A parallelogram has sides with lengths of #15 # and #12 #. If the parallelogram's area is #81 #, what is the length of its longest diagonal?

Question

A parallelogram has sides with lengths of #15  # and #12  #.  If the parallelogram's area is #81  #, what is the length of its longest diagonal?

Avery Allen · Accepted Answer

The length of the longest diagonal is #=26.8u#

The dimensions of the parrallelogram are

#AB=15#

#AD=12#

The #area=81#

But the area of a parallelogram is

#area=AB*AD*sin hatA#

#15*12*sin hatA=81#

#sin hat A=81/(15*21)=0.257#

#hatA=14.9^@#

The longest diagonal is #=AC#

We apply the cosine rule to the triangle #DeltaABC#

#AC^2=AB^2+BC^2-2*A*B*cos(180^@-14.9^@)#

#=15^2+12^2-2*15*12*cos(165.1^@)#

#=716.9#

#AC=sqrt716.9=26.8#

Mason Adams · Answer

undefined To find the length of the longest diagonal of the parallelogram, you can use the formula:

$ \text{Area} = \text{Base} \times \text{Height} $

Given that the area of the parallelogram is 81 and the base is 12, you can rearrange the formula to solve for the height:

$ 81 = 12 \times \text{Height} $

$ \text{Height} = \frac{81}{12} = 6.75 $

Now, to find the length of the longest diagonal (denoted by $ d $), you can use the Pythagorean theorem, since the diagonals of a parallelogram bisect each other:

$ d^2 = 15^2 + 6.75^2 $

$ d^2 = 225 + 45.5625 $

$ d^2 = 270.5625 $

$ d = \sqrt{270.5625} $

$ d \approx 16.45 $

Therefore, the length of the longest diagonal of the parallelogram is approximately 16.45.