An oval indoor running track with semicircular ends has the dimensions of 80' long, 34' wide, and the individual lane width is 5' wide. If the track is to be resurfaced how many square feet of material will be required?
Help Please!
To find the total square footage of material required to resurface the oval indoor running track, we need to calculate the area of the entire track including the lane width.
Step 1: Calculate the area of the main oval track:
- The length of the oval track is given as 80 feet.
- The width of the oval track is given as 34 feet.
- To find the area of the oval track, we can use the formula for the area of an ellipse which is A = πab, where a is the semi-major axis (half the length) and b is the semi-minor axis (half the width) of the ellipse.
- In this case, the semi-major axis is 80 feet/2 = 40 feet, and the semi-minor axis is 34 feet/2 = 17 feet.
- Plugging the values into the formula, we get A_oval = π(40)(17) square feet.
Step 2: Calculate the area of the individual lanes:
- The width of each lane is 5 feet.
- We need to find the combined area of all the lanes. Since there is one lane on the inside and one on the outside, there are two lanes in total.
- To find the area of each lane, we can multiply the length of the oval track by the width of the lane. Thus, each lane has an area of 80 feet × 5 feet = 400 square feet.
- Since there are two lanes, the total area of the lanes is 2 lanes × 400 square feet = 800 square feet.
Step 3: Add the areas together:
- To find the total square footage required, we need to add the area of the oval track to the area of the lanes.
- Total square footage = A_oval + total lane area = A_oval + 800 square feet.
Let's calculate the result:
- The value of π (pi) is approximately 3.14159.
- Plugging in the values, we get total square footage = 3.14159 × (40)(17) + 800 square feet.
- Evaluating this expression, we find the total square footage required to resurface the oval indoor running track.
Please note that due to the rounded dimensions provided and the approximation of pi, the result might be an estimate.