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

Answer: 8.4 inches.

Step-by-step 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):

$C=\frac{2\pi (8in)}{6}\\\\C=8.4in$

Answer 2

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):

$\text { Width }=\frac{2 \pi \times 8}{6}=\frac{\pi \times 8}{3}=\frac{8 \pi}{3}=8.377$

Hence, the cake piece is 8.4 inches wide.