Two opposite sides of a parallelogram have lengths of #12 #. If one corner of the parallelogram has an angle of #pi/4 # and the parallelogram's area is #45 #, how long are the other two sides?

Question

Two opposite sides of a parallelogram have lengths of #12  #.  If one corner of the parallelogram has an angle of #pi/4   # and the parallelogram's area is #45      #, how long are the other two sides?

Claire Akin · Accepted Answer

Assume that the slant side is 12, then solve for #b#, to find #b=sqrt2#

I'm going to assume that the opposite sides of 12 are the "slant" sides. We're being asked for the length of the other 2 opposite sides.

So what do we know? Well, we know the area of the parallelogram is 45. And with the slant sides being known, we're looking for the base sides. The relationship is:

#A=bh#

So if we can figure out #h#, we can find #b#.

We know that the slant side is 12. We also know the angle of the "pointy corner" (the acute angle, not the obtuse one) is #pi/4#. If we can find a right triangle, maybe we can use Pythagoras Theorem or some other method to find #h#.

#pi/4# or #45^o# is a special angle - it tells us that the relationship between the hypotenuse of a right triangle (the slant side of 12) and the bases (which are #h# and a portion of #b#) is:

base = 1, hypotenuse = #sqrt2#

So we can divide the hypotenuse by #sqrt2# to find the base:

#12/sqrt2=12/sqrt2*sqrt2/sqrt2=(12sqrt2)/2=6sqrt2#

We now know that #h=6sqrt2#. We can plug that into our equation for the area of the parallelogram:

#A=bh#

#12=(6sqrt2)(b)#

Let's solve for #b#

#12/(6sqrt2)=b#

#b=2/sqrt2=2/sqrt2*sqrt2/sqrt2=(2sqrt2)/2=sqrt2#