Question

select the correct answer.
rhombus efgh is shown. what is the length of \\(\overline{ef}\\)?
a. 3
b. 6
c. 8
d. 9

Explanation:

Step1: Recall rhombus properties

In a rhombus, diagonals bisect each other, so \(HJ = JE\) and \(EJ = JF\)? Wait, no, diagonals bisect each other, so \(HJ = JF\)? Wait, no, in rhombus EFGH, diagonals EG and FH intersect at J. So \(HJ = JF\) and \(EJ = JG\)? Wait, the sides: all sides of a rhombus are equal, so \(EF = EH = HG = GF\). Also, diagonals bisect each other, so \(EJ = JG\) and \(HJ = JF\). From the diagram, \(HJ = 3x + y\), \(JF = 4x\)? Wait, no, the segments: \(EJ = y + 5\), \(JG = 4x\)? Wait, maybe the diagonals bisect each other, so \(EJ = JG\) and \(HJ = JF\). Wait, the side \(EF\) is \(2x + y + 2\), and \(EH\) is \(3x + y\) (since \(HG = EH\) in rhombus). Also, diagonals bisect each other, so \(EJ = JG\) (so \(y + 5 = 4x\)) and \(HJ = JF\)? Wait, maybe the equations:

From diagonals bisecting each other: \(EJ = JG\) ⇒ \(y + 5 = 4x\) (Equation 1)

From sides of rhombus: \(EF = EH\) ⇒ \(2x + y + 2 = 3x + y\) (Equation 2)

Simplify Equation 2: \(2x + y + 2 = 3x + y\) ⇒ Subtract \(y\) from both sides: \(2x + 2 = 3x\) ⇒ \(x = 2\)

Now plug \(x = 2\) into Equation 1: \(y + 5 = 4(2)\) ⇒ \(y + 5 = 8\) ⇒ \(y = 3\)

Now find \(EF\): \(EF = 2x + y + 2 = 2(2) + 3 + 2 = 4 + 3 + 2 = 9\)? Wait, no, wait the options are 3,6,8,9. Wait maybe I messed up the sides. Wait \(EH\) is \(3x + y\), \(EF\) is \(2x + y + 2\), and in rhombus \(EH = EF\), so \(3x + y = 2x + y + 2\) ⇒ \(3x + y = 2x + y + 2\) ⇒ \(x = 2\), same as before. Then \(x = 2\), then from diagonals: \(EJ = y + 5\), \(JG = 4x\), so \(y + 5 = 4x\) ⇒ \(y + 5 = 8\) ⇒ \(y = 3\). Then \(EF = 2x + y + 2 = 2*2 + 3 + 2 = 4 + 3 + 2 = 9\). Wait but let's check the length of \(EF\). Wait maybe the side is \(EF\), and with \(x=2\), \(y=3\), \(EF = 2*2 + 3 + 2 = 9\)? But let's check the options: D is 9. Wait but let's verify again.

Equation 2: \(2x + y + 2 = 3x + y\) ⇒ \(2x + 2 = 3x\) ⇒ \(x=2\). Correct.

Equation 1: \(y + 5 = 4x\) ⇒ \(y + 5 = 8\) ⇒ \(y=3\). Correct.

Then \(EF = 2x + y + 2 = 2*2 + 3 + 2 = 4 + 3 + 2 = 9\). So the answer is D. 9.

Step2: Solve for x and y

From \(EF = EH\) (rhombus sides equal):
\(2x + y + 2 = 3x + y\)
Subtract \(y\) from both sides: \(2x + 2 = 3x\) ⇒ \(x = 2\).

From diagonals bisecting each other (\(EJ = JG\)):
\(y + 5 = 4x\)
Substitute \(x = 2\): \(y + 5 = 8\) ⇒ \(y = 3\).

Step3: Calculate \(EF\)

Substitute \(x = 2\), \(y = 3\) into \(EF = 2x + y + 2\):
\(EF = 2(2) + 3 + 2 = 4 + 3 + 2 = 9\).

Answer:

D. 9