# Thread: i also need help with this problem

1. ## i also need help with this problem

1. a pizza has a diameter of 16 inches. george has a slice of the pizza. the shape of george's slice is a sector of a circle. george wants to know some information about the size of his slice. he doesn't want to use a protractor to measure the central angle because he'd get cheese on his protractor. he measures the length of the arc formed by the outside crust of his slice. the lenthg of the arc is 6inches. find the central angle in radians and degrees. find the exact perimeter and area of geroge's slice.

2. Originally Posted by Mr. red

1. a pizza has a diameter of 16 inches. george has a slice of the pizza. the shape of george's slice is a sector of a circle. george wants to know some information about the size of his slice. he doesn't want to use a protractor to measure the central angle because he'd get cheese on his protractor. he measures the length of the arc formed by the outside crust of his slice. the lenthg of the arc is 6inches. find the central angle in radians and degrees. find the exact perimeter and area of geroge's slice.

$S=r\theta$ where theta is in radians. We know that the radius is half of the diameter and we know S is 6. so we get

$6=8\theta \iff \frac{3}{4}\mbox{rad}=\theta=\left( \frac{3 \mbox{ rad } }{4}\right)\left( \frac{180^\circ }{\pi \mbox{ rad }}\right)=\left( \frac{135}{\pi} \right)^\circ \approx 42.97^\circ$

The perimeter is going to be the arc of the cirlce plus 2 radii of the circle

$6+8+8=22 in$

The area of a sector of a circle is given by

$A=\frac{1}{2}r^2 \theta =\frac{1}{2}(8 in)^2\left( \frac{3 \mbox{ rad }}{4}\right)=24 in^2$