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This indicates a radius of 4 cm, in which case the area of the circle is: 50.27 square cm

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Q: What is the area of a circle when you open your compass to 4 cm to draw a circle?
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How can you construct a perpendicular bisector of a given segment using only two parallel edges of a straightedge with out usign any mesurements on the ruler and compass?

Do I have a compass to use or not ? It's not clear from your question, but since you mentioned it at the end of the question, I'll assume that I do have a compass, and in that case, I only need one straight-edge. 1). Plant the compass on one end of the line segment, open it to more than half the length of the segment, draw a long arc that crosses the segment. 2). Keep the same opening, pick up the compass. 3). Plant the compass on the other end of the segment, draw another long arc that crosses the segment. 4). Sell the compass. 5). The two arcs intersect at two points on opposite sides of the segment. With your straight edge, draw a line between these two points. That line is the perpendicular bisector of the original segment.


Can you draw an equilateral triangle using a straightedge and compass?

Yes, it is quite simple.Draw a straight line segment, AB. Put the compass point at A and open it so that the pencil point is at B. Then draw an arc. Next, without changing the compass setting move it to B and draw another arc to cut the previous arc at C. [Actually there will be two points, one on either side of AB.] Using the straight edge, join AC and BC. Then ABC is an equilateral triangle.


How do you know whether to use a open circle or a closed circle?

If points on the circumference are excluded from the locus then an open circle, else a closed one.


How do you know whether to use an open circle or a closed circle when graphing an inequality?

If the inequality is > or< then it is an open circle. If it is greater than or equal to or less than or equal to, it is a closed circle.


Open circle at -5 open circle at 4 and shaded between -5 and 4?

Yes, and the question is ... ?

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First draw a horizontal straight line a b Put compass point on a (open compass approximately half the length of line a b - note: ensure the compass is not altered throughout the following). Draw an arc that cuts line a b at c. Put compass point on c and draw an arc from a to cut the first arc at d. Draw a straight line through a d to form a 60 degree angle. For a 120 degree angle, put compass point on dand draw an arc from a to cut the first arc at e. Draw a straight line from a through e to give a second angle of 60 degrees: 60 + 60 = 120 degrees.


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1. first draw a line any measurements 2. then get your compass and place it on the left side of your line remembering that the compass has to be more than halfway open then go up to the middle with ur compass and draw one half of a cross with the compass then draw another half a cross below the line not changing the width of the compass but make sure it is in line with the other half cross now take the compass of your page now place the point on the other side of the line and do the same as i just said then you should get two crosses one above the line and one below the finishing touch is get a ruler and where the crosses meet draw a line strainght the way down and then your done i hope that helped i learnt it in a maths lesson


What is the easiest way to draw a 5 pointed star?

Draw a horizontal line wider than your star, using a ruler to make it absolutely straight. Draw an intersecting vertical line taller than your star and forming a right angle. Label the intersection "A." 2 Open the compass so the measurement is half the size that you want your star. Put the point of the compass on mark "A." Make a circle with the compass that cuts through all the lines. Mark the spot that it intersects on the upper part of the vertical line as "B." 3 Mark any point on the horizontal line between the edge of the circle and point "A." Label this as "C." 4 Adjust the compass so the pencil touches point "B" and the point of the compass touches "C." Make an arc that intersects the horizontal line on the other side of point "A." Mark it as "D." 5 Measure the distance from point "C" to "D." Use a ruler and this measurement. Start at point "B" and draw a line from there to the edge of the circle, on the inside. This is called a chord. Make a second line of the same distance from that point to a other spot on the circle. Do this all the way around until you are back to the original point. You now have a pentagon. 6 Use a straight edge to connect every other corner of the pentagon. Start at point "B" and continue until corners are connected. This forms a pentagram, a perfect five pointed star.


How will you decide when to use an open circle or close circle when graphing?

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How can you construct a perpendicular bisector of a given segment using only two parallel edges of a straightedge with out usign any mesurements on the ruler and compass?

Do I have a compass to use or not ? It's not clear from your question, but since you mentioned it at the end of the question, I'll assume that I do have a compass, and in that case, I only need one straight-edge. 1). Plant the compass on one end of the line segment, open it to more than half the length of the segment, draw a long arc that crosses the segment. 2). Keep the same opening, pick up the compass. 3). Plant the compass on the other end of the segment, draw another long arc that crosses the segment. 4). Sell the compass. 5). The two arcs intersect at two points on opposite sides of the segment. With your straight edge, draw a line between these two points. That line is the perpendicular bisector of the original segment.


What are dividers in sailing?

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Can you draw an equilateral triangle using a straightedge and compass?

Yes, it is quite simple.Draw a straight line segment, AB. Put the compass point at A and open it so that the pencil point is at B. Then draw an arc. Next, without changing the compass setting move it to B and draw another arc to cut the previous arc at C. [Actually there will be two points, one on either side of AB.] Using the straight edge, join AC and BC. Then ABC is an equilateral triangle.


How can you work out the total surface area of an open box?

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