Question

1. if wxyz is a square with wz = 27, find each measure.

zy =
wy =
rx =
m∠wrz =
m∠xyz =
m∠zwy =

Explanation:

Step1: Properties of a square

In a square \(WXYZ\), all sides are equal. So \(ZY = WZ=27\)

Step2: Diagonals of a square

The diagonals of a square are equal. Let the diagonals \(WY\) and \(XZ\) intersect at \(R\). Using the Pythagorean theorem for right - triangle \(WZY\) (where \(WZ = ZY=27\) and \(WY\) is the hypotenuse), \(WY=\sqrt{WZ^{2}+ZY^{2}}=\sqrt{27^{2}+27^{2}}=\sqrt{2\times27^{2}} = 27\sqrt{2}\)

Step3: Length of half - diagonal

Since the diagonals of a square bisect each other, \(RX=\frac{WY}{2}\). Since \(WY = 27\sqrt{2}\), \(RX=\frac{27\sqrt{2}}{2}\)

Step4: Angle formed by diagonals

The diagonals of a square are perpendicular to each other. So \(m\angle WRZ = 90^{\circ}\)

Step5: Angle of a square

Each interior angle of a square is \(90^{\circ}\), so \(m\angle XYZ=90^{\circ}\)

Step6: Angle between side and diagonal

In square \(WXYZ\), the diagonal \(WY\) bisects \(\angle ZWX\). Since \(\angle ZWX = 90^{\circ}\), \(m\angle ZWY=\frac{90^{\circ}}{2}=45^{\circ}\)

Answer:

\(ZY = 27\), \(WY=27\sqrt{2}\), \(RX=\frac{27\sqrt{2}}{2}\), \(m\angle WRZ = 90^{\circ}\), \(m\angle XYZ = 90^{\circ}\), \(m\angle ZWY=45^{\circ}\)