"Pascal triangle" Essays and Research Papers

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    Heron's Formula

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    2012 The event that I chose to discuss is Heron’s method that was written about in the three book series called metrica discovered in the period of about 60 AD. His formula for calculating the area of a triangle was one of the formulas which was written about in the books. Heron of Alexandria was also very well known by Hero. Their has been much debate on the birth date and ear of when Heron of Alexandria was alive. First people thought that it was around

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    The Koch Snowflake

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     This snowflake appeared to be one of the earliest fractal curves. The fractal is built by starting with an equilateral triangle. One must remove the inner third of each side and replace it with another equilateral triangle. The process is repeated indefinitely.     The length of each side is one which will help you determine the perimeter of each triangle. With having the perimeter of each triangle‚ the height can be determined so the area can be defined. The height must be determined because to use the formula A=12bh to find the area of

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    Teeder Todder

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    of tank will be empty in 1hr the entire tank will be empty in 3 hrs. (Marks: 3.0‚ Negative Marks : 0.0). 3. In the above figure‚ "CD " is the diameter‚ PQSR is a square and a of the square‚ if area of the biggest triangle is 36 sq. cm is : is an isosceles triangle. The area (A). (C). 12 sq. cm 24

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    Shapes

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    “Platonic Solids”. Till now many breath-taking huge structures are inspired by these polyhedrons. It is also found in nature in micro-organisms‚ crystals‚ molecules and many other forms. Even these solids are made of 2D shapes‚ especially 3 shapes: the triangle‚ the square‚ and pentagon polygons. These 3 simple shapes give those solids a unique symmetry impossible to find in any other polyhedron. Every shape has a concept behind it that affects the eye and brain when subjected to it‚ and imprints this concept

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    Course Outline

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    Polynomials UNIT III. Geometry A. Points‚ Lines‚ Planes and Space -Points on a Line -Distance Between Two Points -Postulates on Points and Lines -Convex Sets -Theorems on Points and Lines B. Shapes -Triangles Properties of Triangles Triangle Congruence Isoscleles Triangle Congruent Right Triangles -Polygons Polygon Sum Parallelograms Properties of Special Prallelograms Properties of Trapezoid and Kites Midsegments -Circles Arcs and Angles Chord Properties Properties of Chors‚ Secants

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    Geo 07 0236

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    △ABD ≅ △CBD 28. Critical Thinking Draw two triangles that are not congruent but have an area of 4 cm 2 each. 29. � /////ERROR ANALYSIS///// Given △MPQ ≅ △EDF. Two solutions for finding m∠E are shown. Which is incorrect? Explain the error. ��� � � � � � � � ������������������������� ������������������������� ���������� ��������������������������� ������������������������ ���������������������� 30. Write About It Given the diagram of the triangles‚ is there enough information to prove that

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    AQA 43652H QP JUN12

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    b)h h 2 b Volume of prism = area of cross-section × length crosssection h lengt r 4 Volume of sphere = – πr3 3 Surface area of sphere = 4 π r 2 1 Volume of cone = – πr2 h 3 l h Curved surface area of cone = π rl r In any triangle ABC C 1 Area of triangle = 2 ab sin C Sine rule a sin A = b sin B = b a c sin C A c B Cosine rule a 2 = b 2 + c 2 – 2bc cos A The Quadratic Equation The solutions of ax 2 + bx + c = 0‚ where a ≠ 0‚ are given by x= (02) – b ± √ (b2 – 4ac) 2a

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    Ia Math

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    which means that they are equivalent circles. Therefore‚ in the ΔAOP’‚ AO=AP. When a triangle has two equivalent sides‚ it is an isosceles triangle. By that logic‚ ∠O=∠P’. Now‚ I looked at the triangle that is already drawn in the above figure‚ ΔAOP. We know that this triangle is also isosceles because OP=AP. By that logic‚ ∠A=∠O. Using the law of cosines c^2=a^2+b^2-2abcos(C)‚ which works for any triangle‚ I assigned θ to ∠O and determined that cos(θ)=1/(2*OP). Then‚ using the law of sines

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    Sine Law

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    APPLICATION OF SINE LAW The shorter diagonal of a parallelogram is 5.2 m. Find the perimeter of the parallelogram if the angles between the sides and the diagonal are 40o and 30o10’. From the top of a 150 m lighthouse‚ the angles of depression of two boats on the shore are 20o and 50o‚ respectively. If they are due north of the observation point‚ find the distance between them. SOLUTION Solved for a 150/sin⁡50 = a/sin⁡90 a = (150 sin⁡90)/sin⁡50

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    Math and Logic Questions

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    1. The street of a city are arranged like the lines of a chessboard ‚ there are m street running north and south and n east and west . Find the no. ways in which a man can travel from N.W to the S.E corner‚ going by the shortest possible distance. 2. How many different arrangement can be made out of the letters in the ex-pression a^3b^2c^4 when written at full length? Ans. 1260. 3. How many 7-digit numbers exist which are divisible by 9 and whose last but one digit is 5? 4. You continue

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