A parallelogram has sides with lengths of #12 # and #8 #. If the parallelogram's area is #12 #, what is the length of its longest diagonal?

Question

A parallelogram has sides with lengths of #12  # and #8  #.  If the parallelogram's area is #12  #, what is the length of its longest diagonal?

Mason Adams · Accepted Answer

#color(green)("length of Longest diagonal " bar(AC) = 19.97#

#"Given : " AB = b = 12, AD = a = 8, Area = 12, " To find " AC = q#

#bar(EC) = h = (Area) / b = 12 / 12 = 1#

#bar(BE) = sqrt((BC)^2 - (CE)^2) = sqrt(a^2 - h^2)#

#bar(BE) = sqrt(8^2 - 1^2) = 7.94#

#bar(AE) = AB + BE = 12 + 7.94 = 19.94#

#bar(AC) = sqrt((AE)^2 + (EC)^2) = sqrt(19.94^2 + 1^2) = 19.97#

#color(green)("length of Longest diagonal " bar(AC) = 19.97#

Wyatt Anderson · Answer

undefined The length of the longest diagonal of the parallelogram can be found using the formula:

$$ \text{Longest diagonal} = \sqrt{a^2 + b^2 + 2ab\cos(\theta)} $$

Where $a$ and $b$ are the lengths of the sides of the parallelogram, and $\theta$ is the angle between these sides. However, if the area is given, we can use the formula for the area of a parallelogram to find $\theta$, then use it to calculate the longest diagonal.

$$ \text{Area} = ab\sin(\theta) $$

Given: $a = 12$, $b = 8$, and $\text{Area} = 12$.

From the formula for the area of a parallelogram:

$$ 12 = (12)(8)\sin(\theta) $$

Solving for $\theta$:

$$ \sin(\theta) = \frac{12}{12 \times 8} = \frac{1}{8} $$

$$ \theta = \arcsin\left(\frac{1}{8}\right) $$

Now, we can use $\theta$ to find the longest diagonal:

$$ \text{Longest diagonal} = \sqrt{12^2 + 8^2 + 2(12)(8)\cos(\theta)} $$

$$ \text{Longest diagonal} = \sqrt{144 + 64 + 192\cos(\theta)} $$

$$ \text{Longest diagonal} = \sqrt{208 + 192\cos\left(\arcsin\left(\frac{1}{8}\right)\right)} $$

$$ \text{Longest diagonal} ≈ \sqrt{208 + 192 \times 0.992} $$

$$ \text{Longest diagonal} ≈ \sqrt{208 + 190.464} $$

$$ \text{Longest diagonal} ≈ \sqrt{398.464} $$

$$ \text{Longest diagonal} ≈ 19.961 $$

So, the length of the longest diagonal of the parallelogram is approximately 19.961 units.