A parallelogram has sides with lengths of #16 # and #9 #. If the parallelogram's area is #120 #, what is the length of its longest diagonal?

Question

A parallelogram has sides with lengths of #16  # and #9  #.  If the parallelogram's area is #120  #, what is the length of its longest diagonal?

Savannah Aguilar · Accepted Answer

Longest diagonal measures 22.2755

Given #l = 16, w = 9, Area = 120#

Area of the parallelogram = l * h = 120
#:. BE = CF = h = (Area)/l = 120 / 16 = 7.5#

AE = DF = a = sqrt(w^2-h^2) = sqrt(9^2 - 7.5^2) = 4.9749#

AF = l + a = 16 + 4.9749 = 20.9749#

Longest diagonal AC #= sqrt(AF^2 + CF^2) = sqrt(20.9749^2 + 7.5^2) = 22.2755#

Aaliyah Anderson · 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 base times height. In this case, the base could be either of the sides, and the height would be the length of the other side. However, we need to determine which side is the base and which is the height.

Let's denote the sides of the parallelogram as $ a = 16 $ and $ b = 9 $. We know that the area of the parallelogram is 120. So, using the formula for the area $ A = b \times h $, where $ b $ is the base and $ h $ is the height, we can solve for the height:

$$ 120 = 16 \times h $$

Solving for $ h $, we get:

$$ h = \frac{120}{16} = 7.5 $$

Now, since the height of the parallelogram is perpendicular to the base, we can consider the longer side, 16, as the base, and the height as 7.5.

Using the Pythagorean theorem, we can find the length of the longest diagonal (the hypotenuse of the triangle formed by the sides and the diagonal):

$$ c^2 = a^2 + b^2 $$

$$ c^2 = 16^2 + 7.5^2 $$

$$ c^2 = 256 + 56.25 $$

$$ c^2 = 312.25 $$

$$ c = \sqrt{312.25} $$

$$ c \approx 17.66 $$

So, the length of the longest diagonal of the parallelogram is approximately 17.66.