A parallelogram has sides with lengths of #16 # and #9 #. If the parallelogram's area is #96 #, 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 #96  #, what is the length of its longest diagonal?

Savannah Aguilar · Accepted Answer

Length of the longer diagonal is #23.49 (2dp) # unit

We know the area of the parallelogram as #A_p=s_1*s_2*sin theta or sin theta=96/(16*9)=0.67 :. theta=sin^-1(0.67)=41.81^0 # consecutive angles are supplementary #:.theta_2=180-41.81=138.19^0#. Longer diagonal can be found by applying cosine law: #d_l= sqrt(s_1^2+s_2^2-2*s_1*s_2*costheta_2)# #=sqrt(16^2+9^2-2*16*9*cos138.19) ~~ 23.49 (2dp) # unit Length of the longer diagonal is #23.49 (2dp) # unit [Ans]

Noah Atkinson · Answer

#22.76# units

You have two diagonals (#d_1# and #d_2#) You know your sides (#a=9# units and #b=16# units) Now you can compute diagonals by: #d_1 = sqrt(2a^2+2b^2-d_2^2)# #d_2 = sqrt(2a^2+2b^2-d_1^2)# When you solve these equations, you will find #d_1=12.47# units and #d_2=22.76# units. Reference: onlinemschool(dot)com(slash)math(slash)formula(slash)parallelogram/

Aaliyah Anderson · Answer

undefined To find the length of the longest diagonal of the parallelogram, you can use the formula for the area of a parallelogram, which states that the area $A$ is equal to the product of one of its sides and the corresponding altitude (height) to that side.

Given that the area $A$ is 96 and one of the sides has a length of 16, you can find the height of the parallelogram:

$$A = \text{Base} \times \text{Height} \Rightarrow 96 = 16 \times \text{Height}$$

Solve for the height:

$$\text{Height} = \frac{96}{16} = 6$$

Now, consider the other side of the parallelogram, which has a length of 9. The diagonal of the parallelogram can be found using the Pythagorean theorem:

$$ \text{Diagonal}^2 = 16^2 + 6^2$$

$$ \text{Diagonal}^2 = 256 + 36$$

$$ \text{Diagonal}^2 = 292$$

$$ \text{Diagonal} = \sqrt{292}$$

$$ \text{Diagonal} = \sqrt{4 \times 73}$$

$$ \text{Diagonal} = 2\sqrt{73}$$

So, the length of the longest diagonal of the parallelogram is $2\sqrt{73}$.