Given a parallelogram whose acute angle is 6 0 ∘ . Find the ratio of the lengths of the sides of the parallelogram if the squares of the lengths of the diagonals have a ratio of 1 : 3 .
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Very good use of Cosine rule , + 1 !
Solution using complex numbers:
Let one side of the parallelogram be represented by a ( a ∈ C ).
Then the other side is a r e i 3 π ( r ∈ R + ) where r is the ratio of the sides.
The given data is equivalent to the following equation: ∣ a ( 1 − r e i 3 π ) ) ∣ 2 ∣ a ( 1 + r e i 3 π ) ) ∣ 2 = 3
Solving this equation, we get r = 1
AWESOME SOLUTION ! +1 !
A parallelogram whose acute angle is 6 0 ∘ is a Rhombus.
Why is that necessary?
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I tried to minimize my solution. Adding a proof is easy. I can certainly add a proof.
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Ok then please add a proof . Thank you.
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@Rishabh Tiwari – I don't want to spoil the fun of a one liner :P
I'll give an incomplete hint to you to start with
Notice that the diagonals divide opposite angles into two equal halves each. Let ABCD be a Parallelogram with the Diagonals bisecting at O. (Diagonals of a //gm bisect each other)
Consider Triangle AOB with Angle A as the acute angle
Angle OAB is 30 degrees and Angle ABO is 60 degrees.
Angle AOB becomes 90 degrees. Similarly do so for Triangles AOD, DOC and COB.
You'll notice that you're done.
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@Mehul Arora – I agree with @Rishabh Tiwari 's objection. Plus, it is easy to constuct a parallelogram with one angle as 6 0 ∘ which isn't a rhombus. The fact that the square of the ratio of the diagonals is 3 plays a role in this problem.
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@A Former Brilliant Member – Exactly ! Thanx for commenting!
@Mehul Arora – Diagonals of a parallelogram don't bisect opposite angles, its the property of a rhombus! Hence I think u r wrong somewhere in that proof .
@Mehul Arora – Try this problem
can you prove it ?
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