A parallelogram has sides with lengths of #7 # and #16 #. If the parallelogram's area is #32 #, what is the length of its longest diagonal?

Question

A parallelogram has sides with lengths of #7  # and #16  #.  If the parallelogram's area is #32  #, what is the length of its longest diagonal?

Wyatt Anderson · Accepted Answer

#=22.8# Area of Parallelogram #=32=ab sintheta# where #a=7# and #b=16# or #32=7times16timessintheta# or #sin theta=32/16times1/7# or #sin theta=2/7# or #theta=sin^-1(2/7)# or #theta=16.6# To find the longer diagonal #=y=?# we have to get the supplementary of the angle #16.6# So we have Angle #180-16.6=163.4# Using the Law of Cosine we can write #y^2=7^2+16^2-2times7times16cos(163.4)# #=49+256-2times7times16(-0.96)# #=305+214.66# #=519.66# or #y=sqrt519.66# #=22.8#

Theodore Abad · Answer

undefined Using the formula for the area of a parallelogram, A = base × height, we can find the height of the parallelogram. Since the base is 16 and the area is 32, the height is 2.

The longest diagonal of a parallelogram can be found using the Pythagorean theorem. Let's denote the longest diagonal as d, and the sides of the parallelogram as a = 7 and b = 16. The diagonal d, the height h, and one side of the parallelogram form a right-angled triangle.

Using the Pythagorean theorem, $d^2 = a^2 + h^2$, where h = 2 (the height we found earlier).

Substituting the values, $d^2 = 7^2 + 2^2 = 49 + 4 = 53$.

Therefore, the length of the longest diagonal of the parallelogram is √53 units.