A parallelogram has sides with lengths of #18 # and #4 #. If the parallelogram's area is #48 #, what is the length of its longest diagonal?

1 Answer
Aug 21, 2016

#sqrt(340+48sqrt(5)) ~~21.15#

Explanation:

enter image source here

The problem is to find #AC# in the image above, given
#AB=CD=18#
#AD=BC=4#
#"Area"_(ABCD)=48#

As the area of a parallelogram is given by the product of its base and height, we have #AE*CD=48#. Thus

#AE = 48/(CD)=48/18=8/3#

As #triangleAED# is a right triangle, we may apply the Pythagorean theorem to find that #AE^2+ED^2=AD^2#. Substituting in our known values for #AE# and #AD#, we get

#(8/3)^2+ED^2=4^2#

#=> ED=sqrt(16-64/9) = sqrt(80/9)=(4sqrt(5))/3#

As #triangleAEC# is also a right triangle, we can apply the Pythagorean theorem once more to get #AE^2+EC^2=AC^2#. As #EC = ED+CD#, we can once again substitute in our known values:

#AC^2 = AE^2+(ED+CD)^2#

#= (8/3)^2+((4sqrt(5))/3+18)^2#

#=340+48sqrt(5)#

Then, taking the square root, we get our final result:

#AC = sqrt(340+48sqrt(5)) ~~21.15#