Question 1123955
The area of the rhombus is {{{540cm^2}}}; the length of one of its diagonals is {{{d[1]}}} = 4.5 dm = 45 cm. 
What is the distance between the point of intersection of the diagonals and the side of the rhombus? 
~~~~~~~~~~~~~~~~~~~~~~~
 

I will try to produce the solution in more simple form than that by the tutor @MathLover1.



{{{drawing(300,400,-5.5,9.5,-3,3,
triangle(0,0,5,0,3.7,0.624),locate(2.6,0.43,r),
triangle(-5,0,5,0,0,2.4),triangle(0,-2.4,5,0,0,2.4),
line(-5,-2.4,5,-2.4),line(-5,2.4,5,2.4),
line(-5,-2.4,-5,2.4),line(5,-2.4,5,2.4),
red(line(-5,0,0,-2.4)),red(line(5,0,0,-2.4)),
red(line(-5,0,0,2.4)),red(line(5,0,0,2.4)),
locate(5.1,0.1,4.5dm=45cm),arrow(5.5,0.2,5.5,2.4),
arrow(5.5,-0.2,5.5,-2.4),locate(-2.55,0,x),locate(2.45,0,x)
)}}}


<pre>
The area of the rhombus is 4 times the area of the small right-angled triangle formed by its diagonals.


It gives the equation to find "x",  which is half of the second diagonal

{{{4*(1/2)*(45/2)*x}}} = 540,   which implies   x = {{{540/45}}} = 12 cm.



Then the side of the rhombus is equal

{{{sqrt((45/2)^2+12^2)}}} = {{{sqrt(650.25)}}} = 25.5 cm.     (Pythagoras)


The area of each of four small right-angled triangle formed by its diagonals is

{{{(1/2)*r*25.5}}} = {{{540/4}}} cm^2,   


where "r" is the distance under the question,  which gives 


r = {{{(540*2)/(25.5*4)}}} = {{{1080/102}}} = {{{180/17}}} = 10.59 cm (rounded with 2 decimal places).


<U>Answer</U>.  10.59 cm (rounded with 2 decimal places).
</pre>

Solved.