A parallelogram has sides with lengths of 14 and 9 . If the parallelogram's area is 63 , what is the length of its longest diagonal?

1 Answer
Mar 22, 2016

sqrt(277+126sqrt3)~=22.254

Explanation:

Consider the parallelogram of the figure below where a=14 and b=9

I created this figure using MS ExcelI created this figure using MS Excel

In this figure I made
90^@ < alpha<180^@
0^@ < beta<90^@
h is the height relatively to sides AB and CD

Comparing the triangles ABC and ABD it's intuitive that the long diagonal is BD because this segment is opposed to an internal angle (alpha) bigger than the internal angle (beta) which AC (short diagonal) is opposed to.

So we want to find BD

In triangle_(ADE)
cos(alpha-90^@)=h/b => cos alpha*cos 90^@+sin alpha*sin 90^@=h/b
=> sin alpha=h/b

Finding cos alpha (remember that 90^@ < alpha<180^@ => cos alpha<0)
-> cosalpha=-sqrt(1-h^2/b^2)=- sqrt(b^2-h^2)/b

Applying the Law of Cosines to triangle_(ABD) we get

BD^2=a^2+b^2-2abcosalpha
BD^2=a^2+b^2-2a cancel(b)(-sqrt(b^2-h^2)/cancel(b))
BD^2=a^2+b^2+2a*sqrt(b^2-h^2)

And we know that
S_("parallelogram")=base*height
63=14*h => h=9/2

So

BD^2=14^2+9^2+2*14*sqrt(9^2-(9/2)^2)
BD^2=196+81+28*(9/2)*sqrt(4-1)
BD^2=277+126sqrt(3)
BD=sqrt(277+126sqrt3)~=22.254

By the way, in this question
sin alpha=h/b=4.5/9=1/2 => alpha=150^@