Gre Math prep geometry problem 8C

In summary, a right triangle can be extended from one side of a parallelogram with a height of 4 and a base of 2, with the hypotenuse being the side of the parallelogram. The length of the long diagonal can be found using the Pythagorean theorem and the length of the short diagonal can be found by subtracting the length of the new side from the given length of the parallelogram's side. Practice with similar triangles and basic math skills can greatly improve problem solving in geometry.
  • #1
zaldar
21
0

Homework Statement



You have a parallelogram ABCD. You know one side has a length of 12. They have given a right triangle extended from one side of the parallelogram height 4 and base 2 hypotenuse is the side of the parallelogram. There is a diagonal that goes from A to C that they want you to find the length of. (the right triangle is off of the DC side of the parallelogram so I can't see a way to use the Pythagorean theorem here) Direct link to the pdf with the problems
http://www.ets.org/Media/Tests/GRE/pdf/GREmathPractice.pdf"

Homework Equations


The Pythagorean theorem and the area of a triangle (1/2 base times height) are all I can see as being useful here. May be others though that I don't see.


The Attempt at a Solution


The 4 2 right triangle can be used to find the length of the other side of the parallelogram which turns out to be 2[itex]\sqrt{5}[/itex] If this was a rectangle then since the diagonals would be equal this would be easy. Tried also using the area of the parallelogram and the fact that the diagonal makes two similar triangles but that only gives you the length of the side of the parallelogram you all ready know. At a loss now...any help very much appreciated!

Thanks very very much! Any location for geometry problems for more practice (this type especially) would be great. Have a geometry book that I have worked through but it didn't have anything like this.
 
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  • #2
I might be misunderstanding the problem (can't find it in your link) since this seems awfully easy to me. The length of the long diagonal is sqrt(160) and the length of the short diagonal is sqrt(116)

OOPS ... make that long is sqrt(212)
 
  • #3
Info: OP refers to Problem 8 on p. 48 of the document at the link.
 
  • #4
SteamKing said:
Info: OP refers to Problem 8 on p. 48 of the document at the link.

Hm ... how did you get from "8c" to "Prob 8 on p 48" ? You're right of course so I guess I'm missing something.

Anyway, my answer is correct for that figure.
 
  • #5
Sorry it seems easy for you. It is the 8th problem in the geometry section sorry I didn't make that clear enough. Can you explain HOW you got that answer? Got the one for the long diagonal, the short diagonal is what is giving me an issue.
 
  • #6
I tossed my figure but as I recall, the long one is sqrt (4**2 + (10+2+2)**2) and the short one is sqrt(4**2 + 10**2). It's all just simple right triangles and you know the base and side of each so the hypotenuse is just that trivial.
 
  • #7
hmm still missing something. Unless you drew something in I don't see how you have the base and the side for the right triangle formed by the long diagonal. The short diagonal is the only one you have the right triangle drawn in outside of the parallelogram.

Thanks for all your help, sorry it is taking me so long to see this.
 
  • #8
zaldar said:
hmm still missing something. Unless you drew something in I don't see how you have the base and the side for the right triangle formed by the long diagonal. The short diagonal is the only one you have the right triangle drawn in outside of the parallelogram.

Thanks for all your help, sorry it is taking me so long to see this.

Look for similar triangles (draw a couple that are obvious) and if you still can't get it, post again and I'll draw a figure.
 
  • #9
I think that hint did it, and I feel like an idiot now. To make sure I am on the right track, you can draw a straight line down from a to b in the INSIDE of the parallelogram this gives you two right triangles one that you use to find the length of the new side (lets call it b to e). You can find this length using the Pythagorean theorem. Then subtract this from twelve, to get the length of the triangle e to c. Then you can use the Pythagorean theorem again to get the diagonal length.

Doable but not in a minute thirty...at least for me. Thanks for the help guys going to work this now see if I get the right answer then find some more geometry problems where you have to draw stuff in.
 
  • #10
zaldar said:
I think that hint did it, and I feel like an idiot now. To make sure I am on the right track, you can draw a straight line down from a to b in the INSIDE of the parallelogram this gives you two right triangles one that you use to find the length of the new side (lets call it b to e). You can find this length using the Pythagorean theorem. Then subtract this from twelve, to get the length of the triangle e to c. Then you can use the Pythagorean theorem again to get the diagonal length.

Doable but not in a minute thirty...at least for me. Thanks for the help guys going to work this now see if I get the right answer then find some more geometry problems where you have to draw stuff in.

Yes, you've got it. I think whether or not you can do this kind of thing in 10 seconds or 5 minutes is a function of (1) how conversant are you with math basics and (2) how much practice do you have?

For basic stuff like this, I have lots of both so it jumped right out at me, but I see no reason why you should feel bad that it didn't for you. I've been doing this sort of thing for 50+ years.
 
  • #11
Ah I've been teaching chemistry so ratios and algebra and linear equations are more my thing. Geometry (and statistics) seem to be the weak point. Got any good locations for other problems like this? The high school geometry textbook I had didn't have that many this complicated. (though I am going to look again).

What made me especially mad about this one was I could get 11 which was more difficult I thought. (there it is a rectangle so everything is 90 degrees which actually made it easier).

Anyway thanks for all the help guys. Anyone have knowledge about the pHD in chemical education degree from purdue that is what I am going for here.
 

Related to Gre Math prep geometry problem 8C

1. What is the given geometric shape in Gre Math prep geometry problem 8C?

The given geometric shape in Gre Math prep geometry problem 8C is a trapezoid.

2. What is the formula for the area of a trapezoid?

The formula for the area of a trapezoid is 1/2 x (sum of parallel sides) x height.

3. How do you find the height of a trapezoid in Gre Math prep geometry problem 8C?

To find the height of a trapezoid in Gre Math prep geometry problem 8C, you can use the Pythagorean theorem or the formula h = (base1 + base2) / 2 x tan(angle between the bases).

4. What is the given value of angle A in Gre Math prep geometry problem 8C?

The given value of angle A in Gre Math prep geometry problem 8C is 60 degrees.

5. How do you find the area of the shaded region in Gre Math prep geometry problem 8C?

To find the area of the shaded region in Gre Math prep geometry problem 8C, you can subtract the area of the smaller trapezoid from the area of the larger trapezoid. This can be done by finding the height of both trapezoids and using the area formula for each.

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