A parallelogram has sides with lengths of #24 # and #9 #. If the parallelogram's area is #14 #, what is the length of its longest diagonal?

Question

A parallelogram has sides with lengths of #24  # and #9  #.  If the parallelogram's area is #14 #, what is the length of its longest diagonal?

Evelyn Arboleda · Accepted Answer

#color(red) (32.986)# (approx.)

Since the #"Area"# is given as #14# (sq.units)
and (relative to side with length #24#) #"Area" = 24 xx h#

#rarr h= 14/24= 7/12#

By the Pythagorean Theorem
#color(white)("XXX")x=sqrt(9^2-(7/12)^2) ~~8.981076# (with calculator)

and the longest diagonal is
#color(white)("XXX")d=sqrt((24+x)^2+h^2)#

#color(white)("XXXX")=sqrt(32.981076^2 + (7/12)^2)#

#color(white)("XXXX")~~32.98623# (again, with calculator)

Caroline Aucoin · 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 the sides.

Given $a = 24$ and $b = 9$, we first find the angle $\theta$ using the formula for the area of a parallelogram:

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

Thus, $\sin(\theta) = \frac{\text{Area}}{ab} = \frac{14}{24 \times 9}$.

Now, we can find $\cos(\theta)$ using the identity $\sin^2(\theta) + \cos^2(\theta) = 1$.

Then, we can calculate the length of the longest diagonal using the formula above.