the area of the triangles which is 1 (for a total of 2) so the area of the quadrilateral is 2 (4-2) which is larger than 3/2 answer is A.

Is there a trick to seeing it is larger than 3/2 without actually calculating it? If I worked this I could maybe get in done in 1.30 but if I had to think about it at all first not happening.

Thanks for all the help, and glad to see the engineers here know their geometry :).

On the octagon one, the triangles you get look like they MIGHT be isosceles 90 degree but I am not sure you can make that assumption, especially without calculating the angles. The rest could be squares, but again not comfortable just assuming that....

Knowing what the answer is it looks like you are supposed to assume that the triangles are 1/2 the area of the squares formed in your drawing and each of the squares has an area of 1 (the triangles would then give you an area of 1/2). Adding this together you get an area of 7. Looks like if I take the GRE again (taking it today at five) along with memorizing cubes I need to memorize the common right triangle sides.

Last edited: Jul 20, 2011
5. Jul 20, 2011

### Staff: Mentor

Good job. Yeah, memorizing the main triangles is a good idea. Have a good test!

6. Jul 20, 2011

### zaldar

Thanks, having issues in general with data analysis problems so any general tips you can give with those would be great. Would like to get into the 600 range which should be the top 50%.

Any comments on the octagon problem did I approach that one correctly?

7. Jul 20, 2011

### Staff: Mentor

On the octagon problem, after I mentally drew the 2 horizontal and 2 vertical lines, it looked a lot like the triangles could be 1-1-SQRT(2) triangles, which made the whole problem easier.

I didn't look at the answers, but as long as the correct answer was different enough from the other choices, the assumption about the 1-1-SQRT(2) triangles would probably be true.