Question

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?

Answer 1

`sqrt(277+126sqrt3)~=22.254`

Explanation:

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

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^@`