A parallelogram has sides with lengths of #16 # and #8 #. If the parallelogram's area is #32 #, what is the length of its longest diagonal?

Question

A parallelogram has sides with lengths of #16  # and #8  #.  If the parallelogram's area is #32  #, what is the length of its longest diagonal?

Autumn Armstrong · Accepted Answer

#23.830#

As shown in the diagram, #ABCD# is the parallelogram.
Area of a parallelogram #A= bxxH#, where #b# is the base and #H# is the height.
Given that #AB=CD=8, AD=BC=16#, and area #A=32#,
#=> 32=8xxH, => H=4#

#DeltaBCE# is a right triangle.
Using Pythagorean theorem, we know that
#BC^2=CE^2+H^2#
#=> CE=sqrt(16^2-4^2)=sqrt240=4sqrt15#

#DeltaBDE# is also a right triangle.
#DB^2=DE^2+H^2#
#DE=DC+CE=8+4sqrt15=23.492#
#DB=sqrt(23.492^2+4^2)=23.830#

Hence, longest diagonal = #DB=23.830#

Theodore Abad · 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 the product of the base and the height. In this case, the base is given by one of the sides of length 16, and the height can be calculated using the given area.

Area of parallelogram = base × height

Given that the area is 32 and one of the sides (base) has a length of 16:

32 = 16 × height

Solving for the height:

height = 32 / 16
height = 2

Now, we can use the Pythagorean theorem to find the length of the longest diagonal. In a parallelogram, the diagonals bisect each other, so each half of the diagonal forms a right triangle with the base (one of the sides) and the height.

Using the given side lengths of 16 and 8:

$ \text{Longest diagonal}^2 = 16^2 + (2 \times 8)^2 $

$ \text{Longest diagonal}^2 = 16^2 + 2^2 \times 8^2 $

$ \text{Longest diagonal}^2 = 256 + 64 $

$ \text{Longest diagonal}^2 = 320 $

Taking the square root of both sides:

$ \text{Longest diagonal} = \sqrt{320} $

$ \text{Longest diagonal} \approx 17.89 $

So, the length of the longest diagonal of the parallelogram is approximately 17.89 units.