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

The Fencing Problem - Essay ExampleLet us con brassr the square of sides a.The Perimeter of square = Summation of only sides = a + a + a + a = 4 * aThe Perimeter is given as 1000m1000 = 4 * a a = 250m = each side of the square cranial orbit of squareA = a2 A = 2502 A = 62500m2Hence if the shape the of the plot is square then the theater that can be cover with the fencing of 1000m is 62500m2Rect fee Let the two sides of the rectangle be a and bCase I Let side b = 2 * a i.e. the sides are in the ratio of 21The 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 = 333.2mArea of rectangle A = a * b A = 166.6 * 333.2 A = 55511m2Hence if the shape the of the plot is rectangular with sides in the ratio of 21 then the area that can be covered with the fencing of 1000m is 55511m2.Case II Let the sides be in the ratio of 32 i.e. b = 1.5 * aP = 2 * (a + b)1000 = 2 * ( a + 1.5 * a) 1000 = 5 * a a = 200m b = 1.5 * a = 300mArea of rectangleA = a * b A = 200 * 300 A = 60000m2Hence if the shape the of the plot is rectangular with sides in the ratio of 32 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 asTotal length of fencing/ 3 a = 1000/3 a = 333.3mThe area of equilateral triangle is given byA = * mean * HeightA = * a * HThe height of equilateral triangle is given bysin60 = H/ side of triangle (a)H = sin60 * 333.3(Angle 60o is the internal angle of the equilateral triangle)H = 289mA = * 333.3 * 289A = 48098m2Hence 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 wheel spoke 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. thenceforth we shall consider rectangle in this we shall consider the sides of ratios 21 and 32, 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 honorable 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.Pentago n is a type of the polygon with five sides. The sum of total angles inside the polygon is 3600.