Question 182836
A rhombus has one diagonal equal to the sides. 
<pre><font size = 4  color = "indigo"><b>
Here is the rhombus ABCD, with diagonal AC equal to the sides,
we let each side and the diagonal have length s: 
{{{drawing(300,300,-2,2,-2,2, triangle(-1,0,1,0,0,sqrt(3)),
triangle(-1,0,1,0,0,-sqrt(3)), locate(0,0,s),
locate(-1.2,.1,A), locate(0,-sqrt(3),B), locate(1.1,.1,C), locate(0,1.9,D),
locate(.6,1,s),locate(-.7,1,s),locate(.6,-.9,s),locate(-.6,-.9,s)
 )}}} 

The rhombus therefore consiste of two equilateral
triangles with a common side AC. 
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The other diagonal has length 60.
<pre><font size = 4  color = "indigo"><b>
So we draw the other diagonal DB, intersecting
the first diagonal at point E. Since the
diagonals of any parallogram bisect each other,
and since a rhombus is a parallelogram, the top
and bottom halves of diagonal DB are half of 60,
So DE = EB = 30.  Also AE = EC = {{{s/2}}}: 

{{{drawing(300,300,-2,2,-2,2, triangle(-1,0,1,0,0,sqrt(3)),
triangle(-1,0,1,0,0,-sqrt(3)), triangle(0,sqrt(3),0,-sqrt(3),1,0),
locate(-1.2,.1,A), locate(0,-sqrt(3),B), locate(1.1,.1,C), locate(0,1.9,D),
locate(.6,1,s),locate(-.7,1,s),locate(.6,-.9,s),locate(-.6,-.9,s),
locate(-.5,.5,s/2), locate(.4,.5,s/2),locate(.05,.2,E),
locate(.05,.87,30),locate(.05,-.7,30)
 )}}} 

Since the diagonals of a rhombus are also perpendicular,
triangle AED is a right triangle, so we can use the
Pythagorean theorem.  (We could use any one of the
four right triangles since all four are congruent.)

{{{matrix(12,5,
c^2, "=", a^2, "+", b^2,
AD^2, "=", AE^2, "+", ED^2,
s^2, "=", (s/2)^2,"+", 30^2,
s^2, "=", s^2/2^2,"+",900,
s^2, "=", s^2/4, "+", 900,
4s^2, "=", s^2, "+", 3600,
3s^2, "=", 3600, "", "",
s^2, "=",1200, "","",
s, "=", sqrt(1200), "","",
s, "=", sqrt(400*3), "", "",
s, "=", sqrt(400)sqrt(3), "","",
s, "=", 20sqrt(3), "","" )}}}

or each side s is about 34.64101615.

Edwin</pre>