princessgloomy Posts: 31, Reputation: 1 Junior Member #1 Jan 17, 2008, 11:00 AM
Are of square, inside of a circle
A square is inscribed in a circle of radius 3(square root sign)2. Find the area of the square.

First off, Idk how to put a square root symbol up on this question haha.

Second, I get the radius. I just did 18x3.14

But I don't know how to get the area of the square if the only measurement given is the radius of the circle. :eek:
 Capuchin Posts: 5,255, Reputation: 656 Uber Member #2 Jan 17, 2008, 12:29 PM
The radius of the circle is equal to half the diagonal of the square. Use trigonometry to find the length of the side of the square. From there it's easy. Or you canjust use the area of a triangle once you have the length of the diagonal.
 princessgloomy Posts: 31, Reputation: 1 Junior Member #3 Jan 17, 2008, 01:32 PM
So the diagonal of this square is 6?
 Capuchin Posts: 5,255, Reputation: 656 Uber Member #4 Jan 17, 2008, 01:44 PM
no it's $6 \sqrt{2}$
 inthebox Posts: 787, Reputation: 179 Senior Member #5 Jan 17, 2008, 01:59 PM
pythagorean

A squared + B squared = C squared

C = 6,square root of 2

figure it out from there
 antra786 Posts: 1, Reputation: 1 New Member #6 Jul 29, 2008, 02:49 AM
I do agree with this answer and it is absolutely right answer.
_________________________
Antra
Wide Circles
 bert501 Posts: 1, Reputation: 1 New Member #7 Jul 30, 2008, 02:20 AM
The problem is interesting. I tried to solve it. But I got confused that whether the diagonal has any relationship with the side of the square. So if you try to solve this I may proceed with the problem.
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Bert

Wide Circles
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #8 Aug 2, 2008, 03:11 AM
Or, without the pythagoras theorem... a square is composed of two triangles if you cut it through the diagonal, in two. You have the base, $6\sqrt2$, and perpendicular height. Using half base times height, you get one triangle, and multiplying by two, you have the area of the square.

Or simpler, if you 'mix' the two steps, you have the area of the square as: diameter of circle x radius of circle.
 kinjalpanwala Posts: 1, Reputation: 1 New Member #9 Jul 30, 2010, 02:53 AM
A square is inscribed in a circle of radius 3(square root sign)2. Find the area of the square.
using the trigonometry you can find out the area of square as following:
for squares,
a² + b² = c²
but a = b (for squares)
so 2a² = c²

but c = 2r
so 2a² = (2r)²
so a² = 2r²

but r = 3√2
so a² = 36.
 galactus Posts: 2,271, Reputation: 282 Ultra Member #10 Jul 30, 2010, 05:18 AM

Why would you answer a question that is 2 years old? It is likely a moot point by now.
 bamal Posts: 1, Reputation: 1 New Member #11 Dec 15, 2010, 08:37 AM
galactus - I was searching for help with a similar problem when I came across this forum so I'm glad that kinjalpanwala answered a 2 year old question!
 twingleton Posts: 1, Reputation: 1 New Member #12 Oct 1, 2011, 09:01 AM
Well, galactus, since my son has the same question TODAY, it's not moot to me! 8) His question is: the area of the square is 50, what's the area of the circle? We get 12.5 times pi, but the book says it's 25 times pi.
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #13 Oct 1, 2011, 10:46 AM

The area of the square is 50, the length of one side is therefore $\sqrt{50}$

The radius of the circle is then $\frac{\sqrt{50}}{2}$

If the square is inscribed inside the circle instead of the other way round, then, you have the radius of the circle as $\frac{\sqrt{50}\sqrt2}{2} = 5$

And there goes the 25pi.
 kb5uew Posts: 1, Reputation: 1 New Member #14 Oct 26, 2011, 08:33 AM

A REAL practical application: I have a container with a diameter of 2.5" used to etch a circuit board. What is the maximum size of square circuit board that I can fit into the container? (1.78")
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #15 Oct 26, 2011, 08:48 AM
Is the height 1.78"?

If so, you already have know that the square board will have to have a length of 1.78" or shorter. The diameter is even larger that that, so I see no problem for the board to fit in, nor the need to bend the board.

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