Question

A circular cake with a radius of 8 inches is cut from the center into 6 equal pieces. How many inches wide, to the nearest tenth of an inch, is the outer edge of each piece of cake?

Answer 1

The width of the outer edge of each piece of cake 8.4 inches.

SOLUTION:

Given, a circular cake with a radius of 8 inches is cut from the center into 6 equal pieces. We have to find the width of the outer edge of each piece of cake.

We know that the circumference of a circle can be calculated with this formula:
`2\pi r` where "r" is the radius of the circle.

We know that the circular cake (whose radius is 8 inches) is cut from the center into 6 equal pieces.

Then, you need to divide the circumference of this circular cake by 6 to find the width of the outer edge of each piece of cake.

Therefore, this is (to the nearest tenth of an inch):

`\text { Width }=(2 \pi * 8)/(6)=(\pi * 8)/(3)=(8 \pi)/(3)=8.377`

Hence, the cake piece is 8.4 inches wide.

Answer 2

Answer: 8.4 inches.

Explanation:

We know that the circumference of a circle can be calculated with this formula:

`C=2\pi r`

Where "r" is the radius of the circle.

We know that the circular cake (whose radius is 8 inches) is cut from the center into 6 equal pieces. Then, you need to divide the circumference of this circular cake by 6 to find the width of the outer edge of each piece of cake.

Therefore, this is (to the nearest tenth of an inch):

`C=(2\pi (8in))/(6)\\\\C=8.4in`