Question

A parallelogram has sides with lengths of `7` and `15`. If the parallelogram's area is `8`, what is the length of its longest diagonal?

Answer 1

`BD=sqrt(274+2sqrt(10961))=21.986...`

Explanation:

Let the parallelogram `ABCD` , `AB=7`, `BC=15` and `∠ABC` be acute angle.

(I'm not good at drawing figures.The acutual parallelogram would be far flatter.)

Then, `BD` is the longest diagonal.

If the parallelogram's area is `8`, the area of `△BCD` is `4`.
Let `∠BCD=theta`.

`1/2*BC*CD*sin∠BCD=△BCD`
`1/2*15*7*sintheta=4`
`sintheta=8/105`

As `theta` is an obutase angle,
`costheta=-sqrt(1-sin^2theta)`
`=-sqrt(1-(8/105)^2)`
`=-sqrt(10961)/105≒-0.9971`

Then, calculate `BD` with law of cosine.
`BD^2=BC^2+CD^2-2*BC*CD*costheta`
`=15^2+7^2-2*15*7*(-sqrt(10961)/105)`
`=274+2sqrt(10961)`

`BD=sqrt(274+2sqrt(10961))=21.986...`