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

Maverick Smith · Accepted Answer

Length of the longest diagonal AC = 30.9827

Given #l = 16, w = 15, Area = 16#

Area of the parallelogram = l * h = 16
#:. BE = CF = h = (Area)/l = 16 / 16 = 1#

AE = DF = a = sqrt(w^2-h^2) = sqrt(15^2 - 1^2) = 14.9666#

AF = l + a = 16 + 14.9666 = 30.9666#

Longest diagonal AC #= sqrt(AF^2 + CF^2) = sqrt(30.9666^2 + 1^2) = 30.9827#

James Atkins · Answer

undefined To find the length of the longest diagonal of the parallelogram, we can use the formula:

$$ \text{Area} = \text{Base} \times \text{Height} $$

Given that the area of the parallelogram is 16 and one of the sides (which acts as the base) is 15, we can find the height of the parallelogram:

$$ \text{Area} = \text{Base} \times \text{Height} $$
$$ 16 = 15 \times \text{Height} $$
$$ \text{Height} = \frac{16}{15} $$

Now, to find the length of the longest diagonal, we can use the Pythagorean theorem, as the diagonal, base, and height form a right triangle:

$$ \text{Longest diagonal}^2 = \text{Base}^2 + \text{Height}^2 $$
$$ \text{Longest diagonal}^2 = 15^2 + \left(\frac{16}{15}\right)^2 $$
$$ \text{Longest diagonal}^2 = 225 + \frac{256}{225} $$
$$ \text{Longest diagonal}^2 = \frac{225 \times 225 + 256}{225} $$
$$ \text{Longest diagonal}^2 = \frac{50681}{225} $$

Taking the square root of both sides:

$$ \text{Longest diagonal} = \sqrt{\frac{50681}{225}} $$

$$ \text{Longest diagonal} \approx 12.704 $$

Therefore, the length of the longest diagonal of the parallelogram is approximately 12.704 units.