prove that the radius of a circle circumscribed about a triangle ABC=abc/4S, where S is the area of a triangle
552021
Consider this triangle: ( AC "not" equal to BC )\n\n'''''''''''''C\n''''''''''/.|.&#92;\n''''''''/..|..&#92;\n''''''/...|...&#92;\n''''/....|....&#92;\n''/.....|.....&#92;\n/......|......&#92;\nA''''''H''''''B\n                               \n\n\nCH is the hypotenus of the triangle.\nlet : a= BC\n      b= AC\n      c= AB\n\n\nNow we start:\n\n1-)* SinA = CH/AC => CH = b*SinA\n\n2-)* S= (CH*AB)/2\n  => S=(b*c*SinA)/2 [replace CH by the value in 1-))\n\n3-)* we procede the same for SinB :\n     S=(a*c*SinB)/2\n     since we have SinB=CH/a\n\n4-)* this implies for SinC as well, so S=(a*b*SinC)/2.\n cause if u take another hypotenus, for exemple AG, then     S = (AG*a)/2, and AG = bSinC, where do come the result.\n\n5-)* Now for the conlusion:\n  (a/SinA) = (b/SinB) = (c/SinC) = (2*b*c)/(2S)\n [/(this shows that you only need one of the above equations in 2-) , 3-) and 4-))&#92;]\n\nor we know that if a circle circumscribs about a triangle, we have:\n(a/SinA) = (b/SinB) = (c/SinC) = 2R  [verified at the end]\nwhere R is the radius of the circle.\nso we have:\n(a*b*c)/(2S) = 2R\n => R = (abc)/(4S).\n\npart II: why 2R = (a/SinA) = ...\n\nconsider a triangle inside a "demi"-circle. we know that this triangle will have a 90 degree angle. suppose the angle A=90. [i study math in french and i'm not comfortable with english math, so excuse me].\n\nA = 90 degree => SinA = 1.\nsince the triangle is inside a demi-circle, a = diameter\n => a = 2R  (it will be easier to understand if u draw the image of the triangle inside the demi-circle).\nremember that a = BC.\na = 2R => a/1 = 2R\nor SinA=1  =>  [a/SinA = 2R]\nremember again that this is true for any triangle inside a circle and not only inside a demi-circle.\n\nand everything is verified.
610782
A circle has 360 degrees. Now imagine that from the icosagon you can make 20 equally-sized triangles. The angle made by each triangle is 360/20=18 degrees. The other two angles of each triangle are both 81 degrees.\n\nThe basic formula for the area of a triangle is 1/2 base x height. How do we find these? Well, imagine that from each of the 20 triangles we make two additional right triangles. In doing this we know several things:\n\n-The hypotenuse is equal to 6.5 cm (the radius).\n-The angles of the smaller triangle are now 9 degrees, 81 degrees, and 90 degrees.\n-The sine of 9 degrees is equal to the opposite side divided by the hypotenuse. If we multiply this value by 6.6 we get the length of the opposite side ("1/2 base").\n-The sine of 81 degrees multiplied by 6.5 gives us the "height".\n-Once you compute each value (1/2 base and height) you multiply them by each other and you get the area of the triangle.\n-Multiply the area of that triangle by 20, since you have 20 of them. Voila.
141519
It is the length divided by the radius
589996
Assuming that A, B, C are all on the same line.\n\nLarger circle:\nRadius = AC = 6\nArea = PI * r^2 = 36*PI\n\nSmaller circle:\nRadius = AC - AB = 6 - 2 = 4\nArea = PI * r^2 = 16*PI\n\nShaded area:\nArea = (Large A) - (Small A)\n     =  36*PI - 16*PI = 20*PI = 62.8 units^2
662187
There is no work to show, rather theorems to justify the solutions.\n\nInscribed angles of a circle that subtend a diameter are 90 degrees. This is a corollary of a theorem which states that inscribed angles are 1/2 the measure of a sector angle which subtends the same arc. In the diameter case the sector angle is 180 degrees, thus the inscribed angles are 90.\n\nThe quadrilateral that contains angle Z has it's vertices on the circumference and is cyclic. A cyclic quadrilateral has the property that opposite angles are supplimentary (add  up to 180) thus angle Z = 180 - 148 = 32 degrees.\n\nI am not sure how you would show work toher than proving the theorems.\n\nHope it helps
548292
Assuming that an area may be dissected into triangles, then the area of each triangle is given by \n\nArea = (s(s-a)(s-b)(s-c))^(1/2)\n\nwhere s = (1/2)(a + b + c)\n\nand a, b, and c are the lengths of sides of each triangle -\n\nThe sought for area is (finally) the sum of all the areas of the triangles.\n\nGiven sufficient patience, any arbitrary area will be found.\n\n(Drawing a diagram while slogging through the exercise and looping around the separate triangles in the same direction can yield a shortcut whereby all of the internal perimeters eliminate each other, leaving only the perimeter of the original figure.)
315798
I think what you're trying to explain is that all points A,B,C are on the cirumference of a circle, with the line AC being a diameter of the circle.\n\nThis forms a right angled triangle with the right angle at point B. That's just a rule of circle geometry when 1 side of a triangle in a circle is a diameter. By angle sum of a triangle, the angle at C must be 62 degrees.\n\nDraw in the line 0B, forming an iscoceles triangle OBC, since OB and OC are equal radii. In this triangle, angles at B and C are both 62 degrees, so angle at centre is 56 degrees. This is the angle that subtends the arc you want to find.\n\nIf you want the arc length you need the radius and use the formula\nl = 2 * pi * r * 56 / 360
625973
How to do this problem:\n1. Find the Radius. To do this, take the diameter and divide by two. This gives you a radius of 5.\n2. Because the Area of a circle is equal to Pi times the radius squared, and your teacher wants the answer in terms of pi, square the radius and get 25 pi.
385118
the circumference divided by 2 x PI
607226
1.What are the applications for proving 2 Triangles  similar?\n2.What is the formula that is used to calculate the area of a i}Trapezium\n  ii}Hexagon\n3.How can you prove 2 circles are similar.\n\n\nHope this helps..I hope this is what u were asking for?
999585
I knew it as I did the activity.Draw a circle an divide it into 8 equal parts.It should be equal.Then cut it and stick such that the tip of one cone is up and the other is down.See that the edges stick each other close together.Do it until you finish all the 8 parts.\nNow you needed some concentration.You can see that it forms a parallelogram.Now area of parallelogram=base*height and the circumference of circle=Pi*2*radius.The base of the diagram is half the circumference that is Pi*radius,and the height will be the radius.You will also notice that Area of parallelogram=Area of circle.Therefore,Base*height=Pi *Radius *radius.so, now you have got pi*radius square.\nHope you like my answer.
1056342
Hint, try to work out the radius!  Do your homework yourself.
743138
join AC\nnow youhavetwo triangles ABC and ADC\nin triangle ABC <CAB+<B+<ACB=180*.............(1)<s of a triangle\nin triangle ADC <CAD+<D+<ACD=180*............(2)--------do--------\nadding (1) and (2)\n<ABC+<ADC+<B+<D+ACB+<ACD=360*\n=><DAB+<B+<BCD+<D=360*\nhence proved
135768
Area = Pi times the radius squared (A=pr2)\nIf you need to know how to find the radius, go here:\n\nhttp://math.about.com/library/blcircle.htm
858274
of a circle? you multiply 2 * pi * radius = circumference\nor\nmultiply 2 * diameter = circumference
407600
The area of the segment is given by\n\nA = r^2[theta - sin(theta)]/2\n\nSo we need theta, which is the angle that contains the chord, and the radius of the circle.\n\nThe radius of the circle is given by\n\nr = (c^2 + 4h^2)/(8h)\n= (20^2 + 4(6)^2)/(8*6)\n= (400 + 144)/48\n= 11.3 m\n\ntheta is given by\n\ntheta = 2 arcsin(c/[2r])\n= 2 arcsin(20/[2(11.3)])\n= 2.17 radians\n\nSo\n\nA = r^2[theta - sin(theta)]/2\n= (11.3)^2[2.17 - sin(2.17)]/2\n= 85.92 m^2
236956
pi times the radius squared