Question

quadrilateral wxyz is a rectangle. answers and information from one problem does not carry to the next problem.

1. if wy = 19, then zx = —.
2. if wy = 19, then wt = —.
3. if tx = 4.5, then wy = —.
4. if m\angle twz = 70\textdegree, then m\angle tzw = — & m\angle wtz = —.
5. if wy = 3a + 16 & zx = 5a - 18, then a = —

Explanation:

Step1: Properties of rectangle diagonals

In a rectangle, the diagonals are equal and bisect each other. So \(WY = ZX\) and \(WT=\frac{1}{2}WY\), \(TX = \frac{1}{2}WY\) (diagonals bisect each other).

Step2: Solve for \(WY\) in problem 9

Given \(TX = 4.5\), since \(TX=\frac{1}{2}WY\), then \(WY = 2\times TX\).
Substitute \(TX = 4.5\) into the formula: \(WY=2\times4.5 = 9\).

Step3: Solve for \(WT\) in problem 8

Given \(WY = 19\), since \(WT=\frac{1}{2}WY\), then \(WT=\frac{1}{2}\times19 = 9.5\).

Step4: Solve for \(m\angle TZW\) and \(m\angle WTZ\) in problem 10

In \(\triangle TZW\), \(TZ = TW\) (diagonals of a rectangle bisect each other, so \(TZ=\frac{1}{2}ZX\), \(TW=\frac{1}{2}WY\) and \(ZX = WY\)).
If \(m\angle TWZ=70^{\circ}\), then \(m\angle TZW=m\angle TWZ = 70^{\circ}\) (isosceles triangle property).
Using the angle - sum property of a triangle (\(m\angle TWZ+m\angle TZW+m\angle WTZ = 180^{\circ}\)), we have \(70^{\circ}+70^{\circ}+m\angle WTZ=180^{\circ}\).
Then \(m\angle WTZ=180^{\circ}-(70^{\circ}+70^{\circ})=40^{\circ}\).

Step5: Solve for \(a\) in problem 11

Since \(WY = ZX\) (diagonals of a rectangle are equal), \(3a + 16=5a-18\).
Subtract \(3a\) from both sides: \(16=5a - 3a-18\), which simplifies to \(16 = 2a-18\).
Add \(18\) to both sides: \(16 + 18=2a\), so \(34 = 2a\).
Divide both sides by \(2\): \(a = 17\).

Answer:

1. \(19\)
2. \(9.5\)
3. \(9\)
4. \(m\angle TZW = 70^{\circ}\), \(m\angle WTZ=40^{\circ}\)
5. \(17\)