if O is the center of the circle and P is on its circumference, then OP is a radius
Because of one of the Circle Theorems that states that the angle subtended by any arc at the centre of the circle is half that at the circumference. A rough version of the proof follows: Suppose the quadrilateral ABCD is inscribed in a circle with centre O. Join AO and CO. This partitions the circumference into two arcs - both AC but going around different sides of the centre. One of the arcs AC, subtends angle B at the circumference and suppose the angle subtended by the same arc at O is X. Then 2B = X The other arc AC subtends angle D at the circumference and suppose the angle subtended by the same arc at O is Y. Then 2D = Y So 2B + 2D = X + Y But X + Y = 360 degrees. So B + D = 180 that is, B and D are supplementary. And then, since A+B+C+D = 360, A + C = 180. The converse can be proved similarly.
Join the centre of the circle O and the point A .Extend it to both sides to form a line.This is the required locus
all the points in the inscribed circle are also in the bigger circle. If all the points in the outside circle are the set O ... and all the points in the inner circle are the set I... Then I is a subset of O, just like a venn diagram
the number is spelled pi, not pie. Pie is a food, pi is different.Pi is an irrational number (number which doesn't end) which is the ratio to the circumference o f a circle to its diameter.Some beginning numbers of pi are: 3.14159265358979323846264338327950288419716939937510...and so on
if O is the center of the circle and P is on its circumference, then OP is a radius
A line segment extending from the center of a circle or sphere t o the circumference or bounding surface.
I presume you mean the circumference of a circle. If P and Q are two points on the circumference of a circle with center O, the number of degrees in the arc PQ is defined as the number of degrees in the angle POQ.
O = pi*D cm = 3,14 * 3,5 cm = 11 cm The circumference from the inside square is Li = 4*(Root from (2*(D/2)square))= 4*Root 2,47 =9,89 cm. The circumference from the outside square is Lo = 4* D = 14 cm.
Because of one of the Circle Theorems that states that the angle subtended by any arc at the centre of the circle is half that at the circumference. A rough version of the proof follows: Suppose the quadrilateral ABCD is inscribed in a circle with centre O. Join AO and CO. This partitions the circumference into two arcs - both AC but going around different sides of the centre. One of the arcs AC, subtends angle B at the circumference and suppose the angle subtended by the same arc at O is X. Then 2B = X The other arc AC subtends angle D at the circumference and suppose the angle subtended by the same arc at O is Y. Then 2D = Y So 2B + 2D = X + Y But X + Y = 360 degrees. So B + D = 180 that is, B and D are supplementary. And then, since A+B+C+D = 360, A + C = 180. The converse can be proved similarly.
an O another name for a circle is an O and a spherical
C-O-C-O-C-O-L-A
It depends on what instruments you have or are allowed to use. That is NOT specified and it is assumed that a compass and ruler are available for use. To construct a hexagon with sides of length L, draw a circle with centre at a point O and radius L. Pick any point on the circumference of the circle (B). Draw two arcs using the compass with the same radius as before so as to intersect the circumference of the circle at two other points, A and C - one on either side of B. Let the line AO meet the circumference on the far side of the circle at D. Similarly, BO at E and CO at F. Then ABCDEFA is a hexagon with sides of length L.
Well an o is a boy and a circle looks like a o... so its a boy!!!
Short instructions:Construct the diameter of the circle at the tangent point Construct a line at right angles to the diameter at the tangent point. this is a tangent to the circle at that point.Detailed instructions with compass and straight edge:Given: circle C with a point T on the circumference Sought: Tangent to C at TFind the center circle CPlace the needle of the compass on the (circumference of) circle C (anywhere), draw a circle [circle 1] (I think circle 1 has to be smaller than twice the diameter of circle C).Without changing the compass size, place the needle of the compass on the intersection of circles C and circle 1, draw a circle (circle 2)Without changing the compass size, place the needle of the compass on the other intersection of circles C and circle 1, draw a circle (circle 3)Connect the intersections of circle 1 and circle 2 (one is outside and one inside circle A) this we call [ line 1]Connect the intersections of circle 2 and circle 3 (also here one is outside and one inside C) [line 2]The intersection of line 1 and Line 2 is [O]. This is the center of circle CDraw a line [line 3] from [O] through [T] and beyondConstruct the diameter of the circle at [T] (the point for the tangent) and extend it beyond the circumference of circle C With your compass needle at [T] mark off equal distances on [line 3] inside and outside circle C. We call these points [4] & [5]Increase the compass size somewhat and with the needle at [4] draw a circle [circle 4]Without changing the compass draw [circle 5] centered on [5]Construct a line perpendicular to line 3 at [T]The line through the intersections of circle 4 and circle 5 is the sought tangent at [T]Note: although the instructions say "draw a circle" often it is sufficient to just mark a short arc of the circle at the appropriate place. This will keep the drawing cleaner and easier to interpret.
3.14159
Pi.