The problem asks to find the perimeter of kite $WXYZ$. We are given $WX = XY = 14$ and $ZU = 28$. We already know that $YZ = 14\sqrt{5}$. Since $ZU$ is a perpendicular bisector of $WY$, we know that $U$ is the midpoint of $WY$, so $WU = UY = 7$. We are also given $WZ = YZ$.

1. Problem Description

The problem asks to find the perimeter of kite $WXYZ$. We are given $WX = XY = 14$ and $ZU = 28$. We already know that $YZ = 14\sqrt{5}$. Since $ZU$ is a perpendicular bisector of $WY$, we know that $U$ is the midpoint of $WY$, so $WU = UY = 7$. We are also given $WZ = YZ$.

2. Solution Steps

Since $WXYZ$ is a kite, we have $WX = XY$ and $WZ = YZ$. We are given $WX = XY = 14$ and $UZ = 28$. We need to find the length of $WZ$.

We know that $\triangle WUZ$ is a right triangle, with legs $WU = 7$ and $UZ = 28$. We can use the Pythagorean theorem to find $WZ$.

$WZ^2 = WU^2 + UZ^2$

$WZ^2 = 7^2 + 28^2$

$WZ^2 = 49 + 784$

$WZ^2 = 833$

$WZ = \sqrt{833} = \sqrt{49 \times 17} = 7\sqrt{17}$

Since $WZ = YZ$, we have $YZ = \sqrt{833}$. The problem provides $YZ=14\sqrt{5}$ but it is inconsistent with the given image, so using Pythagorean theorem we find $YZ=WZ=\sqrt{WU^2+UZ^2} = \sqrt{7^2+28^2} = \sqrt{49+784} = \sqrt{833}=7\sqrt{17}$.

The perimeter of kite $WXYZ$ is $WX + XY + YZ + WZ$.

Perimeter $= 14 + 14 + \sqrt{833} + \sqrt{833}$

Perimeter $= 28 + 2\sqrt{833}$

Perimeter $= 28 + 2(7\sqrt{17})$

Perimeter $= 28 + 14\sqrt{17}$

3. Final Answer

$28 + 14\sqrt{17}$