The Fencing Problem - Essay Example

The Fencing Problem Aim: A farmer has exactly 1000 meters of fencing and wants to fence a plot of a level land. Our aim is to find out the shape inwhich maximum area of the plot will be covered. For this we shall calculate areas of the plots of various shapes. Introduction: The farmer has a fence of 1000 meters and wants to fence of the plot. She is not concerned about the shape of the plot but has fixed perimeter. For her the only important thing is that the maximum area of the plot should be covered. Different shapes have different surface area. This depends upon the dimensions of the sides. 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. 1) Circle: The perimeter of fencing = 1000m = circumference of circle (C) Now C = 2r Where r - radius of the circle. 1000 = 2 * 3.14 * r r = 159.2m Area of circle A = r2 A = 3.14 * (159.2) 2 A = 79615m2 Hence if the shape the of the plot is round then the area that can be covered with the fencing of 1000m is 79615m2 Square: Square is a quadrilateral having all the four sides of equal dimensions. 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 plot is equilateral triangle then the area that can be covered with the fencing of 1000m is 48098m2. Isosceles Triangle: In the isosceles triangle two sides are equal while base is different. Here the total perimeter of the triangle is 1000m. Sides (a) of triangle = 350m Base of triangle (b) = 300m The area of isosceles triangle is given by: A = * Base * Height A = * b * H The height of isosceles triangle is given by: a2 = (b/2) 2 + H2 H2 = (350) 2 - (150) 2 H = 316.2m A = * 300 * 316.2 A = 47434m2 Hence if the shape the of the plot is isosceles triangle then the area that can be covered with the fencing of 1000m is 47434m2. Right Angled Triangle: Let us consider the right-angled triangle with two equal sides 'a' and a hypotenuse 'H'. In this case H2 = a2 + a2 H = (2 * a2) 1/2 H = 1.414 * a The total length of the fencing is 1000m a + a + H = 1000 2 * a + 1.414 * a = 1000 a = 290.7m H = 1.414 * 290.7 H = 419m The area of right-angled triangle is given by: A = * Base * Height A = * a * a A = * 290.7 * 290.7 A = 42253m2. Hence if the shape the of the plot is right angled triangle then the area that can be covered with the fencing of 1000m is 42253m2 Regular Pentagon: Pentagon is a type of the polygon with five sides. The sum of total angles inside the polygon is 3600. Since Pentagon is made up of five sides the angle corresponding to each side is given by 360/5 = 720. The sides (a) of the pentagon are given by: a = 1000/ total number of sides a = 1000/5 a = 200m The height of the pentagon H is given by: tan36 = (a/2) / H H = 100 / 0.73 H = 138m Area of pentagon is given by: A = H2 * n * tan (/n) Where n = number of sides of the polygon A = 1382 * 5 * tan (180/5) A = 69511m2 Hence if the shape the of the plot is pentagonal then the area that can be covered with the fencing of 1000m is 69511m2 Regular Hexagon: A regular hexagon is made up of six equal sides. The angle corresponding to each side is 360/6 = 600. Since the total length of the fencing is 1000m length of each side (a) of the hexagon is given by: a = 1000/6 = 166.7m The height of the hexagon H is given by: tan30 = (a/2) / H H = (166.7/2) / 0.58 H = 144m Area of hexagon is given by: A = H2 * n * tan (/n) Where n = number of sides of the polygon A = 1442 * 6 * tan (180/6) A = 72161m2 Hence if the shape the of the plot is hexagonal then the area that can be covered with the fencing of 1000m is 72161m2 Regular Octagon: A regular octagon is made up of eight equal sides. The angle corresponding to each side is 360/8 = 450. Since the total length of the fencing is 1000m length of each side (a) of the octagon is given by: a = 1000/8 = 125m The height of the octagon H is given by: tan22.5 = (a/2) / H H = (125/2) / 0.40 H = 151m Area of octagon is given by: A = H2 * n * tan (/n) Where n = number of sides of the polygon A = 1512 * 8 * tan (180/8) A = 72963m2 Hence if the shape the of the plot is octagonl then the area that can be covered with the fencing of 1000m is 72963m2 Parallelogram: Let us consider the parallelogram with four equal sides. The two opposite angles are 450 each Length of each side (a) = 1000/4 = 250m Height (H) = a* sin45 H = 250 * 0.70 H = 177m Area A = a * H A = 250 * 177 A = 44176m2 Hence if the shape the of the plot is parallelogram then the area that can be covered with the fencing of 1000m is 44176. Evaluation: Calculations have carried for the various shapes as shown above. The summary of these calculations is shown below: 1) If the shape of the plot is circular, the area of the plot covered by fencing with 1000m fence is 79615m2 2) If the shape of the plot is square, the area of the plot covered by fencing with 1000m fence is 62500m2 3) If the shape of the plot is rectangular, the area of the plot covered by fencing with 1000m fence is 55511m2 (for 2:1) and 60000m2 (for 3:2). 4) If the shape of the plot is triangular, the area of the plot covered by fencing with 1000m fence is 48098m2 for equilateral triangle, 47434m2 for isosceles triangle and 42253m2 for right-angled triangle. 5) If the shape of the plot is polygonal, the area of the plot covered by fencing with 1000m fence is 69511m2 for pentagonal shape, 72161m2 for hexagonal shape and 72963m2 for octagonal shape. 6) If the shape of the plot is parallelogram, the area of the plot covered by fencing with 1000m fence is 44176m2 Conclusion: As seen from the above evaluations if the plot is circular in shape then 79615m2 of area is covered by the fence of 1000m, this is the maximum area covered by any shape. Amongst the other shapes the octagonal shape covers the area of 72963m2, which is closest to the circular area. This is also considering the fact that polygon approaches the circular shape as the number of the sides of the polygon are increased. The square covers the area of 62500m2, which is more than that of both the rectangles. The least area is covered by the triangles. Amongst the three triangles considered the least area is covered by the right-angled triangle. The maximum area of the plot covered by the fence of 1000m is by the circular shape. Read More