Given a circle and a square next to the circle . How can I construct a rectangle with the same surface as the square in the circle? Only using a compass and ruler.

Thanks

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- Feb 9th 2011, 10:56 AMbram kierkelsConstruction Rectangle in Circle
Given a circle and a square next to the circle . How can I construct a rectangle with the same surface as the square in the circle? Only using a compass and ruler.

Thanks - Feb 10th 2011, 09:02 AMUnknown008
Measure the length of one side of the square.

The length of the rectangle will be that length, x, multiplied by a certain constant k you don't know.

So, l = kx

The width will be w = x/k

So that the area of the square is x^2 and the area of the rectangle is (kx)(x/k) = x^2

In your circle of radius r that you measure, you will find that:

Substitute l and w.

Plug in the values of x and r to find k.

From there, you can get l and w, the dimensions of the rectangle to be drawn in your circle. - Feb 13th 2011, 11:27 AMbram kierkels
Thanks Unknown008. But I ment a ruler without numbers. So this should be a question in the topic "Euclidean Geometry". Can you help?

- Feb 14th 2011, 03:37 AMUnknown008
Oh, then no, sorry, I don't know about Euclidean Geometry... I'd be most happy to help otherwise.

- Feb 14th 2011, 03:51 AMProve It
- Feb 14th 2011, 06:02 AMbjhopper
the square is the maximum area quadrilateral which can be inscribed in a circle so the problem is impossible

bjh - Feb 14th 2011, 06:23 AMArchie Meade
The original post doesn't give any values for the diameter of the circle

and the side of the square.

The circle certainly needs to be of a minimum size in relation to the square.

Also, may we assume "surface" refers to surface area ? - Feb 14th 2011, 04:13 PMbjhopperconstruction of rectangle in circle
Followup to prior post

Given a square side lenght a and a circle diameter b.

the square when inscribed in a circle produces the largest area of the 4sided polygons.

A square with side a can be inscribed in a circle with diameter a*rad2. No rectangle with area a^2 can fit in this circle.An infinite number of smaller area rectangles will fit with areas ranging from close to 0 to a max close to a^2.These rectangles have diagonal lengths a*rad2

bjh