Question 1060917
{{{drawing(300,300,-5,0,-1,4,
triangle(-3.74,0,0,0,-3.74,2.45),
triangle(-1.74,3.46,0,0,-3.74,2.45),
locate(-2,-0.1,sqrt(14)),locate(-3.7,1.5,sqrt(6)),
locate(-2.74,2.95,sqrt(5)),rectangle(-3.74,0,-3.54,0.2),
line(-1.95,3.35,-1.84,3.15),line(-1.64,3.25,-1.84,3.15),
locate(-1,1.75,x)
)}}}
The line segment in the middle of the figure is the hypotenuse of both right triangles.
According to the Pythagorean theorem it's length is
{{{sqrt((sqrt(14))^2+(sqrt(6))^2)=sqrt(14+6)=sqrt(20)}}} .
It is also
{{{sqrt(20)=sqrt(x^2+(sqrt(5))^2)=sqrt(x^2+5)}}}
That gives us the equation
{{{sqrt(20)=sqrt(x^2+5)}}} and we can solve that equation for {{{x}}} :
{{{20=x^2+5}}}
{{{20-5=x^2}}}
{{{15=x^2}}}
{{{highlight(x=sqrt(15))}}}