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A compass and straight edge.
A pair of compasses and straight edge will suffice.
No single tool will suffice. You will need a pair of compasses and a straight edge.
"Having the same center" is the meaning of concentric. A regular hexagon in a concentric circle can be viewed by using the link and looking at the "active construction" window where there is a short animation of the construction of a hexagon inside a circle. It is posted by our friends at Wikipedia, where knowledge is free.
It depends on the diameter of the circle and the width of the square, if they are the same then the answer is no. If you draw yourself a square then inscribe a circle with a radius of half the length of a side of the square, the circle will fit inside the square but the corners of the square will be outside the circle. Thus by inspection the area of the square is larger than the area of the circle.
An equilateral triangle has 60 degree angles. 60 degrees x 6 = 360 degrees. A hexagon has 6 sides so....
Perpendicular bisector.
A compass and straight edge.
A pair of compasses and straight edge will suffice.
No single tool will suffice. You will need a pair of compasses and a straight edge.
Then you would draw the polygon inside of the circle, or in other words, "inscribe" the polygon.
"Having the same center" is the meaning of concentric. A regular hexagon in a concentric circle can be viewed by using the link and looking at the "active construction" window where there is a short animation of the construction of a hexagon inside a circle. It is posted by our friends at Wikipedia, where knowledge is free.
The artist decided to inscribe a meaningful message on the statue's base before it was displayed in the park.
It is not. If you draw yourself a square then inscribe a circle with a radius of half the length of a side of the square, the circle will fit inside the square but the corners of the square will be outside the circle. Thus by inspection the area of the square is larger than the area of the circle.
It depends on the diameter of the circle and the width of the square, if they are the same then the answer is no. If you draw yourself a square then inscribe a circle with a radius of half the length of a side of the square, the circle will fit inside the square but the corners of the square will be outside the circle. Thus by inspection the area of the square is larger than the area of the circle.
You can draw 9 diagonals inside a hexagon.
Here is one way of approaching this formula.We need to know two things:* the area of a triangle is half the base times the altitude; * the circumference of a circle is 2 pi r. Draw a circle and inscribe a hexagon inside the circle. Then draw the radii from the centre of the circle to each of the six vertices of the hexagon. (Sorry, I don't have a diagram.)The hexagon has been divided into six triangles. Look at one of these triangles: it has base one side of the hexagon and altitude a bit less than the radius of the circle.The area of all six triangles is6 times (1/2) times (side of hexagon) times (altitude of triangle).Shuffle this slightly to get(1/2) times 6 times (side of hexagon) times (altitude of triangle).Now 6 times (side of hexagon) is the perimeter of the hexagon. Soarea of hexagon = (1/2) times (perimeter of hexagon) times (altitude of triangle).Do this again with a 12-sided figure instead of a hexagon, then a 24-sided figure, and so on. We getarea = (1/2) times (perimeter of many-sided figure) times (altitude of triangle).If we take a figure with a lot of sides, its area will be very close to that of the whole circle, its perimeter will be very close to the circumference of the circle, and the altitude of one of the (very thin) triangles will be very close to the radius.So (waving my hands a bit here),area of circle = (1/2) times (perimeter of circle) times (radius of circle).If we know that the perimeter of the circle is 2 pi r, we getarea of circle = (1/2) time 2 time pi times r times r = pi times r squared.This isn't quite a precise proof, because of the hand-waving bit. But it could be made into one. See Archimedes' proof in http://en.wikipedia.org/wiki/Area_of_a_disk.