+738 Hi guest!
So for this problem, we can connect WY so that we have 2 right triangles, \(\triangle WXY \text { and } \triangle WYZ\).
(Just in case you don't know why \(\triangle WYZ\) is a right triangle, it's because \(\triangle WXY\) is a 45-45-90 triangle (since it has equal legs) and 135-45=90.)
First, let's find WY. Triangle WXY is a right triangle, so using the Pythagorean theorem, \(WY=4\sqrt{2}\).
Now we know the 1 leg and the hypotenuse of \(\triangle WYZ\)! From here, we can just use the Pythagorean theorem.
\(a^2+(4\sqrt2)^2=9^2\)
\(a^2+32=81\)
\(a^2=49\)
\(\boxed{a=7}\)
I hope this helped you, guest!
:)
