Answer:
B. y = 4 root under 2, x = 7
B. 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.
now
triangle ABE is a right angled triangle so
taking 45 as reference angle,
sin45 = p/h
or, sin45 = AB/AE
or, sin45 = 4/y (AB = CD = 4)
or,y = = 4/sin45
[tex]y = \frac{4}{ \frac{1}{ \sqrt{2} } } \\ y = 4 \sqrt{2} [/tex]
now
ABCD is a rectangle
BD = AC = 3
in triangle ABE
[tex]y = 4 \sqrt{2} [/tex]
AB = 4
using pythagoras theorem
h² = p²+b²
or, y² = AB²+ EB²
or,
[tex] {(4 \sqrt{2} )}^{2} = {4}^{2} + b[/tex]
or, 32 = 16 + EB²
EB² = 16
so, EB = 4
Now
x = EB+BD
= 4+3
so, x = 7
