Answer:
D
Explanation:
The diagonals of a rhombus bisect each other, thus
FJ = JH, that is
y + 5 = 4x ( subtract 5 from both sides )
y = 4x - 5 → (1)
In a rhombus all sides are congruent, thus
HG = EF, substitute values
3x + y = 2x + y + 2 ( subtract y from both sides )
3x = 2x + 2 ( subtract 2x from both sides )
x = 2
Substitute x = 2 into (1)
y = 4(2) - 5 = 8 - 5 = 3
Thus
EF = 2x + y + 2 = 2(2) + 3 + 2 = 4 + 3 + 2 = 9 → D