Solve Geometry Applications: Circles and Irregular Figures: example 9.41

A circular sandbox has a radius of $2.5$ feet. Find the ⓐ circumference and ⓑ area of the sandbox.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

A circular mirror has radius of $5$ inches. Find the ⓐ circumference and ⓑ area of the mirror.

How did it go?

A circular sandbox has a radius of $2.5$ feet. Find the ⓐ circumference and ⓑ area of the sandbox.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

A circular mirror has radius of $5$ inches. Find the ⓐ circumference and ⓑ area of the mirror.

How did it go?

ⓐ Step 1. Read the problem. Draw the figure and label it with the given information.
Step 2. Identify what you are looking for.	the circumference of the circle
Step 3. Name. Choose a variable to represent it.	Let c = circumference of the circle
Step 4. Translate. Write the appropriate formula Substitute	$C = 2 π r$ $C = 2 π (2.5)$
Step 5. Solve the equation.	$C \approx 2 (3.14) (2.5)$ $C \approx 15 ft$
Step 6. Check. Does this answer make sense? Yes. If we draw a square around the circle, its sides would be 5 ft (twice the radius), so its perimeter would be 20 ft. This is slightly more than the circle's circumference, 15.7 ft.
Step 7. Answer the question.	The circumference of the sandbox is 15.7 feet.

ⓑ Step 1. Read the problem. Draw the figure and label it with the given information.
Step 2. Identify what you are looking for.	the area of the circle
Step 3. Name. Choose a variable to represent it.	Let A = the area of the circle
Step 4. Translate. Write the appropriate formula Substitute	$A = π r^{2}$ $A = π {(2.5)}^{2}$
Step 5. Solve the equation.	$A \approx (3.14) {(2.5)}^{2}$ $A \approx 19.625 sq. ft$
Step 6. Check. Yes. If we draw a square around the circle, its sides would be 5 ft, as shown in part ⓐ. So the area of the square would be 25 sq. ft. This is slightly more than the circle's area, 19.625 sq. ft.
Step 7. Answer the question.	The area of the circle is 19.625 square feet.

ⓐ Step 1. Read the problem. Draw the figure and label it with the given information.
Step 2. Identify what you are looking for.	the circumference of the circle
Step 3. Name. Choose a variable to represent it.	Let c = circumference of the circle
Step 4. Translate. Write the appropriate formula Substitute	$C = 2 π r$ $C = 2 π (2.5)$
Step 5. Solve the equation.	$C \approx 2 (3.14) (2.5)$ $C \approx 15 ft$
Step 6. Check. Does this answer make sense? Yes. If we draw a square around the circle, its sides would be 5 ft (twice the radius), so its perimeter would be 20 ft. This is slightly more than the circle's circumference, 15.7 ft.
Step 7. Answer the question.	The circumference of the sandbox is 15.7 feet.

ⓑ Step 1. Read the problem. Draw the figure and label it with the given information.
Step 2. Identify what you are looking for.	the area of the circle
Step 3. Name. Choose a variable to represent it.	Let A = the area of the circle
Step 4. Translate. Write the appropriate formula Substitute	$A = π r^{2}$ $A = π {(2.5)}^{2}$
Step 5. Solve the equation.	$A \approx (3.14) {(2.5)}^{2}$ $A \approx 19.625 sq. ft$
Step 6. Check. Yes. If we draw a square around the circle, its sides would be 5 ft, as shown in part ⓐ. So the area of the square would be 25 sq. ft. This is slightly more than the circle's area, 19.625 sq. ft.
Step 7. Answer the question.	The area of the circle is 19.625 square feet.