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elleboedefeld

Find the area of a regular pentagon with a side length of 10m. Round to nearest tenth. PLEASE HELP!

  • 2 years ago
  • 2 years ago

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  1. elleboedefeld
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    The study guide is due tomorrow and I have a lot of questions on it! :(

    • 2 years ago
  2. Study23
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    So, the area of a regular pentagon, given side length is \(\ \huge \frac{5s^2}{4tan(\frac{180}{5})} \). The side length is s, so replace s with 10. The 5 can be replaced with the number of sides of the shape. A pentagon has 5 sides, so I inserted a 5. Now, its a matter of solving the equation... Do you need help with that?

    • 2 years ago
  3. elleboedefeld
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    Let me try solving it first.

    • 2 years ago
  4. Study23
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    Okay. I'll solve it, and post your answer when you get it to verify

    • 2 years ago
  5. alexwee123
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    |dw:1334722501665:dw| divide pentagon into 5 equalaterial triangles this way is easier for me

    • 2 years ago
  6. Study23
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    @elleboedefeld Did you get an answer?

    • 2 years ago
  7. elleboedefeld
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    688.19? :(

    • 2 years ago
  8. Study23
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    Okay, so @elleboedefeld, I got \(\ \huge \approx 172.1m^2 \).

    • 2 years ago
  9. elleboedefeld
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    Im so stupid.

    • 2 years ago
  10. Study23
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    @elleboedefeld No you're not, it takes time to learn new things. Here's what I did:

    • 2 years ago
  11. Study23
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    \(\ \huge \frac{5 \bullet 10^2}{4tan(\frac{180}{5})} \) = \(\ \huge \frac{500}{4tan(\frac{180}{5})} \) = Type this into a calculator. I think this is where you might have made a mistake. Type this into your calculator like this: 500/4tan((180/5)). If you don't the calculator will misinterpret what you typed in. Then, after rounding to the nearest tenth, I got \(\ \huge \approx 172.1m^2 \). Do you understand?

    • 2 years ago
  12. elleboedefeld
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    Yes! Thank you! :)

    • 2 years ago
  13. Study23
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    @elleboedefeld No problem! Good Luck! :D

    • 2 years ago
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