A running track, 5 meters wide, has the dimensions shown in the diagram. The ends of the track are semicircles with diameter 20 meters. What is the surface area of the track? -100 meters - 20 meters
A running track, 5 meters wide, has the dimensions shown in the diagram. The ends of the track are semicircles with diameter 20 meters. What is the surface area of the track? -100 meters - 20 meters
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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![**12. A running track, 5 meters wide, has the dimensions shown in the diagram. The ends of the track are semicircles with a diameter of 20 meters. What is the surface area of the track?**
**Diagram Explanation:**
- The diagram shows a top-down view of a running track.
- The track has straight sections of 100 meters in length, connected by semicircular sections at each end.
- Each semicircle has a diameter of 20 meters.
- The width of the track around its length is 5 meters.
**Steps to Solve:**
To find the surface area of the track, we need to consider both the rectangular and semicircular parts of the track and then account for the overall width across its dimensions.
**Calculate the area of the rectangular part:**
- Length: 100 meters
- Width: 20 meters (as the diameter of semicircles determines the rectangle's width)
\[ \text{Area of rectangle} = \text{Length} \times \text{Width} = 100 \, \text{meters} \times 20 \, \text{meters} = 2000 \, \text{square meters} \]
**Calculate the area of the semicircular parts:**
- The total length of the semicircles is equivalent to one full circle since there are two semicircles.
- The radius of the semicircles is 10 meters (since the diameter is 20 meters).
\[ \text{Area of a full circle} = \pi \times (\text{radius})^2 = \pi \times (10 \, \text{meters})^2 = 100\pi \, \text{square meters} \]
**Calculate the total area of the track:**
- Sum the area of the rectangular part and the area of the semicircular parts:
\[ \text{Total Surface Area} = \text{Area of rectangle} + \text{Area of full circle} = 2000 \, \text{square meters} + 100\pi \, \text{square meters} \]
Let's take π ≈ 3.14 for an approximate numerical result:
\[ 100\pi \, \text{square meters} ≈ 100 \times 3.14 = 314 \, \text{square meters} \]
\[ \text{Total Surface Area} ≈ 2000 \,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff8311114-a7cb-49e3-b44e-1e05de4bc29a%2F7e030914-427a-4594-8f97-a6a0e760393a%2Fd37no6j.png&w=3840&q=75)
Transcribed Image Text:**12. A running track, 5 meters wide, has the dimensions shown in the diagram. The ends of the track are semicircles with a diameter of 20 meters. What is the surface area of the track?**
**Diagram Explanation:**
- The diagram shows a top-down view of a running track.
- The track has straight sections of 100 meters in length, connected by semicircular sections at each end.
- Each semicircle has a diameter of 20 meters.
- The width of the track around its length is 5 meters.
**Steps to Solve:**
To find the surface area of the track, we need to consider both the rectangular and semicircular parts of the track and then account for the overall width across its dimensions.
**Calculate the area of the rectangular part:**
- Length: 100 meters
- Width: 20 meters (as the diameter of semicircles determines the rectangle's width)
\[ \text{Area of rectangle} = \text{Length} \times \text{Width} = 100 \, \text{meters} \times 20 \, \text{meters} = 2000 \, \text{square meters} \]
**Calculate the area of the semicircular parts:**
- The total length of the semicircles is equivalent to one full circle since there are two semicircles.
- The radius of the semicircles is 10 meters (since the diameter is 20 meters).
\[ \text{Area of a full circle} = \pi \times (\text{radius})^2 = \pi \times (10 \, \text{meters})^2 = 100\pi \, \text{square meters} \]
**Calculate the total area of the track:**
- Sum the area of the rectangular part and the area of the semicircular parts:
\[ \text{Total Surface Area} = \text{Area of rectangle} + \text{Area of full circle} = 2000 \, \text{square meters} + 100\pi \, \text{square meters} \]
Let's take π ≈ 3.14 for an approximate numerical result:
\[ 100\pi \, \text{square meters} ≈ 100 \times 3.14 = 314 \, \text{square meters} \]
\[ \text{Total Surface Area} ≈ 2000 \,
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