Question

4 rectangle efgh
eq =
hf =
in
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 we assume the given \(6.6\) in is related to \(EQ\) (assuming it's a typo - maybe the length from \(E\) to a point on the diagonal part). But if we consider the standard property of rectangle diagonals bisecting each other. If we assume \(EQ\) is half of \(EG\). But wait, if we consider the triangle formed. Wait, no - wait, in rectangle \(EFGH\), diagonals \(HF\) and \(EG\) intersect at \(Q\). So \(EQ=\frac{1}{2}EG\). But if we assume the given \(6.6\) is \(EQ\) (maybe mis - labeled). But wait, no - wait, in a rectangle, diagonals are equal. Let's re - think.

Wait, another approach:

• For \(EQ\): In rectangle \(EFGH\), diagonals bisect each other. If we assume \(EQ = 6.6\) in (maybe the segment from \(E\) to \(Q\) is given as \(6.6\) in.
• For \(HF\): Since diagonals are equal and bisect each other. If \(EQ = 6.6\), then \(EG=2\times EQ = 13.2\) in. And \(HF = EG=13.2\) in.
• For \(m\angle FEH\): \(\angle FEH = 90^{\circ}- 21^{\circ}=69^{\circ}\) (because \(\angle FEH\) is composed of a \(21^{\circ}\) angle and another angle, and \(\angle FEH = 90^{\circ}\) in a rectangle, so \(m\angle FEH=90 - 21=69^{\circ}\))
• For \(m\angle1\): \(\angle1 = 2\times21^{\circ}=42^{\circ}\) (using the exterior angle property of a triangle. The triangle formed has an exterior angle \(\angle1\) and the non - adjacent interior angles. Since the diagonals bisect each other and we have isosceles triangles. Let's explain: In \(\triangle EHQ\), \(EQ = QH\) (diagonals bisect each other in a rectangle). Let \(\angle EHQ=\angle HEQ = 21^{\circ}\). Then \(\angle1\) (exterior angle of \(\triangle EHQ\)) is \(\angle1=\angle EHQ+\angle HEQ = 21 + 21=42^{\circ}\))

Answer:

\(EQ = 6.6\) in
\(HF=13.2\) in
\(m\angle FEH = 69^{\circ}\)
\(m\angle1 = 42^{\circ}\)