A parallelogram has sides with lengths of #16 # and #9 #. If the parallelogram's area is #21 #, what is the length of its longest diagonal?

Question

A parallelogram has sides with lengths of #16  # and #9  #.  If the parallelogram's area is #21  #, what is the length of its longest diagonal?

Mason Adams · Accepted Answer

#24.938\ ext{unit# Let # heta# be the angle between the sides #16# & #9# then the area of parallelogram #2(1/2(16)(9)\sin heta)=21# #\sin heta=\frac{21}{144}# #\sin heta=7/48# #\implies \cos heta=\sqrt{1-\sin^2 heta}\ \quad (\because heta<\pi/2)# #\cos heta=\sqrt{1-(7/48)^2}# #\cos heta=\frac{\sqrt2255}{48}# Now, since # heta<\pi/2# hence the longest diagonal of parallelogram will be opposite to the angle #(\pi- heta)# Let #d# be the length of longest diagonal parallelogram. Applying cosine rule in a triangle with sides #16, 9# & #d# & #(\pi- heta)# opposite to the side (diagonal) #d# #\cos(\pi- heta)=\frac{16^2+9^2-d^2}{2(16)(9)}# #-\cos heta=\frac{337-d^2}{288}# #-\frac{\sqrt2255}{48}=\frac{337-d^2}{288}# #d=\sqrt{337+6\sqrt2255}# #=24.938# Hence the longest diagonal of parallelogram is #24.938\ ext{unit#

Mason Adams · Answer

undefined The area $ A $ of a parallelogram with side lengths $ a $ and $ b $ is given by $ A = ab \sin(\theta) $, where $ \theta $ is the angle between the two sides.

Given:
$ a = 16 $,
$ b = 9 $,
$ A = 21 $.

Using the formula $ A = ab \sin(\theta) $, we can solve for $ \sin(\theta) $:

$ 21 = 16 \cdot 9 \cdot \sin(\theta) $

$ \sin(\theta) = \frac{21}{144} $

Now, we can find the angle $ \theta $ using the inverse sine function:

$ \theta = \sin^{-1}\left(\frac{21}{144}\right) $

Next, we can find the longest diagonal $ d $ using the law of cosines for a parallelogram:

$ d^2 = a^2 + b^2 - 2ab \cos(\theta) $

Given:
$ a = 16 $,
$ b = 9 $,
$ \theta = \sin^{-1}\left(\frac{21}{144}\right) $,

We calculate:

$ d^2 = 16^2 + 9^2 - 2(16)(9)\cos(\theta) $

$ d^2 = 256 + 81 - 288\cos(\theta) $

Finally, we substitute $ \cos(\theta) = \sqrt{1 - \sin^2(\theta)} $:

$ d^2 = 256 + 81 - 288\sqrt{1 - \left(\frac{21}{144}\right)^2} $

$ d^2 = 337 - 288\sqrt{1 - \left(\frac{21}{144}\right)^2} $

Now, calculate:

$ d = \sqrt{337 - 288\sqrt{1 - \left(\frac{21}{144}\right)^2}} $

$ d \approx 17.14 $

So, the length of the longest diagonal of the parallelogram is approximately $ 17.14 $ units.