A parallelogram has sides with lengths of #18 # and #4 #. If the parallelogram's area is #12 #, what is the length of its longest diagonal?

Question

A parallelogram has sides with lengths of #18  # and #4  #.  If the parallelogram's area is #12  #, what is the length of its longest diagonal?

Evelyn Arboleda · Accepted Answer

Length of the longest diagonal AC = 21.9541

Given #l = 18, w = 4, Area = 12#

Area of the parallelogram = l * h = 12
#:. BE = CF = h = (Area)/l = 12 / 18 = 0.6667#

AE = DF = a = sqrt(w^2-h^2) = sqrt(4^2 - 0.6667^2) = 3.944#

AF = l + a = 18 + 3 .944= 21.944#

Longest diagonal AC #= sqrt(AF^2 + CF^2) = sqrt(21.944^2 + 0.6667^2) = 21.9541#

James Atkins · Answer

undefined The length of the longest diagonal of the parallelogram can be found using the formula: $ \text{Diagonal} = \sqrt{(a^2 + b^2 + 2ab)} $, where $ a $ and $ b $ are the lengths of the sides of the parallelogram.

Given that the sides of the parallelogram are 18 and 4, and the area is 12, we can use the formula for the area of a parallelogram, which is $ \text{Area} = \text{base} \times \text{height} $, to find the height of the parallelogram.

We know that $ \text{Area} = 12 $ and the base is 18. So, we can find the height:
$ 12 = 18 \times \text{height} $
$ \text{height} = \frac{12}{18} = \frac{2}{3} $

Now, we can use the Pythagorean theorem to find the length of the longest diagonal:
$ \text{Diagonal} = \sqrt{(18^2 + (\frac{2}{3})^2 + 2(18)(\frac{2}{3}))} $
$ \text{Diagonal} = \sqrt{(324 + \frac{4}{9} + 24)} $
$ \text{Diagonal} = \sqrt{(348 + \frac{4}{9})} $
$ \text{Diagonal} = \sqrt{\frac{3136}{9}} $
$ \text{Diagonal} = \sqrt{\frac{3136}{9}} $
$ \text{Diagonal} = \frac{56}{3} $

So, the length of the longest diagonal of the parallelogram is $ \frac{56}{3} $ units.