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?

1 Answer
Mar 13, 2018

Assuming a bit of basic Trigonometry...

Explanation:

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.