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--- ## Question 1 **Question content**: In rhombus \( EFGH \), the diagram shows side \( EH = 4x \), side \( HG = 3x + y \), side \( EF = 2x + y + 2 \), and segments of the diagonals \( EJ = y + 5 \) and \( JG = 3x + y \) (where \( J \) is the intersection of the diagonals). Determine the length of \( EH \). The options are: A. 3, B. 6, C. 8, D. 9. **Discipline**: Mathematics **Grade**: High School (Geometry, focusing on properties of rhombuses) ### Correct answer \(\boldsymbol{8}\) (Option C) ### Detailed problem-solving steps 1. **Recall the property of a rhombus**: All sides of a rhombus are equal in length. Thus, \( EH = HG \) and \( EH = EF \). 2. **Set up equations from equal sides**: - From \( EH = HG \): \( 4x = 3x + y \). Simplify: \( 4x - 3x = y \) ⟹ \( x = y \). - From \( EH = EF \): Substitute \( y = x \) into \( EF = 2x + y + 2 \), so \( 4x = 2x + x + 2 \). 3. **Solve for \( x \)**: Simplify \( 4x = 3x + 2 \) ⟹ \( 4x - 3x = 2 \) ⟹ \( x = 2 \). 4. **Find the length of \( EH \)**: Substitute \( x = 2 \) into \( EH = 4x \): \( EH = 4(2) = 8 \). ### Knowledge points involved 1. **Properties of a rhombus** - A rhombus is a quadrilateral with all four sides equal in length. This means \( EH = HG = EF = FG \) in rhombus \( EFGH \). 2. **Solving linear equations** - Using the equality of sides to set up linear equations (e.g., \( 4x = 3x + y \) and \( 4x = 2x + y + 2 \)) and solving for variables \( x \) and \( y \). 3. **Substitution method** - Substituting \( y = x \) (from \( 4x = 3x + y \)) into the second equation to solve for \( x \), then using \( x \) to find the length of \( EH \). ---
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