A parallelogram has sides with lengths of #15 # and #8 #. If the parallelogram's area is #12 #, what is the length of its longest diagonal?

To find the length of the longest diagonal of a parallelogram, we can use the formula:

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

Given that the area of the parallelogram is 12, we can use the formula for the area of a parallelogram:

(A = ab\sin(\theta))

Substituting the given values, we have:

(12 = 15 \times 8 \times \sin(\theta))

Solving for ( \sin(\theta) ), we get:

(\sin(\theta) = \frac{12}{120} = \frac{1}{10})

Using the given sides of the parallelogram, we can find the length of the longest diagonal using the Law of Cosines:

(d^2 = 15^2 + 8^2 + 2(15)(8)\cos(\theta))

Substituting the values, we have:

(d^2 = 225 + 64 + 240\cos(\theta))

Now, we need to find the value of ( \cos(\theta) ). We know that ( \sin(\theta) = \frac{1}{10} ). Using the identity ( \sin^2(\theta) + \cos^2(\theta) = 1 ), we can find ( \cos(\theta) ):

(\cos^2(\theta) = 1 - \sin^2(\theta))

(\cos^2(\theta) = 1 - \left(\frac{1}{10}\right)^2 = 1 - \frac{1}{100} = \frac{99}{100})

(\cos(\theta) = \pm \sqrt{\frac{99}{100}} = \pm \frac{3\sqrt{11}}{10})

Since cosine is positive in the first and fourth quadrants, we take the positive value.

(d^2 = 225 + 64 + 240 \times \frac{3\sqrt{11}}{10})

(d^2 = 289 + 72\sqrt{11})

Finally, taking the square root of both sides, we get:

(d = \sqrt{289 + 72\sqrt{11}} \approx \sqrt{289} \approx 17)

So, the length of the longest diagonal of the parallelogram is approximately 17 units.