Question

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rectangle efgh
5 in 24°
eq = in
hf = in
m∠feh =
m∠1 =

Explanation:

Step1: Properties of rectangle diagonals

In a rectangle, diagonals are equal and bisect each other. So \(EQ = QG\), \(FQ=QH\), and \(HF = EG\).

Step2: Calculate \(EQ\)

Since \(EQ = QG\) and \(EG\) is the diagonal. But if we assume the given \(5\) in is related to some side (maybe wrong assumption, but if we consider the fact that in a rectangle diagonals bisect each other. If we assume \(EQ = QG\) and if we consider the triangle formed. Wait, no, in a rectangle \(EQ=\frac{1}{2}HF\). But if we consider the right - angled triangle property. Wait, no, actually in a rectangle \(EQ = QG\), \(FQ = QH\), and diagonals are equal. If we assume \(EQ\) is half of the diagonal. Wait, no, let's start over.
In rectangle \(EFGH\), diagonals \(EG\) and \(HF\) bisect each other at \(Q\). So \(EQ=\frac{1}{2}EG\). But if we consider the triangle \(EFH\) (assuming the given \(5\) in is a red - herring, no, wait no. Wait, in a rectangle all angles are \(90^{\circ}\). The diagonals of a rectangle are equal. Let's first find \(m\angle FEH\). Since \(EFGH\) is a rectangle, \(\angle FEH = 90^{\circ}\).
For \(m\angle1\): In rectangle \(EFGH\), \(EH\parallel FG\), so \(\angle1=\angle EHG\). Also, in \(\triangle EHG\), if we consider the angle given \(24^{\circ}\). Wait, no, in rectangle \(EFGH\), \(\angle FEH = 90^{\circ}\). If there is an angle of \(24^{\circ}\) at \(E\) (say \(\angle AEH = 24^{\circ}\), assuming the non - marked line creates an angle of \(24^{\circ}\) with \(EH\)), then \(\angle FEH=90^{\circ}\).
For \(HF\): If we assume that in right - triangle (using the property of rectangle diagonals). Wait, no, in a rectangle \(EFGH\), diagonals \(EG\) and \(HF\) are equal. If we consider the fact that in a rectangle \(EQ = QG\), \(FQ = QH\). If we assume \(EQ = 5\) in (maybe mis - labeled, but if we go with the problem's structure). Then \(HF=EG = 2\times EQ\). So \(HF = 10\) in.
For \(m\angle1\): Since \(EH\parallel FG\), \(\angle1=\angle EHG\). In \(\triangle EHG\), if \(\angle HEQ = 24^{\circ}\), and \(EQ = QH\) (diagonals bisect each other in a rectangle), so \(\angle EHG=\angle HEQ = 24^{\circ}\), then \(\angle1 = 66^{\circ}\) (because \(\angle FEH = 90^{\circ}\), \(\angle FEH=\angle FEQ+\angle QEH\), if \(\angle QEH = 24^{\circ}\), then \(\angle FEQ=66^{\circ}\), and \(\angle1=\angle FEQ\) (alternate interior angles as \(EH\parallel FG\)).

Answer:

\(EQ = 5\) in, \(HF = 10\) in, \(m\angle FEH=90^{\circ}\), \(m\angle1 = 66^{\circ}\)