A parallelogram has sides with lengths of #12 # and #8 #. If the parallelogram's area is #72 #, what is the length of its longest diagonal?

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Answer 1 · 2018-01-07T23:33:57

#sqrt(208+48sqrt(7))~~18.303#

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#Area# #of# #ABCD=(DF*BC)=72#

#=>(DF*8)=72#

#=>DF=9#

Using the Pythagorean Theorem, I know that
#DF^2+CF^2=DC^2#

#=> CF^2=DC^2-DF^2=12^2-9^2=63#

#=> CF=sqrt(63)=3sqrt(7)#

#BF=CF+BC=3sqrt(7)+8#

Again, using the Pythagorean Theorem,
#DF^2+BF^2=DB^2#

#DB^2=9^2+(8+3sqrt(7))^2=81+64+48*sqrt(7)+63=208+48sqrt(7)#

#=>DB=sqrt(208+48sqrt(7))~~18.303#

We can also find the length of the other diagonal using the same method and to verify which diagonal is the longer one.

#Area# #of# #ABCD=AE*DC=72#

#=>AE=6#

Using the Pythagorean Theorem,

#DE^2+AE^2=AD^2#

#=>DE^2=AD^2-AE^2=8^2-6^2=64-36=28#

#=>DE=sqrt(28)=2sqrt(7)#

#CE+DE=CD#

#=>CE=CD-DE=12-2sqrt(7)#

Again, using the Pythagorean Theorem,
#AE^2+CE^2=AC^2#

#=>AC^2=6^2+(12-2sqrt(7))^2=36+144-48sqrt(7)+28=208-48sqrt(7)#

#=>AC=sqrt(208-48sqrt(7))~~9.0002#

Since DB>AC, DB is the longest diagonal.