Question 1113653
By definition, all sides of a rhombus have the same length.
Also, the diagonals are perpendicular bisectors of each other.
Because of that, the diagonals divide the rhombus into 4 congruent right triangles.
For each of those triangles, the hypotenuse is a rhombus side, of length {{{BC=5cm}}} ,
and one leg length is {{{OA=OC=3.9cm}}} .
Applying the Pythagorean theorem we can find the other length, {{{x=OB=OD}}} :
{{{x^2+(3.9cm)^2=(5cm)^2}}} .
Solving for {{{x}}} ,
{{{x^2=(5cm)^2-(3.9cm)^2}}}
{{{x^2=25cm^2-15.21cm^2}}}
{{{x^2=9.79cm^2}}}
{{{x=sqrt(9.79cm^2)=approximately}}}{{{3.129cm}}}
 
Taking one leg as the base and the other leg as the height,
the area of each triangle can be calculated as
{{{(3.9cm)*(3.129cm)/2}}} ,
and the area of the rhombus (4 triangles) is
{{{4*(3.9cm)*(3.129cm)/2=2*(3.9cm)*(3.129cm)=approximately}}}{{{highlight(24.4cm^2)}}}
 
The length of diagonal BD is approximately
{{{BD=OB+OD=3.129cm+3.129cm=highlight(6.258cm)}}} ,
or we could say it is approximately {{{highlight(6.26cm)}}} .