We have two right isosceles triangles. Hypotenuse of that triangle is a [tex] \sqrt{2} [/tex] or you can use the Pythagorean theorem: 11² + 11² = c²
121 + 121 = c²
242 =c² c= [tex] \sqrt{242}= \sqrt{121*2}=11 \sqrt{2} [/tex] Answer is the same: B) 11 square root 2
