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?

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Answer 1 · 2017-05-10T10:44:03

#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/

Answer 2 · 2017-05-10T10:56:22

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]