Two opposite sides of a parallelogram have lengths of #3 #. If one corner of the parallelogram has an angle of #pi/12 # and the parallelogram's area is #14 #, how long are the other two sides?

Question

Two opposite sides of a parallelogram have lengths of #3  #.  If one corner of the parallelogram has an angle of #pi/12   # and the parallelogram's area is #14      #, how long are the other two sides?

Noah Atkinson · Accepted Answer

Assuming a bit of basic Trigonometry...

Let x be the (common) length of each unknown side. If b = 3 is the measure of the base of the parallelogram, let h be its vertical height. The area of the parallelogram is #bh = 14# Since b is known, we have #h = 14/3#. From basic Trig, #sin(pi/12) = h/x#. We may find the exact value of the sine by using either a half-angle or difference formula. #sin(pi/12) = sin(pi/3 - pi/4) = sin(pi/3)cos(pi/4) - cos(pi/3)sin(pi/4)# #= (sqrt6 - sqrt2)/4#. So... # (sqrt6 - sqrt2)/4 = h/x# # x(sqrt6 - sqrt2) = 4h# Substitute the value of h: # x(sqrt6 - sqrt2) = 4(14/3)# # x(sqrt6 - sqrt2) = 56/3# Divide by the expression in parentheses: # x = 56/(3(sqrt6 - sqrt2))# If we require that the answer be rationalized: # x = 56/(3(sqrt6 - sqrt2))*((sqrt6 + sqrt2)/(sqrt6 + sqrt2))# #= 56(sqrt6 + sqrt2)/(3(4))# #= (14(sqrt6 + sqrt2))/(3)# NOTE: If you have the formula #A = ab sin(theta)#, you may use it to arrive at the same answer more rapidly.

Maverick Smith · Answer

undefined The lengths of the other two sides of the parallelogram are 7 units each.