Answer:B. y = 4 root under 2, x = 7B. is the correct answer.Step-by-step explanation:I drew AB such that it is equal and parallel to CD and perpendicular to ED. So, ABCD is a rectangle.nowtriangle ABE is a right angled triangle sotaking 45 as reference angle,sin45 = p/hor, sin45 = AB/AEor, sin45 = 4/y (AB = CD = 4)or,y = = 4/sin45[tex]y = \frac{4}{ \frac{1}{ \sqrt{2} } } \\ y = 4 \sqrt{2} [/tex]nowABCD is a rectangle BD = AC = 3in triangle ABE[tex]y = 4 \sqrt{2} [/tex]AB = 4using pythagoras theoremh² = p²+b²or, y² = AB²+ EB²or, [tex] {(4 \sqrt{2} )}^{2} = {4}^{2} + b[/tex]or, 32 = 16 + EB²EB² = 16so, EB = 4Nowx = EB+BD = 4+3so, x = 7