Two perpendicular chords intersect in a circle...?
149565
cool! I'll think about it\n\n---\n\nhere I am.\nUse the two perpenducular segments as a coordinate system with O on their intersection.\n\nSay: the 4-3 segment defines the x axis, and the 2-6 segment defines the y axis.\nYou know four points belonging to the circle: (-4;0),(0;-2),(3;0) and (0,6)\nThe center of the circle has the mean value between -4 and 3 as its x and the mean value between -2 and 6 as its y (property of chords: the center is on the chord's axis)\n\nso the center's coordinates are C(-1/2;2)\n\nNow you can easily calculate the distance between C and anyone of the four points you know to get the radius.\n\nThen finish finding the diameter\n\n----
663303
it would be called tangent...if it touches the outside of the circle once.\n\nthe radius of the circle is a line that goes through the cirlce half way, the diamiter is a line that goes through the circle and cuts it right in half, so the radius is half the diamiter.\n\nsecant intersects a cirlce at two points.\n\nSO the answer is tangent.\n\nhope this helps.
1107584
|\n  | \n  |  A\n  |\n  |_______\n......B\n\nThis is what "A" perpendicular to "B" looks like.  I hope this helps!
832720
let line 1 make an angle x with the x-axis\nso the line perpendicular toit will make an\nangle of 90*+x with the xaxis\nm1=tanx\nm2=tan(90+x)\n=-cotx\nm1m2=tanx*(-cotx)=-1\nhence proved
1165642
One way to reason is this.  Let n be any even integer greater than or equal to 4.  Divide the circle into n equal wedges (by drawing n/2 diameter lines with angles between successive lines equal to 360/n degrees).  Now, cut the wedges out and make a "rectangle" with them: n/2 wedges with the points downward and the arcs on top of adjacent pairs of wedges touching each other, then the other n/2 wedges with the points upward and adjacent arcs touching, filling the spaces of the first n/2 wedges.  What you end up with is almost a rectangle with top and bottom each made of n/2 360/n-degree arcs, and the lateral sides slanting slightly, made from the long edges of the wedges on each end.\n\nYou can estimate the area of this figure (which is the same as the area of the original circle of radius r) by pretending it IS a rectangle.  The length is the circumference of the circle, or 2*pi*r/2 (since half the circumference is the top, half is the bottom) (by definition of pi).  The width is the length of a wedge, or r.  So the area is about pi*r^2.  \n\nTo turn the estimate into an exact value, consider what happens as n goes to infinity, that is, it increases without bound.\n\nAs n goes to infinity, the slanted edges aproach being verticle and the number of arcs making up the top and bottom approaches infinity, and the arcs themselves approach straight horizontal lines.  Thus, the figure approaches a rectangle of exactly area pi*r^2.  However, for each n, you have the same circle of the same area, so the area of that circle is pi*r^2.
61692
two faces of a polyhedron intersect along an edge
467313
ok , so you should have the equation for the right circular cone -\n\nthe area of the side of the cone — equals the circumference of the base circle multiplied by the slant height of the cone (the distance from the apex to any point on the outside of the base).\n\nNow - if the Slant height is 10 and the total area is 144 pi (square units) then it is simply a matter of substituting into the equation\n\nArea = Circumference * Slant Height\nArea of a Circle = Pi*(Radius^2)\n\nSo, what do you know?\n144 = (3.1415926* Radius^2)*10\nYou ought to be able to solve this equation for R^2.\n\nI do not believe in giving all of the answer to the homework assignment - I hope this helps you.
524608
Imagine the square of string:\n\nWhat will you do to make it a circle? You would have to bend the sides to form an arc like the way you would bend an archer's bow. For it to become a circle, you have to bend the four sides equally so that the corners will no longer be present.\n\nSo your new circle now will have an area bigger than the square. If you overlap the 2 figures together, you would have a big circle and inside it is a square whose 4 corners touch 4 equidistant points on the circle.\n\nArea of the circle: 64/pi OR 20.3718
450988
The angle is 36 degrees (10%). The chord length is 30.901699 cm = 2*50*cos(90-36/2) cm.
105143
Draw a force diagram to help you figure this out.\n\nRemember, perpendicular vectors are independent.\n\nF=ma
472218
I'm assuming that A and B are on the circle and O is the center of the circle.  Since OB=6, AB=6 as well, since OB and AB are radii.\n\n1.  Area of a circle = pi*r^2\n= pi*(6)^2\n= 36pi\n\n2.  Area of a sector (pie wedge) is easiest if you consider that the pie wedge is a certain fraction of the whole circle.  In this case, your sector is 60 degrees, and the whole circle is 360 degrees.  So the sector is 1/6 of the total area, which we found in part 1.  So the area = 6pi.\n\n3.  The segment is trickier.  If you look at the triangle, the segment is the curved part - almost like the top of the ice cream cone, if you can follow my description.\n\nThe area of a segment is given by\n\nA = r^2[theta - sin(theta)]/2\n\nwhere theta is the angle in radians.  The angle we have is 60 degrees.  To convert to radians, multiply by pi/180.  So theta is pi/3.\n\nNow since we know that the triangle has a 60 degree angle, and that OA and OB are both 6, through some reasoning we can see that this is an equilateral triangle, and AB (the chord) is also 6.\n\nSo the area of the segment then is\n\nA = r^2[theta - sin(theta)]/2\n= (6)^2[pi/3 - sin(pi/3)]/2\n= 3.26\n\n4.  Circumference = 2*pi*r\n= 2*pi*6\n= 12pi\n\nI don't think that X served any purpose except to indicate the region you're supposed to be considering for this problem.
638243
1) think of  a series of disks stacked on top of eachother, so thier centers are aligned.  Each disk has a slightly smaller radius then the one under it.  If the disks are very thin (infinately so), and you stack enough of them so that the top disk has a radius of zero (a point), then you have a  cone. \n\n2) Take a circle, and remove some portion of it- like a pizza with one or more pieces removed. Take the remaining portion of the circle, and roll it, so that the straight edges adjoining the removed section are connected together along their entire lengths, and you have a cone.
1112988
I don't think these two curves intersect at P because for r1 to cross the point P(1,1,3), t has to be 1, but if you plug t = 1 into r2, you get the point (0,1/5,4). Unless you meant to choose a different variable for r2, say for example s. Then r1 and r2 would intersect when t=1 and s =2.\nthe tangent vector for r1 is <2t, 1, 9t^2> = <2,1,9> when t=1.\nthe tangent vector for r2 is <1,(1/2)s,-1> = <1,1,-1> when s=2.\nthe angle between then the two vectors is then \narccos((a.b)/(|a||b|) )= \narccos((<2,1,9>.<1,1,-1>)/(|<2,1,9>||<1,1,-1>|))=\narccos ((2+1-9)/(16)) = 112
38417
here you assume there is almost no point of intersection...or at least that it is exactly a point therefor no measureable space so you must determine the two circumfrences...use the equation\nc=pi * diameter\nc1 = pi *6\nc2 = pi * 18\n\ntotal distance is 24pi\n\nhope that helps!\n\nactually, that is a good point, when i first read it i assumed that we would want to put the rope around the circle...meaning that we would have the rope touch the circle at all times...the problem with stating that there is a line from one circle to the other will be slightly inaccurate, the line from the two circles will be different from the horizontal by several degrees, that would be the need for teh trig function.  I would say that for this point in the school year for a high school geometry class they probably havnt gotten to trig yet...\n\nhope this helps!!
723693
I'm pretty sure you are correct.  I am also working thru Calc III (at the moment, actually).  It helps to picture the gradient as perpendicular to the tangent plane.
746093
It's straight ahead geometry.\nDraw a circle with center O.  Draw a diameter horizontally through the circle. Label the left intersect as A and the right intersect as B.  Now draw a vector from O to a point P on the circle.  What tdo you know about triangles AOP and BOP in terms of the angle the vector makes with the diameter?  Use law of sines, law of cosines, trig, etc. to show that angle APO and angle BPO add up to 90 degrees and your done.\n\n\nDoug
288646
Your question is a little bit generic. You can draw a lot of "angles and triangles" with a compass.\nThe simplest procedure that comes to my mind is this:\n1) Trace an horizontal segment from a point A to another point B.\n2) Now point your compass in A, with radius equal to AB and trace an arc over the segment.\n3) Then point in B with the same radius and trace another arc. The two arcs will intersect in a third point C.\n4) If now you join A with C and C with B, you have a triangle with all segments and angles equal.\nBe more specific if you need more help.