A parallelogram has sides with lengths of #16 # and #15 #. If the parallelogram's area is #48 #, what is the length of its longest diagonal?

Question

A parallelogram has sides with lengths of #16  # and #15  #.  If the parallelogram's area is #48  #, what is the length of its longest diagonal?

Wyatt Anderson · Accepted Answer

#=31# Area of Parallelogram #=48=ab sintheta# where #a=16# and #b=15# or #48=16times15timessintheta# or #sin theta=48/16times1/15# or #sin theta=1/5# or #theta=sin^-1(1/5)# or #theta=11.54# To find the longer diagonal #=y=?# we have to get the supplementary of the angle #11.54# So we have Angle #180-11.54=168.46# Using the Law of Cosine we can write #y^2=16^2+15^2-2times16times15cos(168.46)# #=256+225-2times16times15(-1)# #=481+480# #=961# or #y=sqrt961# #=31#

Claire Akin · Answer

undefined To find the length of the longest diagonal of the parallelogram, we can use the formula for the area of a parallelogram, which is given by $ \text{Area} = base \times height $. Given that the area is 48 and one of the sides (base) is 16, we can rearrange the formula to solve for the height, which gives us $ height = \frac{Area}{base} = \frac{48}{16} = 3 $. Now, we can use the Pythagorean theorem to find the length of the longest diagonal. The diagonal splits the parallelogram into two congruent triangles. The length of the diagonal, the base (16), and the height (3) form a right triangle. So, using the Pythagorean theorem, $ \text{longest diagonal} = \sqrt{16^2 + 3^2} = \sqrt{256 + 9} = \sqrt{265} $. Therefore, the length of the longest diagonal of the parallelogram is $ \sqrt{265} $ units.