Answer:
[tex]\displaystyle A_l=6.97\ ft^2[/tex]
Step-by-step explanation:
Subtractive Areas
When we want to find areas of surfaces that are inside of greater surfaces, we must subtract the greater minus the rest of it. We want to find the shaded area which is part of the circular sector, along with a right triangle shown in the figure. The area of a circle is
[tex]\displaystyle A=\pi \ r^2[/tex]
The area of a circular sector of center angle \theta is
[tex]\displaystyle A_s=\frac{\theta \ r^2}{2}[/tex]
The area of a right triangle with base b and height h is
[tex]\displaystyle A_t=\frac{b.h}{2}[/tex]
We have the value of [tex]\theta[/tex] is
[tex]\displaystyle \theta =100^o=\frac{100.\pi }{180}[/tex]
[tex]\displaystyle \theta =\frac{5\pi }{9}rad[/tex]
The radius is half the diameter, thus
[tex]\displaystyle r=\frac{6}{2}=3\ ft[/tex]
The area of the circular sector is computed now:
[tex]\displaystyle A_s=\frac{5\pi }{9}.3^2[/tex]
[tex]\displaystyle A_s=5\pi\ ft^2[/tex]
We also have the length of the base as double of the given value (because of symmetry)
[tex]\displaystyle b=4.6\ ft*2[/tex]
[tex]\displaystyle b=9.2\ ft[/tex]
The height is
[tex]\displaystyle h=1.9\ ft[/tex]
The area of the triangular surface is
[tex]\displaystyle A_t=\frac{(9.2)(1.9)}{2}[/tex]
[tex]\displaystyle A_t=8.74\ ft^2[/tex]
Finally, the shaded area that Geneva will lose from the table is
[tex]\displaystyle A_l=A_s-A_t[/tex]
[tex]\displaystyle A_l=5\pi -8.74[/tex]
[tex]\boxed{\displaystyle A_l=6.97\ ft^2}[/tex]
