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Thread: 8th grade math...

20050331, 02:01 #1
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8th grade math...
This is not nearly as much a puzzle as it is two math problems that me an my 8th grader need some help with...
 Ricky

20050331, 03:27 #2
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Re: 8th grade math...
1) x = <span style="backgroundcolor: #FFFF00; color: #FFFF00; fontweight: bold">32/45*24 = 17.067</span hide>
y = <span style="backgroundcolor: #FFFF00; color: #FFFF00; fontweight: bold">25/45*24 = 13.333</span hide>
2) Assuming <span style="backgroundcolor: #FFFF00; color: #FFFF00; fontweight: bold"> he inside is a square (it is not explicity stated), but it can't be solved otherwise...
2 sides = 2 *60 = 120
Around cricles = pi * 60 = 188.50
Total circumference = 308.50 yds</span hide>
3) <span style="backgroundcolor: #FFFF00; color: #FFFF00; fontweight: bold">Area of square = 60*60 = 3600 sq yds
Area of cirlce = pi (60/2)^2 = 2827.43 sq yds
Total area = 6427.33 sq yds</span hide>
Steve

20050331, 04:57 #3
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Re: 8th grade math...
Interesting, I had just decided that there wasn't a solution to 2)
StuartR

20050331, 11:07 #4
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Re: 8th grade math...
As you can see I assumed we just lost some of the info in the posting...
I figure when I answer questions if I state my assumptions, my answer is less likely to be wrong, just my assumptions...
Steve

20050331, 13:49 #5
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Re: 8th grade math...
One really handy assumption to state would therefore be that your answer is correct.
Then you can't really go wrong! <img src=/S/grin.gif border=0 alt=grin width=15 height=15>
Alan

20050331, 16:10 #6
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Re: 8th grade math...
But that goes "without saying..." <img src=/S/grin.gif border=0 alt=grin width=15 height=15>
Steve

20050331, 16:20 #7
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Re: 8th grade math...
I usually assume that my answers are wrong, and sadly the assumption is usually correct <img src=/S/sad.gif border=0 alt=sad width=15 height=15>
StuartR

20050331, 22:30 #8
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Re: 8th grade math...
In Q1, we came up with the same solution as you but to me it didn't seem like a "clean enough" answer for 8th grade work. There were other problems of similar type that all resulted with whole number answers. So I thought I must be missing something...Ends up we (and you) were correct.
And from the "Why should I consider a quick simple solution when I can spend hours on a complex solution?" department...
In Q2 and Q3, I scoured the web looking for a formula that would give the area and circumference of an oval. The solution required the long diameter as well as the short diameter. And it all seemed to be beyond what an 8th grader should know. Should've known I was heading down the wrong path. Now after a good nights rest and taking a fresh look (and seeing your response), I can clearly see the figure doesn't even resemble an oval. And, if all the information needed to provide a solution is present, then the interior shape must be a square. Meaning that the shapes other than the square represent two halves of a circle with diameter 60.
Thanks and <img src=/S/cheers.gif border=0 alt=cheers width=30 height=16> Ricky

20050331, 23:12 #9
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Re: 8th grade math...
For an oval you would have needed to know 2 pieces of info that were not given. The other side of the rectangle, and the other radius of the oval.
To solve it, it had to be a square and semicircle or more info was needed. I assumed if I my assumption was incorrect, I could get more info...
Steve

20050408, 14:36 #10
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Re: 8th grade math...
same answers as steve... very similar methods
1) <span style="backgroundcolor: #FFFF00; color: #FFFF00; fontweight: bold">i used the first side to find a factor of 1.875 and then just divided the other sides by it</span hide>
2) <span style="backgroundcolor: #FFFF00; color: #FFFF00; fontweight: bold">60 + 60 + 2pi*r : if it was outside of the track i would have added 2 feet extending in each direction (i think thats the standard for a running track)  so s = 64 instead of 60</span hide>
3) <span style="backgroundcolor: #FFFF00; color: #FFFF00; fontweight: bold"> s^2 + pi * r^2 </span hide>
sidenote: <span style="backgroundcolor: #FFFF00; color: #FFFF00; fontweight: bold"> i think it is explicitly stated (mathematically) that you are to use the given perimeter information as a basis of your  because the distance around it would be the limit just outside the given perimeter, meaning .000000000000000000001 millimeters outside of the track is technically outside of the given track and the difference from an infinitely small change is negligible in eigth grade math </span hide> <img src=/S/cool.gif border=0 alt=cool width=15 height=15><img src=/w3timages/blueline.gif width=33% height=2>
<big>John</big>