KimJaFo
  • KimJaFo
Jason designed an arch made of wrought iron for the top of a mall entrance. The 11 segments between the two concentric circles are each 1.25 m long. Find the total length of wrought iron used to make the structure. Round the answer to the nearest meter. Show your work. I need some hwlp with this, Ill try and post the picture in the comments
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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KimJaFo
  • KimJaFo
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KimJaFo
  • KimJaFo
@jagr2713 Could you help?
anonymous
  • anonymous
so we have 11 segments of 1.25 m each a semicircle with a diameter of 15 m and a semicircle with diameter (15 - 1.25 - 1.25) m|dw:1438747266213:dw|

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anonymous
  • anonymous
let's just add up the lengths. The circumference of a circles is πd, so for a semicircle it's \[C=\frac{ \pi d }{ 2 }\]
anonymous
  • anonymous
Find the circumference for the 15 m semicircle using that formula
KimJaFo
  • KimJaFo
Thanks!
KimJaFo
  • KimJaFo
So, the formula, once I plug in everything, would be C = pi15/2?
anonymous
  • anonymous
yes, but we write 15pi/2. Then do the same with the smaller semicircle
KimJaFo
  • KimJaFo
What do you mean? I'm kinda getting a lil confused. Sorry '-'
anonymous
  • anonymous
they're asking for the total length of all the lines/curves in the picture. 15π/2 is just the top semicircle. We also have to account for the bottom semicircle and all 11 line segments between them.
anonymous
  • anonymous
Does that make sense?
KimJaFo
  • KimJaFo
A little bit, let me try it.
anonymous
  • anonymous
to find the diameter of the smaller circle, subtract the 1.25 m on each end from the diameter of the larger. d = 15 - 2(1.25) = 12.5 m So use 12.5 in the circumference formula to get the length of this part
KimJaFo
  • KimJaFo
how?
anonymous
  • anonymous
C = πd/2 where d = 12.5
KimJaFo
  • KimJaFo
Ah ok, thank you
anonymous
  • anonymous
|dw:1438748775575:dw|
anonymous
  • anonymous
The last thing you have to find is the length of the segments. There are 11 and the are each 1.25 m. What's the total length?
KimJaFo
  • KimJaFo
I think I got it, thank you
anonymous
  • anonymous
you're welcome

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