The formulas listed below are commonly required in geometry to calculate lengths, areas and volumes. The above formulas can be summarised in the below table Suppose we want to calculate the volume of a cone having a radius of 6 cm and a height of 14 cm. It has three dimensions as illustrated in the figure below: The base of a rectangular prism is a rectangle. It is considered to be a prism because of its cross-section along the length. Volume of Rectangular PrismĪ polyhedron containing two pairs of congruent parallel bases is called a rectangular prism. While the units of area are always in square units in the case of perimeter they are always in the standard units of length such as m, cm, dm, km, etc. Note the units of perimeter and the area. Hence, circumference of this circle = 2πr = 2 x (22/7) x 7 = 44 cm. In order to find its circumference, we need to use the formula 2πr. Let us consider a circle that has a radius of 7 cm. In other words, the circumference of a circle is what a perimeter is for other geometrical figures such as a rectangle of a square. It is given by 2πr, where r is the radius. The length that equals the boundary of a circle is called its circumference. Now, area of this circle = πr 2 = (22/7) x 4.2 x 4.2 = 55.44 cm 2 Circumference of a Circle Suppose we are required to find the area of a circle having a diameter of 4.2 cm.īy the relation between radius and diameter, we have, r = d/2. So, we half the given diameter and obtain the radius. R = d/2, where ‘d’ is the diameter and ‘r’ is the radius. We know that in a circle, the radius is half of the diameter. On Wolfram|Alpha Quadratic Equation Cite this as:įrom MathWorld-A Wolfram Web Resource.Suppose, instead of the radius we are given the diameter of a circle, how do we calculate the area? "The Quadratic Function and Its Reciprocal." Ch. 16 in AnĪtlas of Functions. Cambridge, England:Ĭambridge University Press, pp. 178-180, 1992. Recipes in FORTRAN: The Art of Scientific Computing, 2nd ed. "Quadratic and Cubic Equations." §5.6 in Numerical Oxford,Įngland: Oxford University Press, pp. 91-92, 1996. Is Mathematics?: An Elementary Approach to Ideas and Methods, 2nd ed. "Quadratic Equations."Īnd Polynomial Inequalities. Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. Viète was among the first to replace geometric methods of solution with analytic ones, although he apparently did not grasp the idea of a general quadratic equation (Smith 1953, pp. 449-450).Īn alternate form of the quadratic equation is given by dividing (◇) through by : The Persian mathematiciansĪl-Khwārizmī (ca. 1025) gave the positive root of the quadratic formula, as statedīy Bhāskara (ca. 850) had substantially the modern rule for the positive root of a quadratic. Of the quadratic equations with both solutions (Smith 1951, p. 159 Smithġ953, p. 444), while Brahmagupta (ca. (475 or 476-550) gave a rule for the sum of a geometric series that shows knowledge The method of solution (Smith 1953, p. 444). Solutions of the equation, but even should this be the case, there is no record of ![]() It is possible that certain altar constructions dating from ca. 210-290) solved the quadratic equation, but giving only one root, even whenīoth roots were positive (Smith 1951, p. 134).Ī number of Indian mathematicians gave rules equivalent to the quadratic formula. ![]() In his work Arithmetica, the Greek mathematician Diophantus The Greeks were able to solve the quadratic equation by geometric methods, and Euclid's (ca.
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