The Fencing Problem Essay Example | Topics and Well Written Essays - 2500 words

The Fencing Problem - Essay Example Let us consider the square of sides 'a.' The Perimeter of square = Summation of all sides = a + a + a + a = 4 * a The Perimeter is given as 1000m 1000 = 4 * a a = 250m = each side of the square Area of square: A = a2 A = 2502 A = 62500m2 Hence if the shape the of the plot is square then the area that can be covered with the fencing of 1000m is 62500m2 Rectangle: Let the two sides of the rectangle be 'a' and 'b' Case I: Let side b = 2 * a i.e. the sides are in the ratio of 2:1 The Perimeter of rectangle = sum of all sides = 2 * (a + b) Here b = 2a P = 2 * (a + 2a) 1000 = 2 * (3 * a) a = 166.6m and b = 2 * 166.6 = 333.2m Area of rectangle A = a * b A = 166.6 * 333.2 A = 55511m2 Hence if the shape the of the plot is rectangular with sides in the ratio of 2:1 then the area that can be covered with the fencing of 1000m is 55511m2. Case II : Let the sides be in the ratio of 3:2 i.e. b = 1.5 * a P = 2 * (a + b) 1000 = 2 * ( a + 1.5 * a) 1000 = 5 * a a = 200m b = 1.5 * a = 300m Area of rectangle A = a * b A = 200 * 300 A = 60000m2 Hence if the shape the of the plot is rectangular with sides in the ratio of 3:2 then the area that can be covered with the fencing of 1000m is 60000m2. Equilateral Triangle: The equilateral triangle has three sides of the equal lengths. Here the three sides of triangle (a) will have length as: Total length of fencing/ 3 a = 1000/3 a = 333.3m The area of equilateral triangle is given by: A = * Base * Height A = * a * H The height of equilateral triangle is given by: sin60 = H/ side of triangle (a) H = sin60 * 333.3 (Angle 60o is the internal angle of the equilateral triangle) H = 289m A = * 333.3 * 289 A = 48098m2 Hence if the shape the of the...In this essay we shall first study the circle considering the perimeter as the circumference and from that finding the radius of the circle which than gives the area of the circle which can be covered with 1000m of the fence. Then we shall consider the square shape, for which first we shall find the sides of the square and then the area of the square. Thereafter we shall consider rectangle; in this we shall consider the sides of ratios 2:1 and 3:2, with the procedure same as that of the square. Then further we shall consider the triangle; first equilateral triangle is considered. For this the sides and the height of the triangle are found out and from that we get the area of the triangle. Then we have considered other two triangles; isosceles triangle and right angled triangle with the similar calculations. Thereafter various polygons are considered. Beginning with the pentagon its sides and the height are found and from that the area of the pentagon is found out. Similar approach is followed for the hexagon and the octagon. In the essay detailed calculations are shown for the various areas. The shape, which gives the maximum area, is also found and then the recommendations accordingly have been made. The calculations carried out are simple mathematical calculations. Pentagon is a type of the polygon with five sides. The sum of total angles inside the polygon is 3600.