Rhombus EFGH is shown. What is the length of EF ? A. 3 B. 6 C. 8 D. 9

D

Step-by-step 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

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step-by-step explanation:

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