```Question 333740
<pre><font size = 3 color = "indigo"><b>
Here is the rhombus drawn twice, once with each diagonal.  The four
sides of a rhombus have equal measures. Let each side have length x.
Then the perimeter of the rhombus will be 4x.

The drawings below are to scale:

{{{drawing(225,400,-9,9,-16,16,
green(line(-8,0,8,0), locate(0,1.5,16cm)),
locate(-8.9,0.6,A), locate(-.2,-15.1,B), locate(8.3,.6,C), locate(0,15.98,D),
locate(-5.3,8,x),locate(-4.6,-8,x),locate(4.4,-8,x),locate(4.8,8,x),
line(-8,0,0,15),line(0,15,8,0),line(8,0,0,-15),line(0,-15,-8,0))}}}{{{drawing(225,400,-9,9,-16,16,
green(line(0,15,0,-15), locate(.5,0,30cm)),
locate(-8.9,0.6,A), locate(-.2,-15.1,B), locate(8.3,.6,C), locate(0,15.98,D),
locate(-5.3,8,x),locate(-4.6,-8,x),locate(4.4,-8,x),locate(4.8,8,x),
line(-8,0,0,15),line(0,15,8,0),line(8,0,0,-15),line(0,-15,-8,0))}}}

Looking at the lower triangular half of the left drawing we use the
law of cosines on triangle ABC:

{{{AC^2}}}{{{""=""}}}{{{AB^2}}}{{{""+""}}}{{{BC^2}}}{{{""-""}}}{{{2*AB*BC*cos(B)}}}

{{{16^2}}}{{{""=""}}}{{{x^2}}}{{{""+""}}}{{{x^2}}}{{{""-""}}}{{{2*x*x*cos(B)}}}

{{{256}}}{{{""=""}}}{{{x^2}}}{{{""+""}}}{{{x^2}}}{{{""-""}}}{{{2*x^2*cos(B)}}}

{{{256}}}{{{""=""}}}{{{2x^2}}}{{{""-""}}}{{{2x^2cos(B)}}}

Divide every term by 2

{{{128}}}{{{""=""}}}{{{x^2}}}{{{""-""}}}{{{x^2cos(B)}}}

-------------------
Now we look at the left triangular half of the right drawing, and
we use the law of cosines again, this time on triangle ABD:

{{{30^2}}}{{{""=""}}}{{{x^2}}}{{{""+""}}}{{{x^2}}}{{{""-""}}}{{{2*x*x*cos(A)}}}

{{{900}}}{{{""=""}}}{{{x^2}}}{{{""+""}}}{{{x^2}}}{{{""-""}}}{{{2*x^2*cos(A)}}}

{{{900}}}{{{""=""}}}{{{2x^2}}}{{{""-""}}}{{{2x^2cos(A)}}}

Divide every term by 2

{{{450}}}{{{""=""}}}{{{x^2}}}{{{""-""}}}{{{x^2cos(A)}}}

---------

Here are the two equations we have found above:

{{{128}}}{{{""=""}}}{{{x^2}}}{{{""-""}}}{{{x^2cos(B)}}}
{{{450}}}{{{""=""}}}{{{x^2}}}{{{""-""}}}{{{x^2cos(A)}}}

A rhombus is a parallelogram, and the adjacent angles in a
parallelogram are supplementary.  Therefore the sum of the
measures of angles A and B is 180°.  Therefore

B = 180° - A

Therefore we use the identity:  {{{cos("180°"-theta)=-cos(theta)}}},

and get:

cos(B) = cos(180°-A) = -cos(A)

We substitute -cos(A) for cos(B) in the first equation and simplify:

{{{128}}}{{{""=""}}}{{{x^2}}}{{{""-""}}}{{{x^2cos(B)}}}
{{{128}}}{{{""=""}}}{{{x^2}}}{{{""-""}}}{{{x^2(-cos(A))}}}
{{{128}}}{{{""=""}}}{{{x^2}}}{{{""+""}}}{{{x^2cos(A)}}}

Now we put that together with the other equation and we have this system:

{{{system(128=x^2+x^2cos(A), 450=x^2-x^2cos(A))}}}

When we add those two equations term-by term, the terms on the right cancel

{{{system(128=x^2+cross(x^2cos(A)), 450=x^2-cross(x^2cos(A)))}}}

and we get:

{{{578=2x^2}}}

Divide both sides by 2

{{{289=x^2}}}

Taking positive square roots of both sides:

{{{sqrt(289)=sqrt(x^2)}}}

{{{17=x}}}

So the perimeter is 4x or 4(17) or 68.

Edwin</pre>```