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There is a parallelogram whose acute angle is 60 degrees. Find the ratio of the lengths of the sides of the parallelogram if the squares of the lengths of the diagonals are related as 1:3 .

Posted by - Rohan


The answer is 1.

Rohan Chandra by Rohan Chandra , 24, India

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1 solution

Rohan Chandra
Jan 9, 2014

Please draw the figure yourself as here, I am not able to post the figure. Soln. -> Assume that the sides of the parallelogram are a & b, the smaller diagonal is d1 and the larger diagonal is d2. -> Then, 2 a 2 + 2 b 2 = d 1 2 + d 2 2 2a^{2} + 2b^{2} = d1^{2} + d2^{2} But, by the hypothesis, d 2 2 = 3 d 1 2 d2^{2} = 3d1^{2} -> Therefore, a 2 + b 2 = 2 d 1 2 a^{2} + b^{2} = 2d1^{2} -> By using the cosine rule, d 1 2 = a 2 + b 2 − 2 a b . ( c o s 60 d e g r e e s ) d1^{2} = a^{2} + b^{2} - 2ab.(cos 60 degrees) = 2 d 1 2 − a b , i . e , a b = d 1 2 2d1^{2} - ab , i.e, ab = d1^{2} -> Consequently, a 2 + b 2 + 2 a b a^{2} + b^{2} + 2ab = 2 d 1 2 + 2 d 1 2 2d1^{2} + 2d1^{2} = 4 d 1 2 4d1^{2} or, ( a + b ) 2 (a+b)^ {2} = 4 d 1 2 4d1^{2} i.e., a + b = 2 d 1 2 2d1^{2}

We got a system : 1) a + b = 2d1 ; 2) a . b = d 1 2 a.b = d1^{2}

whose solution is a = b = d1 , i.e., a/b = 1 \boxed{1}

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