Question 1209054
<pre>
The four sides and one diagonal of a rhombus each have sides 3√6 cm long. Find the area of the rhombus, in cm^2.


Since one of its diagonals is equal to its sides, then 2 EQUILATERAL triangles will emerge, to form the rhombus

Each angle of each equilateral triangle = {{{180/3}}} = 60<sup>o</sup>
Area of any NON-RIGHT triangle = {{{1/2}}} the PRODUCT of 2 of its CONSECUTIVE sides, TIMES the sine of the INCLUDED angle 
So, AREA of one of this rhombus' 2 equilateral triangles = {{{(1/2)3sqrt(6) * 3sqrt(6) * sin 60^o}}} = {{{(1/2)(3sqrt(6))^2*sin 60^o}}} 
                                                                                  {{{matrix(3,1, (1/2)(9(6)(sqrt(3)/2)), (1/2)(54(sqrt(3)/2)), 27(sqrt(3)/2))}}} --- Substituting {{{sqrt(3)/2)}}} for sin 60<sup>o</sup>
As there are 2 of these equilateral triangles, AREA of both equilateral triangles/RHOMBUS = {{{2(27(sqrt(3)/2)))}}} = {{{cross(2)(27(sqrt(3)/cross(2))))}}} ={{{highlight(highlight_green(highlight(matrix(1,2, 27sqrt(3), cm^2))))}}}</pre>