Question

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

Answer 1

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

Answer 2

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?

Answer 3

Let me try solving it first.

Answer 4

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

Answer 5

divide pentagon into 5 equalaterial triangles
this way is easier for me

Answer 6

@elleboedefeld Did you get an answer?

Answer 7

688.19? :(

Answer 8

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

Answer 9

Im so stupid.

Answer 10

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

Answer 11

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

Answer 12

Yes! Thank you! :)

Answer 13

@elleboedefeld No problem! Good Luck! :D