anonymous 4 years ago Find the area of a regular pentagon with a side length of 10m. Round to nearest tenth. PLEASE HELP!

1. anonymous

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

2. anonymous

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?

3. anonymous

Let me try solving it first.

4. anonymous

Okay. I'll solve it, and post your answer when you get it to verify

5. alexwee123

|dw:1334722501665:dw| divide pentagon into 5 equalaterial triangles this way is easier for me

6. anonymous

@elleboedefeld Did you get an answer?

7. anonymous

688.19? :(

8. anonymous

Okay, so @elleboedefeld, I got $$\ \huge \approx 172.1m^2$$.

9. anonymous

Im so stupid.

10. anonymous

@elleboedefeld No you're not, it takes time to learn new things. Here's what I did:

11. anonymous

$$\ \huge \frac{5 \bullet 10^2}{4tan(\frac{180}{5})}$$ = $$\ \huge \frac{500}{4tan(\frac{180}{5})}$$ = 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?

12. anonymous

Yes! Thank you! :)

13. anonymous

@elleboedefeld No problem! Good Luck! :D

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