an equilateral triangle is inscribed in a circle of radius r. expresss the circumference C as function of the?
966174
a triangle with angles 30, 60,90\nhas sides in ratio 1: (SR 3) : 2\nThe equilateral triangle can be formed from\ntwo 30,60,90 triangles.\neach of those can be formed from 3 smaller 30,60,90 triangles\nso the equilateral triangle can be formed from six 30,60,90 triangles.\nIf radius of circle is 1 unit,\neach of the six small 30,60,90 triangles has \nsides (1/2) ((SR 3)/2) 1\nso the equilateral triangle has side SR(3) units\nso if the equilateral triangle has side x, the circle has radius x /(SR(3)),\nand the circle circumference is 2 pi x /((SR 3))
204855
Looks to me like you are trying to figure out a "PERIMETER" rather than an "AREA".  If so, you need to ADD the lengths of all the sides together.  If you ARE looking for an area, it is simply Length x Width (for a rectangle or square), 1/2 the base x height (for a triangle), or Pi x radius squared (Pi is approximately =3.1415926) for a circle.  If the "shape" of the object you are trying to find the area of is irregular, then you must divide it up into these familiar shapes first, find EACH individual area, then add those areas together for the total area.
764538
For any given triangle ABC with sides a, b and c.\n\nCos C = (a^2 + b^2 - c^2)/2ab  This is the cosine rule.\n\nSin C =sqrt(1 - Cos^2 C). You may need to adjust the sign of the sign to match the triangle.
937438
180 degrees = π radian    ok!\n\nThe sum of the three angles of a triangle = 180 degree\nor, x + y + z = 180 deg.\nor, x + y = (180 - z)deg.\nor, x + y = 180(1 - z/180)deg.\nor, x + y = 180(1 - z/π) radian (pvd)
889089
hi\nthe details which you gave have the answer for this question.\n\nangle E= angle C and angle A= angle B and DA=DB\n\nfrom this given data i will not assume that the triangles are equilateral, I will prove they are.\n\ncalculating the area of the 2 triangles with the formula (1/2)*base*height we can prove these 2 triangles are congruent and AC=BE\n\nsince the angle E=angle C and DA=DB then DB and DA must be the sides opposite these angles E and C respectively. \nhence  the line drawn(at angle of 90 degrees to base DB) from angle E to DB and line drawn(at and angle of 90 degrees to DA) from angle C to DA must be of equal magnitude (otherwise the angles cant be equal and/or sides cant be equal) ,  \n\nhence the length of this line drawn which is called as the height of the triangle is proved to be equal\n\nnow calculating the area with above formula (1/2)*base*height\nheights of 2 triangles are proved to be equal equal and bases are \ngiven to be equal hence areas of these 2 triangles must be equal.\n\nwhen area of 2 trianlges are equal and 2 sets of angles are equal and one side is equal then the 2 triangles must be absolutely congruent. hence AC must be equal to BE
418030
Take a unit circle and a point p on the circle. Place the circle so that it is centered at the origin in a Cartesian xy-plane. The cosine of p, cos(p), is defined to be the x-coordinate, or abscissa, of p. The sine of p is the y-coordinate. These functions are the just the projection functions for the unit circle by the standard embedding in the Cartesian plane as described above.\nIt is common to give the points p on the 1-dimensional circle the coordinate referring to the arclength between p and the positive x-axis (as measured clockwise). Thus, the point (1, 0) on the unit circle is given the native coordinate (0), or 0 for short. Inversely, the point 0 in the unit circle has Cartesian coordinates (cos(0), sin(0)). Since the circumference of a circle is 2Pi, a point that is a quarter of the way around the circle from the positive x-axis is given the coordinate 2Pi/4 = Pi/2 and so forth. This system of coordinates is called radians and is the native way of measuring angles, as each angle corresponds to a unique point on the unit circle.\nThis relationship can be seen by drawing a radius from the origin to mark the angle from the positive x-axis and assigning that angle to the radian measure of wherever it intersects the unit circle. There are more analytical reasons as well, as you go on to study calculus.\nThe radius drawn above also gives the relationship between these circular projection functions, and the way one first encounters trigonometric functions, as ratios of the sides of right triangles. Now you should be able to visualize the triangles inside the unit circle (or any larger circle, dependent on the triangle's hypotenuse), and see why the trigonometric functions are cyclic in 2Pi, and derive some properties of the trigonometric functions that are not easily apparent from just looking at right triangles.\nIt is now however easy to see why all the basic trigonometric functions have a neighbor with prefix co-. The cosine is short for complementary sine. This is because the cosine of a point is obviously the same as the sine of the complementary angle.\nThe tangent function is so named because it is the length of a certain tangent to the circle; a tangent drawn from the positive x-axis to where it meets an extended radius of the circle. You may be able to work out the equation tan(p) = sin(p)/cos(p) from the geometry involved.\nThe arc trigonometric functions are so named because they measure arc length; they give you back the radian measure of the sine or cosine when you restrict these functions to one hemicircle (otherwise you get two possible answers).\nThe secant function does not give you an arclength; it is another function of points on the unit circle (and thus by our unit circle coordinate system, it is a function of arc length). In the right triangle formed by the tangent to the circle, the secant is the length of the hypotenuse. The arcsecant function returns the arc length on the unit circle that the secant intersects. See the sources below for illustrations.
966422
The area of an equilateral triangle is\nA = (1/4)sqrt(3) * a^2, where a is the length of the side.\nSince the triangle is equilateral, you know that the length of each side is 10/3.\nThe area is\nA = (1/4)sqrt(3) (10/3)^2\nA = (1/4)sqrt(3) (100/9)\nA = (25/9)sqrt(3)
763047
Angle 2*pi is one revolution, so this arc is the circumference!\n2*pi*radius = 18*pi\n\nThere is also the general formula \narc length = radius * angle (in radians), which of course gives exactly the same result here.  But after 2 hours I guess you're not there to see this answer!  Sorry no-one responded earlier.
237156
Pi * r^2\n\n(Pi R Squared)\n\nwhere R is the radius (which is also equivalent to half the diameter)
351138
You got it.\n\nThe equation of a circle in standard cartesian from is:\n\n(x-a)^2+(y-b)^2=r^2\n\nThe circle is centered at (a,b) and r is the radius. For you, (a,b) is (0,0) and your r^2 term is 49, hence your r is the square root of 49 which is 7.
366178
Take the average between the side of an inscribed square\nand that of an inscribed hexagon. With a hexagon you\nget an isosceles triangle with base as a side and a 60\ndegree angle at the apex, leaving 120 degrees to be split\nbetween the equal pair which are then also each 60 degrees.\nSo you have an equilateral triangle whose base is equal\nto the radius of the circle and of course the base is a side\nwhich has a length of 1. \n  Now inscribe a square. There are now 4 isosceles\ntriangles with a 90 degree angle at the apex and a pair\nof 45's at the base. Draw an altitude from the apex, \nperpendicular to the base, thus bisecting the 90 at\nthe apex making another 90, 45, 45 triangle whose\nside is 1/2 the side of that of the square. The hypotenuse\nis a radius of the circle so the side of the last mentioned\ntriangle is 1/2^2 (.707). This makes the side of the\nsquare 1.414. So averaging 1 and 1.414 you get 1.207.\n Using trig we find the real answer to be 1.1755.\nSo the percentage of error is 2.67 percent from 1.207.\nI wouldn't win an engineering prize with that kind of\naccuracy, but I think I could build a pretty good\ncatapult.
492563
This cannot be the whole question.  Since you are using the letters r and d, I will assume you are talking about radius and diameter.\nAre you trying to find the area? then use the formula\nA = Pi*r^2\nIf you are trying to find circumference, then use\nC = 2*Pi*r\nRemember that d = 2r (diameter is twice the radius)
613272
could be circle because it doesn't have any sides or angles\n\ncould be rectangle because it is not proportional (side-wise.. and only if the triangle is equilateral)\n\ncould be triangle because if all shapes are placed right-side and divided into four equal parts, like a cross-hair, the triangle is the only one without equal portions.
456513
A=pi*x*y \n\nwhere x is small radius from centre to circumference and y is large radius from centre to circumference.
367687
an angle that is inside a circle or any other figure
804994
Let the Length of the side of the rectangle be x and the length of the side of the triangle is L\n\nThen when the rectangle is inscribed it is symmetrically placed (otherwise you would not get a rectangle). So the distance from its base vertices to the vertices of the triangle are each ½(L - x)\n\nThus the width of the rectangle = ½(L - x) tan 60°\n= √(3)/2 * (L - x)\n\nThus the area of the rectangle A = length * width\n= x * √(3)/2 * (L - x)\n= √(3)/2 * (Lx - x²)\n\ndA/dx = √(3)/2 * (L - 2x)\n= 0 for stationary points\nWhence L - 2x = 0 ie x = ½L\nd²A/dx² = -√(3) < 0 for all x Thus the curve is always concave down So the stationary point is a maximum.\n\nThus the length of the rectangle is ½L and the width is √(3)L/4\n(and its area is √(3)L²/8)
595970
draw a line maesuring 2.5 cm..this is bc.then using a protractor make an angle of 60 degrees n on the angle measure the other side of 3.5 cm.join the end points and u get ur triangle.The circuncircle is a circle that touches all 3 vertices.then draw the perpendicular bisector of any two sides..the intersectin point becomes ur centre for the circle.now with the radius equal to the distance between the centre n the vertex draw a circle..tada!