Question 801765
{{{10}}}= length of one side of the rhombus (because the perimeter, 40, is the sum of 4 such side lengths)
{{{2x}}}= length of the shortest diagonal 
{{{2x+4=2(x+2)}}}= length of the longer diagonal (because it is 4 more than {{{2x}}})
{{{drawing(300,300,-9,9,-9,9,
line(-6,0,0,8),line(0,-8,6,0),
triangle(-6,0,0,0,0,-8),
green(triangle(6,0,0,0,0,8)),
green(rectangle(0,0,0.5,0.5)),
locate(2.5,0,x),locate(0.2,3.8,x+2),
locate(3,5,10)
)}}} The diagonals dplit the rhombus into 4 congruent triangles.
Applying the Pythagorean theorem to one of those 4 triangles, we get
{{{x^2+(x+2)^2=10^2}}}
{{{x^2+x^2+4x+4=100}}}
{{{2x^2+4x+4=100}}}
{{{2x^2+4x-96=0}}}
Dividing both sides of the equal sign by 2, the equation simplifies to
{{{x^2+2x-48=0}}}
Solving by factoring is easy.
Factoring we get
{{{(x+8)(x-6)=0}}} with solutions {{{x=-8}}} and {{{x=6}}}
Since a negative length does not make sense the solution is
{{{x=6}}}, which makes {{{x+2=8}}}
The area of the rhombus is the area of the 4 triangles
{{{2*6*8=highlight(96)}}}