Two opposite sides of a parallelogram have lengths of #8 #. If one corner of the parallelogram has an angle of #pi/8 # and the parallelogram's area is #12 #, how long are the other two sides?

2 Answers
Apr 7, 2017

Nasty answer. Here's a walkthrough.

Explanation:

Let h be the (perpendicular) height extending from one of the bases that measures length b = 8 to the other base having the same length.

since A = bh = 8h is the area of the parallelogram, we have

8h = 12
#h = 12/8 = 3/2# is its height.

Let c be the length of one of the unknown sides.
From basic trigonometry,

#sin(pi/8) = h/c#.

We may obtain the value of #sin(pi/8)# from the half-angle formula for sine.

That value is
#sin(pi/8) = sqrt(2 -sqrt(2))/2#

So that...

#sqrt(2 -sqrt(2))/2 = (3/2)/c#

Cross multiply:

#csqrt(2 -sqrt(2)) = 3#

#c = 3/(sqrt(2 -sqrt(2)))#

If we demand that the expression be rationalized, then
multiply first by
#sqrt(2 -sqrt(2))/sqrt(2 -sqrt(2))# and next by
#(2 + sqrt(2))/(2 +sqrt(2))#.

Apr 7, 2017

Picture

Explanation:

enter image source here

If you also have the formula

#A = bcsin(alpha)#

where #alpha# is the angle included between sides b and c, you may get there more quickly.

#12 = 8csin(pi/8)#.

Use the value of the sine, and it comes out quickly.