A parallelogram has sides with lengths of #15 # and #12 #. If the parallelogram's area is #99 #, what is the length of its longest diagonal?

Question

A parallelogram has sides with lengths of #15  # and #12  #.  If the parallelogram's area is #99  #, what is the length of its longest diagonal?

James Atkins · Accepted Answer

The length of the longest diagonal is #=25.9#

Let #AB=15#

and #AD=BC=12#

#AC# is the longest diagonal

Then,

the area of the parallegram is

#AB*AD*sin(hat(BAD))=99#

Therefore,

#sin(hat(BAD))=99/(15*12)=0.55#

#hat(BAD)=33.37º#

Then,

the angle #hat(ABC)=180-33.37=146.63º#

We apply the cosine rule to the triangle #DeltaABC#

#AC^2=AB^2+BC^2-2*AB*BC*cos(hat(ABC))#

#=15^2+12^2-2*15*12*cos146.63º#

#=669.65#

So,

#AC=sqrt669.65=25.9#

Sadie Allen · 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 those sides. Given that the area of the parallelogram is 99 and the lengths of the sides are 15 and 12, we can find the angle between the sides using the formula for the area of a parallelogram:

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

Substituting the given values:

$$ 99 = 15 \cdot 12 \cdot \sin(\theta) $$

From this, we can solve for $\sin(\theta)$, then find $\cos(\theta)$ since $\sin(\theta)^2 + \cos(\theta)^2 = 1$.

Once we have $\cos(\theta)$, we can substitute all the values into the formula for the longest diagonal to find its length.