Find the area of an equilateral triangle (regular 3-gon) with the given measurement. 3-inch radius A = sq. in.
Area= 1/2ap
I draw an auxiliary triangle off of the radius, the radius being the hypotenuse
By 3 inch radius, do you mean that it is inscribed in a circle of radius 3 inches?
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The apothem is half of the radius
|dw:1348254528658:dw| So it looks like both are drawings are similar?
So its 1.5
I multiply 1.5 times √3 to get the long side
Than i multiply that answer "3√3" by 2 to get the side of the big triangle
I multiply that by 3 to get the perimeter. So i have the perimeter and the apothem.
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.
The answer i got the first time is: 6 3/4√3 but they told me its wrong.
height = 1.5 x apothem = 1.5 x (r/2) = 3r/4.
I don't need the height to find the area though.
All i need is the apothem and the perimeter
You're right, you don't, but you can take advantage of some easy trigonometry to get the answer easier.
Ok, well i'm only in Geometry.. So i'm not suppose to be doing trig
*Well i don't know any
Ok, then I'll stop with that approach.
Lol, if it's fairly easy, I don't mind learning it
It must be easier than the book's method because everyone I talk to on here starts using trig
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.
I'll explain the methodology and you can do the work. It won't be that hard, really.
I got most of my questions right, but this one i got wrong and i'm not sure why.
We'll use apothem and perimeter and derive some cool measurements from them.
Groovy
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!
r= radius?
9/4 is the height?
yes.
Now, here's the trick, and it's really cool conceptually, but a little hard to explain. maybe I can draw a picture.
Ok.
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For an equilateral triangle circumcentre, incentre , centroid are all at the same point use this property to solve the problem
I forgot to put a letter at the top, call it c and call the, here, I'll draw again
@tcarroll010 , has given you the correct diagram, just name the angles and measures of sides
|dw:1348255924927:dw| There.
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.
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