Moved - Oval Track

T

tynshawnta

New member
Joined
Jan 24, 2013
Messages
8


Track for assignment.jpg

Where do I start?


To host the Summer Olympics, a city plans to build an eight-lane track on a rectangular lot just large enough to accommodate it. The track will consist of parallel 100-meter straight-aways with semicircular turns on either end as shown in the figure above. The total distance around the outside edge of the oval track is 451.2 meters. The inside radius of the turns is 30 meters. Grass is planted inside and outside the track on the rectangular lot. In the figure above, the grassy areas are colored green and the track itself is colored light yellow.

1. What is the outside radius of the turns?
(Hint: Use the total distance around the outside edge of the track and the fact that the straight-aways are 100 m long to find the outer circumference of the semicircular turns. Then use the circumference to solve for the outside radius, r.)

2. Find the area of the rectangular lot, in square meters.
3. How many hectares of land are needed for the rectangular lot? Note: 1 hectare = 10,000 m2.
4. Find the area of the track.
(Hint: First find the area of the track plus grassy section inside the track. Then subtract the area of the grassy section inside the track to obtain the area of the track alone.)
5.How many square meters of the rectangular lot will be planted in grass?
 
Last edited:
1. You posted this in the "news" section which is not intended for questions like this.

2. At several points you are given hints. Have you tried to use them? You are told that the " total distance around the outside edge of the oval track is 451.2 meters" and that the two straight parts are 100 meters long. That tells you that the outside length of the curved portions are 451.2- 2(100)= 251.2 meters long. Now those two semicircles make a circle with circumference 251.2 m. Do you know that the circumference of a circle is "2 times pi times the radius". You can use that to find the outside radius of the curved portions and everything else will follow from that.
 
Thanks for the information!

Maybe I should have asked about what I already had. I know that to get the area of a rectangle you multiply the lentgh times the width. What am I mulitplying to get the area of the reactangle in square meters. Is it the 200m for the straight-aways multiplied by the 251.2 for the turns?
start with the total outside edge of 451.2 meters.
minus both 100m straightaways.
451.2m-2(100m)=251.2m
now to find the circumference of the circle left over.
C=2pi*r
now re-arange the formula
C/(2pi)=r
251.2/(2pi)=40m
 
tynshawnta said:
Maybe I should have asked about what I already had. I know that to get the area of a rectangle you multiply the lentgh times the width. Correct

What am I mulitplying to get the area of the reactangle in square meters. Is it the 200m for the straight-aways multiplied by the 251.2 for the turns?

No. You already know the formula for the area of a rectangle. The rectangular lot does not have a length of 200 meters and does not have a width 251.2 meters. Look at the lot. Its length is equal to 100 meters, the straightaways, plus something else. What is that something else. It may be easier to see if you think about in terms of 100 plus two something elses, one on each end.

start with the total outside edge of 451.2 meters. Now you are thinking.
minus both 100m straightaways. OK
451.2m-2(100m)=251.2m
now to find the circumference of the circle left over. This wording makes me nervous.
C=2pi*r Correct
now re-arange the formula
C/(2pi)=r Good
251.2/(2pi)=40m And what exactly does the 40 meters measure? And what next?
Click to expand...
You seem to be on the right track (no pun intended), but do you understand what you have done? Please answer my questions if you can before we go further.
 
I understand how I got my 40 meters. The 40 meters is the outside radius of the turns for question #1. I don't understand what I am multiyplying to get my area of the reactangle. Question #2 is find the area of the rectangular lot, in square meters.
 
tynshawnta said:
I understand how I got my 40 meters. The 40 meters is the outside radius of the turns for question #1. I don't understand what I am multiyplying to get my area of the reactangle. Question #2 is find the area of the rectangular lot, in square meters.
Click to expand...
Imagine a line drawn through the middle of the rectangle parallel to the straightaways. The length of that line is a sum with three summands. What are they? So what is the length of that line?

Now draw a second line that is perpendicular to the first line you drew and joins the endpoints of the straightaways (left or right end makes no difference. The length of that line is also a sum of two summands. What are they? So what is the length of that line.

Now you have width and length and the area of the rectangle is a formula.
 
I think I got the length. Add 30m to each side of the straight-away, 30m + 100m + 30m will give me 160m total for my length. The perpindicular line summands are 30m + 30m which equals 60m meters, so the length is 160m and the width is 60m. Am I correct or am I looking at this wrong?
 
tynshawnta said:
I think I got the length. Add 30m to each side of the straight-away, 30m + 100m + 30m will give me 160m total for my length. The perpindicular line summands are 30m + 30m which equals 60m meters, so the length is 160m and the width is 60m. Am I correct or am I looking at this wrong?
Click to expand...
By the way, next time you have a geometry question, please put it into the Geometry forum, not News. One reason is not all tutors look at News. The second one is that I forgot 90% of my geometry decades ago so it is better to get someone who remembers it.

OK. Back to the problem. You are going in the right direction, but why did you decide to use the inside radius? That leaves out the track itself. The second question asks about the area of the lot on which the track is built. Obviously that area must include the track itself. To include the track, you must use the outside radius. Do you see that now?

I think you can get the second question now. How about the fourth question?
 
Thanks!

I realized I had put it in the wrong spot after the first person replied, I don't think I see it. Getting confused again. I have to use 40m +20m which is 60m for the width and 100m + what for my length.
 
tynshawnta said:
I realized I had put it in the wrong spot after the first person replied, I don't think I see it. Getting confused again. I have to use 40m +20m which is 60m for the width and 100m + what for my length.
Click to expand...

No.... you have to use the diameter of the outer circle as the width (of the outer rectangle). Do you see that?

For the length (of the outer rectangle) - you have to add the radius of the outer circle - twice (once for each end) - to the length of the inner rectangle. Do you see that?
 
tynshawnta said:
I realized I had put it in the wrong spot after the first person replied, I don't think I see it. Getting confused again. I have to use 40m +20m which is 60m for the width and 100m + what for my length.
Click to expand...
I wish I knew how to sketch at this site. You will have to use a lot of imagination.

\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ xxxxxxxxxxxxxxxxxx\)

\(\displaystyle \ \ \ \ \ \ \ \ xxxxxxxxxxxxxxxxxxxxxx\)

\(\displaystyle xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx\)

\(\displaystyle \ \ \ \ \ \ \ \ xxxxxxxxxxxxxxxxxxxxxx\)

\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ xxxxxxxxxxxxxxxxxx\)

The above is not close to scale, but imagine that the x's in the top and bottom lines represent points along the outside of the track, that the x's in the second line from the top and the second line from the bottom represent points along the inside of the track, and the x's in the middle line represent the first line that I asked you to draw running from the two most distant points on the track, which clearly are on the outside of the track. From the end points of the top and bottom lines there is a semicircle on which the final x's of the midline lie. From the endpoints of lines 2 and 4 runs a a smaller semicircle that intersects with the mid line inside its final x's. I hope you understand what I admit is a horrible diagram. Best I can do. Sorry.

Now draw in your imagination two vertical lines that join the ends of the top and bottom lines.

Clearly the length of the lot equals the length of the middle line. Just as clearly the width of the lot equals the length of either of the vertical lines. Do you see why? Now the point of intersection of a vertical line with the midline is the center of one of the semicircles at the ends of the straightaway. Do you now see why the outer radius is what is relevant, not the inner radius?

Have I cleared up the confusion or made it worse?
 
Okay I see what you are talking about for the length. I have to add 40m to both sides of the straight-away to get my length, so the length will be 40m + 40m +100m which is 180m. I don't see what you are talking about for the width.
 
I think I see. The radius of the outer circle is 40m so the diatmeter would be 80m,so the length would be 180m and the width 80m. To get the area of the recatangle lot I would mutiply 180 times 80 which is 14,400 square meters.
 
tynshawnta said:
I think I see. The radius of the outer circle is 40m so the diatmeter would be 80m,so the length would be 180m and the width 80m. To get the area of the recatangle lot I would mutiply 180 times 80 which is 14,400 square meters.
Click to expand...
There you go. A good diagram always helps.

Edit: Not implying that mine was good. Only implying that you need to build your own diagrams: a good one will always be of some help. The one in the book was deliberately set up so as not to give away the answer.
 
Last edited:
Thanks!

I have not completed any math classes in the last five years, so its confusing until I start to get the hang of things. The little pointers that you are giving me are helping. I know it took me a few trys before I got the answers. This is a good website to get help and give it.
 
You must log in or register to reply here.
Top