#### Answer

The area of the portion of the circle bounded by the arc and the chord is $~~619.7~ft^2$

#### Work Step by Step

We can find the radius $r$ of the curve:
$sin~21^{\circ} = \frac{50~ft}{r}$
$r = \frac{50~ft}{sin~21^{\circ}}$
$r = 139.5~ft$
We can convert the angle to units of radians:
$42^{\circ}\cdot~\frac{\pi~rad}{180^{\circ}} = 0.733~rad$
We can find the total area $A_t$ of the sector:
$A_t = \frac{r^2~\theta}{2},~~~$ where $\theta$ is measured in radians
$A_t = \frac{(139.5~ft)^2~(0.733~rad)}{2}$
$A_t = 7132.2~ft^2$
We can find the length $L$ of the line from the center of the circle which bisects the chord:
$\frac{50~ft}{L} = tan~21^{\circ}$
$L = \frac{50~ft}{tan~21^{\circ}}$
$L = 130.25~ft$
We can find the area $A_2$ of half of the triangular section:
$A_2 = \frac{1}{2}(50~ft)(130.25~ft) = 3256.25~ft^2$
The total area of the triangular section is double this area.
The total area of the triangular section is $(2)(3256.25~ft^2) = 6512.5~ft^2$
We can find the area of the portion of the circle bounded by the arc and the chord:
$Area = 7132.2~ft^2 - 6512.5~ft^2 = 619.7~ft^2$
The area of the portion of the circle bounded by the arc and the chord is $~~619.7~ft^2$