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Diagonals of parallelogram?
I've been trying to solve this problem that's in the College Board's SAT review guide.
Imagine a parallelogram. What's the ratio of the diagonals? The answers are:
a) sqrt(2)/1
b) sqrt(3)/1
c) sqrt(2)/2
d) sqrt(3)/2
e) sqrt(3)/sqrt(2)
I fiddled with using the law of cosines. The angles opposing those 2 diagonals are related. I didn't get very far. Thanks for any guidence.
I just realized that there must be some other detail that I've not included. A very "flat" parallelogram would have a ratio that approaches infinity. Hmm. I ran into this while tutoring a student today. I don't have the book. I'll have to contact her and see what detail I've omitted.
Let's assume that it's a rhombus.
Thanks husoki. I see that just saying rhombus doesn't nail it down enough, I actually need some information about the angles. I'm still waiting to hear back from my tutoring student regarding the statement of the question.
2 Answers
- husoskiLv 710 years agoFavorite Answer
The diagonals of a rhombus divide it into four congruent right triangles, where each leg is half of one of the the diagonals. So, the ratio of the diagonals is the same as the ratio of the legs. That's the tangent of an acute angle of the right triangle. The diagonals also bisect the angles of the rhombus, so that acute angle is half of one of the rhombus angles. So in a rhombus ABCD, the ratio BD/AC = tan (A/2).
A little bit of confirmation. You can get two different ratios BD/AC or AC/BD, but they're related by being reciprocals. The two different angles (adjacent angles) add up to 180. That's related by trig:
Since tan(90 - x) = cot(x) = 1/tan x, then tan[(180 - x)/2] = tan(90 - x/2) = 1/tan(x/2).
So, the tangents of the adjacent angles are reciprocals.of each other.
