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Monday, June 22, 2009
Hi TianChong here =] Today i am going to talk about the areas of triangles.Ok lets first start off today by asking.. How many formulas do you all know? Well the most basic ones that any student who studied mathematics up to high school will probably know : Area = 1/2 x base x height or by using another commonly used method, trigonometry: Area = 1/2 absinC So now, i am going to talk about some other methods that are not commonly used such as by using coordinates : If vertex A is located at the origin (0, 0) of a Cartesian coordinate system and the coordinates of the other two vertices are given by B = (xB, yB) and C = (xC, yC), then the area S can be computed as ½ times the absolute value of the determinant Hence, Area = 1/2 det [abs( xB xC )] ___________________[abs( yB yC )] or for three general vertices, the formula is : Area = 1/2 det [abs( xA xB xC )] _____________[abs( yA yB yC )] _____________[abs( 1 1 1 )] In addition, there is also a method by using vectors: The area of triangle ABC can also be expressed in terms of dot products Area = 1/2{sqr root[(AB.AB)(AC.AC) - (AB.AC)sqr]} Lastly, there is the Heron's formula : The shape of the triangle is determined by the lengths of the sides alone. Therefore the area also can be derived from the lengths of the sides. By Heron's formula: Area = {sqr root[s(s - a)(s - b)(s - c)]} where s = ½ (a + b + c) is the semiperimeter, or half of the triangle's perimeter. Yup these formulas are pretty useful huh =D thats all happy mathematic-ing acknowledgement : wikipedia Saturday, June 13, 2009
Hi, im back to post more fun facts about MATHS =)
A. Bogomolny, Did you know that... Random interesting math facts from Interactive Mathematics Miscellany and Puzzles
http://www.cut-the-knot.org/do_you_know/index.shtml, Accessed 14 June 2009 Wednesday, June 10, 2009
Hi everyone, this is Seany--- Firstly for today, i came across this really interesting "proving" of an equation and i'll be here to share it with you. The following is a "proof" that one equals zero. Consider two non-zero numbers x and y such that Then x2 = xy. Subtract the same thing from both sides: x2 - y2 = xy - y2. Dividing by (x-y), obtain x + y = y. Since x = y, we see that 2 y = y. Thus 2 = 1, since we started with y nonzero. Subtracting 1 from both sides, 1 = 0. This equation seems to be flawless, doesn't it? -- Also.. Consider the following integral: INTEGRAL (1/x) dx Perform integration by parts: let u = 1/x , dv = dx du = -1/x2 dx , v = x Then obtain: INTEGRAL (1/x) dx = (1/x)*x - INTEGRAL x (-1/x2) dx = 1 + INTEGRAL (1/x) dx which implies that 0 = 1. This also seems flawless right? -- However, for the first "proving", The problem with this "proof" is that if x=y, then x-y=0. Notice that halfway through our "proof" we divided by (x-y). and as for the second one, it is a common mistake in integration by parts. Students often forget about the constant of integration for indefinite integrals. In this case, the constants on both sides will differ by 1. acknoledgemtns : Su, Francis E., et al. "One Equals Zero!." Mudd Math Fun Facts. -- have a nice day Tuesday, June 2, 2009
![]() But what about the formula for area of an Ellipse? Consider the picture shown on the right, where A and B are semi-major axis of the large and small diameter of the ellipse respectively. A standard equation for such an ellipse centered at origin:(x2/A2) + (y2/B2) = 1. The area of ellipse is PiAB , so similar to that of circle! Facts behind it The above diagram can be seen as a unit circle that has been stretched by a factor A in the x-direction, and a factor B in the y-direction. Hence the area of the ellipse is just AB times the area of the unit circle. However, the perimeter of an ellipse is not at all generalize with that of a circle. :) Su, Francis E., et al. "Area of an Ellipse." Mudd Math Fun Facts. ![]() ![]()
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Sean Tan Xiao Yuting Wan Tian Chong |