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 Post subject: Need Help - Geometry (Located in Advanced Word Problems)Posted: Wed Jun 29, 2011 3:49 pm

Joined: Wed Jun 29, 2011 3:45 pm
Posts: 1
Good Afternoon,

NEED help with this question (Volume XIV):

#19, Page 25:

An equilateral triangle with a side of 26 is perfectly inscribed in a circle. What is the area of the circle?

Okay, we went over this one in class. I understand that equilateral means that all sides are equal, and all sides are 60 degrees.

However, when we went over this in class, we drew the triangle from the bottom of the main equilateral triangle. I am guessing we were trying to draw a triangle within a triangle to somehow assess the radius (since we needed the area of the circle).

My confusion is simply this: How did we know to do that in such a manner? What is the rule and or method here? Many in my class were confused on this, but the teacher kind of glossed over it. If you could give me a step by step method, I would REALLY appreciate it.

Thanks.

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 Post subject: Re: Need Help - Geometry (Located in Advanced Word Problems)Posted: Mon Jul 04, 2011 6:38 am

Joined: Thu Feb 12, 2009 6:32 pm
Posts: 497
we need to get to the radius of the circle.

draw the equilateral triangle within the circle.

draw a right triangle within this triangle as follows:

- the hypoteneuse goes from the lower left vertex of the equilateral triangle to the center of the circle. (this is the radius of the circle.)

- the longer leg of the right triangle is half of the bottom of the equilateral triangle along the bottom. (1/2)*6 = 3

- the short leg of the right triangle is from the center of the circle to the midpoint of the bottom of the equilateral triangle.

per our right triangle rules, the sides are the the following proportions:

x, x*sqrt(3), and x*2

where x = short leg, x*sqrt(3) = long leg, and x*2 = hypoteneuse

we know that the long leg x * sqrt(3) = 3

x = 3/(sqrt(3)

hypoteneuse = radius of circle = 2 * 3/(sqrt(3)
=( 6 * sqrt(3) )/3
= 2 * sqrt(3)

Area = pi * (2 * sqrt(3) )^2 = 12* pi

- Veritas Help

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