Thread: Help with a special right triangles problem

hey can someone show me how to do this problem please?

thanks

Originally Posted by ashleynicolexox

hey can someone show me how to do this problem please?

thanks

The hypotenuse of such a right triangle is sqrt(2) times the length of one of
the other sides.

So going from the side of length x clockwise the sides radiating from the
central point have lengths:

x,
x sqrt(2),
sqrt(2) (x sqrt(2)),
sqrt(2) (sqrt(2) (x sqrt(2))),
sqrt(2) (sqrt(2) (sqrt(2) (x sqrt(2))))

but this last one we are told is 4, so:

sqrt(2) (sqrt(2) (sqrt(2) (x sqrt(2)))) = 4x = 4,

so x=1.

RonL