Question

Please help last question (10 POINTS)

Answer 1

The area
`a` of the smaller sector is such that

`\frac a{64\pi}=(40^\circ)/(360^\circ)\implies a=\frac{64\pi}9`

Then the area of the larger sector is

`64\pi-\frac{64\pi}9=\frac{512\pi}9\approx\boxed{56.9}\pi`