Q58. - Algebra 1

Question

In 1998, Winchell's House of Donuts in Pasadena, California, made the world's largest donut. It weighed 5000 pounds and had a circumference of 298.3 feet. What was the donut's diameter to the nearest tenth?

Hint: $C=\pi d$

Step-by-Step Solution

Answer

The diameter of the donut, rounded to the nearest tenth is 95.0 feet.

Step 1. State the formula.

The circumference of a circle is given by $C=\pi d$, where $C$ is the circumference and $d$ is the diameter of the circle.

 

Then, $d=\frac{C}{\pi }$.

Step 2. List the given data.

The circumference of the donut is 298.3 feet. So, $C=298.3$.

Step 3. Evaluate the diameter.

Put $C=298.3$ in $d=\frac{C}{\pi }$ to get,

$$\begin{gathered} d=\frac{298.3}{\pi } \\ \approx \text{94.9518 }\left(\text{rounded to 4 decimal places}\right) \end{gathered}$$

 

So, $d\approx 94.9518$

Step 4. Rounding off.

It has been obtained that $d\approx 94.9518$.

 

Rounding off to the nearest tenth, $d\approx 95.0$.

Step 5. Determine the unit.

Since the unit of circumference of the donut is given as feet, the unit of diameter must also be feet.

 

So, the diameter of the donut, rounded to the nearest tenth is 95.0 feet.