Question

quadrilateral efgh is a rhombus. what is m∠ehi?
(there is a rhombus efgh with diagonals intersecting at i, and ∠hei is 55°)
m∠ehi = \boxed{ }°

Explanation:

Step1: Recall properties of a rhombus

In a rhombus, the diagonals bisect the angles. Also, the diagonals are perpendicular bisectors of each other, and adjacent sides are equal. In triangle \( \triangle EHI \), we know that \( \angle HEI = 55^\circ \) and \( \angle HIE = 90^\circ \) (since diagonals of a rhombus are perpendicular).

Step2: Calculate \( m\angle EHI \)

In a triangle, the sum of interior angles is \( 180^\circ \). So, in \( \triangle EHI \), we have \( m\angle EHI + m\angle HEI + m\angle HIE = 180^\circ \). We know \( m\angle HEI = 55^\circ \) and \( m\angle HIE = 90^\circ \). Let \( x = m\angle EHI \). Then \( x + 55^\circ + 90^\circ = 180^\circ \). Solving for \( x \), we get \( x = 180^\circ - 55^\circ - 90^\circ = 35^\circ \).

Answer:

\( 35 \)