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QUESTION IMAGE

m∠efg = m∠gdh = m∠feg = m∠dhg =

Question

m∠efg =
m∠gdh =
m∠feg =
m∠dhg =

Explanation:

Step1: Properties of a square

In square \(DEFG\), each angle is \(90^{\circ}\). So, \(m\angle EFG = 90^{\circ}\).

Step2: Diagonals of a square

The diagonals of a square bisect the angles. In square \(DEFG\), diagonal \(DG\) bisects \(\angle EDG\) and \(\angle EFG\). Since \(\angle EDG = 90^{\circ}\), \(m\angle GDH=\frac{1}{2}\times90^{\circ} = 45^{\circ}\).

Step3: Diagonals of a square

Diagonals of a square are equal and bisect each other at right - angles. In \(\triangle FEG\), \(EF = FG\) (sides of a square) and \(\angle EFG = 90^{\circ}\). So, \(\angle FEG=\angle FGE\). Using the angle - sum property of a triangle (\(\angle EFG+\angle FEG+\angle FGE = 180^{\circ}\)), \(2\angle FEG=180^{\circ}-\angle EFG\). Substituting \(\angle EFG = 90^{\circ}\), we get \(\angle FEG = 45^{\circ}\).

Step4: Diagonals of a square

The diagonals of a square are perpendicular to each other. So, \(DG\perp EF\) at \(H\), and \(m\angle DHG = 90^{\circ}\).

Answer:

\(m\angle EFG = 90^{\circ}\)
\(m\angle GDH = 45^{\circ}\)
\(m\angle FEG = 45^{\circ}\)
\(m\angle DHG = 90^{\circ}\)