Question

A parallelogram has sides with lengths of `12` and `8`. If the parallelogram's area is `36`, what is the length of its longest diagonal?

Answer 1

Before going into calculation details it is necessary to understand the problem analytically

Analysis
In the problem we have been given the lengths of two sides `(a =12 and b = 8 )` of a parallelogram and its area ,`A=36`. If `theta` be the angle between these sides,then we can write `A=a*b*sintheta`
OR, `36=12xx8xxsintheta`
`=>sintheta=36/(12xx8)=3/8`
The value of `sintheta` is such that `theta` may be acute (<90) or may be obtuse (>90) .
But we are to take the obtuse value of `theta` to have its opposite diagonal longest.
By properties of triangle the length of the diagonal (d)

`d=sqrt(a^2+b^2-2abcostheta`

Calculation of longest diagonal

`theta` being obtuse `costheta <0`
So `costheta =-sqrt(1-sin^2theta`

Hence the length oflongest diagonal
`d_l=sqrt(a^2+b^2-2ab(-sqrt(1-sin^2theta))`
`=>d_l=sqrt(a^2+b^2+2ab(sqrt(1-sin^2theta))`
`=>d_l=sqrt(12^2+8^2+2xx12xx8xx(sqrt(1-(3/8)^2))`
`=>d_l=sqrt(208+192xx(sqrt(55/8^2))`
`=>d_l=sqrt(208+192xx0.93)~~19.66`