# 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?

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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.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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