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Mathematics 90 Online
OpenStudy (anonymous):

Find the area of an equilateral triangle (regular 3-gon) with the given measurement. 3-inch radius A = sq. in.

OpenStudy (anonymous):

Area= 1/2ap

OpenStudy (anonymous):

I draw an auxiliary triangle off of the radius, the radius being the hypotenuse

OpenStudy (anonymous):

By 3 inch radius, do you mean that it is inscribed in a circle of radius 3 inches?

OpenStudy (anonymous):

|dw:1348254489061:dw|

OpenStudy (anonymous):

The apothem is half of the radius

OpenStudy (anonymous):

|dw:1348254528658:dw| So it looks like both are drawings are similar?

OpenStudy (anonymous):

So its 1.5

OpenStudy (anonymous):

I multiply 1.5 times √3 to get the long side

OpenStudy (anonymous):

Than i multiply that answer "3√3" by 2 to get the side of the big triangle

OpenStudy (anonymous):

I multiply that by 3 to get the perimeter. So i have the perimeter and the apothem.

OpenStudy (anonymous):

Yes, and to make matters easier, the angle and segment bisectors will meet 2/3 of the way to the other side, making the height = 1.5 x apothem.

OpenStudy (anonymous):

The answer i got the first time is: 6 3/4√3 but they told me its wrong.

OpenStudy (anonymous):

height = 1.5 x apothem = 1.5 x (r/2) = 3r/4.

OpenStudy (anonymous):

I don't need the height to find the area though.

OpenStudy (anonymous):

All i need is the apothem and the perimeter

OpenStudy (anonymous):

You're right, you don't, but you can take advantage of some easy trigonometry to get the answer easier.

OpenStudy (anonymous):

Ok, well i'm only in Geometry.. So i'm not suppose to be doing trig

OpenStudy (anonymous):

*Well i don't know any

OpenStudy (anonymous):

Ok, then I'll stop with that approach.

OpenStudy (anonymous):

Lol, if it's fairly easy, I don't mind learning it

OpenStudy (anonymous):

It must be easier than the book's method because everyone I talk to on here starts using trig

OpenStudy (anonymous):

No, that's ok. I can do it the other way. That's the way you're supposed to be doing it and it will blow others away.

OpenStudy (anonymous):

I'll explain the methodology and you can do the work. It won't be that hard, really.

OpenStudy (anonymous):

I got most of my questions right, but this one i got wrong and i'm not sure why.

OpenStudy (anonymous):

We'll use apothem and perimeter and derive some cool measurements from them.

OpenStudy (anonymous):

Groovy

OpenStudy (anonymous):

We can use one thing I said above, that height = 3r/4 because that does come from the apothem to radius relationship and the fact that angle and side bisectors meet in the middle and make that height 3r/4. Stop and draw yourself that on paper to convince yourself. Because that's going to be key. And grooviness is good here!

OpenStudy (anonymous):

r= radius?

OpenStudy (anonymous):

9/4 is the height?

OpenStudy (anonymous):

yes.

OpenStudy (anonymous):

Now, here's the trick, and it's really cool conceptually, but a little hard to explain. maybe I can draw a picture.

OpenStudy (anonymous):

Ok.

OpenStudy (anonymous):

|dw:1348255695035:dw|

OpenStudy (anonymous):

For an equilateral triangle circumcentre, incentre , centroid are all at the same point use this property to solve the problem

OpenStudy (anonymous):

I forgot to put a letter at the top, call it c and call the, here, I'll draw again

OpenStudy (anonymous):

@tcarroll010 , has given you the correct diagram, just name the angles and measures of sides

OpenStudy (anonymous):

|dw:1348255924927:dw| There.

OpenStudy (anonymous):

We know ac = 9/4. We know oa is 3/4. We know ob is 3/4, so we can get ab and you can do 1/2 of h x b for oab. Then double for the whole triangle.

OpenStudy (anonymous):

|dw:1348255963318:dw|

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