Fencing problem.

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GCSE Maths

Problem specification

A farmer has exactly 1000 metres of fencing. With this 1000 meter of fencing she wishes to fence off a plot of level land.

She is not concerned with the shape of the plot, but must have a perimeter of 1000 m. her requirements are that the fence off the plot of land should contain the maximum area.

Plan

At the beginning of this experiment I shall begin experimenting with the simplest of all shapes. Triangles shall be investigated first. These triangles shall include equilateral triangle, an isosceles triangle, a right-hand triangle and a scalene triangle. I would find the area and plot the results upon a table.

After investigation triangles I shall start with the regular shapes with four sides, a square, rectangles, a trapezium, a parallelogram and a rhombus. Once again areas of these shapes shall be found and recorded on a table. Now I shall continue experimenting different shapes by increasing the number of sides of each shape. These will include: triangles, quadrilaterals, a pentagon, a hexagon, a heptagon, and octagon, a nonagon and a decagon.

I would take the whole investigation one step further and experiment with polygons shapes such as letters. For example a polygon shaped as an H, a polygon shaped as an E, and so on. These areas shall also be found and recorded.

Finally I shall scrutinize a circle. The results of the area shall also be observed and noted down on a table.

After doing all the above stages I shall then devise a term that would allow be to calculate any regular shape. The term would be entered into a spreadsheet program such as Microsoft Excel. I will do this so to make sure that my term is correct and also to list the area of many more shapes.

I consider that in a shape the more the sides the more the area will be achieved/. I have considered this because before conducting this investigation I had drew many regular shapes, and I noticed with my visual observation that the more the sides a shape had, it seemed to occupy a larger area.

        

Triangles

I shall first look at triangles. Four different triangles shall be investigated. The following are the four triangles that shall be investigated in part 1:

  • Equilateral triangle: this triangle always has equal sides and equal angles.
  • Isosceles triangle: this triangle always has two equal sides and two equal angles.
  • Right-hand triangle: this triangle always consists of a right angle.
  • Scalene triangle: this triangle always has no equal sides or angles.

Equilateral triangle

The first shape that shall be tested is a regular triangle, also known as an equilateral triangle. An equilateral triangle has three axes of symmetry. If the triangle were folded on any axes of symmetry, it would produce two identical right-angled triangles. I shall find the length of side of the triangle by dividing the total perimeter by the number of sides that the shape consists of. This has been shown below:

Length of each side = Total perimeter ÷ Number of sides

Length of each side = 1000 ÷ 3 = 333.3 m

We know that the exterior angle of the triangle would add up to 3600, this figure is obtained, as the sum of interior angles is always equal to 1800. To find out the exterior angles we would as following:

Exterior angles = 3600 ÷ Number of sides

Exterior angles = 3600 ÷ 3

Exterior angles = 1200

Now we can find out the interior angles by doing the following steps:

Interior angles = 1800 – Exterior angles

Interior angles = 1800 – 1200

Interior angles = 600

Now to find the area of the triangle that has been shown above the formula that has been represented below shall be used:

Area of triangle = ½ × Base × Height

We now are familiar with the base but not the height. As the height is unknown, we shall discover the height before the investigation is continued. I have decided that shall use Pythagoras’s theorem to discover the height. This theorem works with right hand triangles only. Therefore I shall have to half the triangle that has been shown above, to what has been shown below:

 

 

Pythagoras’s theorem can now be used. This will provide me to discover the height. The following is how the height is to be found:

AC2 = AD2 + DC2

(333.3) 2 = AD2 + (166.65) 2

111108.89 = AD2 + 27772.2225

AD2 = 111108.89 – 27772.2225

AD2 = 83316.67

AD = √83316.67

AD = 288.6 m

Now we know that the height of the triangle is 288.6 m. by using the area of a triangle formula we can now discover the area of this triangle. The following shows the formula:

Area of triangle = ½ × Base × Height

Area of triangle = ½ × 333.3 × 288.6 = 48095.2 m2

Hence, the area of an equilateral triangle with a perimeter of 1000m is equal to 48095.2m2

Isosceles triangle

The second shape that is to be investigated is a different type of triangle. This time it is an isosceles triangle.

An isosceles triangle has two sides and two angles equal. It has one axis of symmetry. If the triangle is folded on the axis of symmetry, the two identical right angles will also be formed.

I shall be ascertaining the area of this triangle in the same method as I had done in the equilateral triangle. The same formula shall be used:

Area of triangle = ½ × Base × Height

I shall use Pythagoras theorem to discover the height and then the figure shall be substituted in the formula above and the area shall be found.

AB2 = AD2 + DB2

(325) 2 = AD2 + (157) 2

105625 = AD2 + 30625

AD2 = 105625 – 30625

AD2 = 75000

AD = √75000

AD = 273.86m

Now we know that the height is equal to 273.86m. I shall now substitute this figure into the formula below:

Area of the isosceles triangle = ½ × Base × Height

Area of the isosceles triangle = ½ × 350m × 273.86m

Area of the isosceles triangle = 47925.5m2

Right-angled triangle

The third triangle that has to be explored is a right-angled triangle I shall consider taking the same procedures that have been taken into mind as before.

I shall not need to use the Pythagoras theorem, as I shall know the height of the triangle.  But to prove that the above triangle is a right-angled triangle I shall use the Pythagoras’s theorem.

AC2 = AB2 + BC2

(416.67) 2 = (333.33) 2 + (250) 2

173608.9 = 111108.9 + 62500

173608.9 = 173608.9

Now it is certain that the triangle is a right hand triangle the area can be found.

Area of triangle = ½ × Base × Height

Area of triangle = ½ × 250 m × 333.33 m

Area of the right angled triangle = 41666 m2

Hence the area of this right-angled triangle is equal to 41666 m2

Scalene triangle

The final triangle that I shall be investigating is a scalene triangle. In a scalene triangle all three sides and all three angles are different. I shall not be using the area of a triangle formula during this part of the investigation. The following formula shall be used:

Area of a scalene triangle = √s (s-a)(s-b)(s-c)

The figure ‘s’ shall be found by the following formula:

Area of a scalene triangle (s) = the total perimeter of the scalene triangle ÷ 2

I shall now discover the area of the triangle below:

Area of a scalene triangle (s) = the total perimeter of the scalene triangle ÷ 2

Area of a scalene triangle (s) = 357 m + 349 m + 294 m ÷ 2

Area of a scalene triangle (s) = 1000 ÷ 2

Area of a scalene triangle (s) = 500 m

NOW

Area of a scalene triangle = √s (s-a)(s-b)(s-c)

Area of a scalene triangle = √500 (500-357)(500-349)(500-294)

Area of a scalene triangle = √500 (143 m × 151 m × 206 m)

Area of a scalene triangle = √2224079000

Area of a scalene triangle = 47160.14 m2

Hence, the area of a scalene triangle with dimensions of 349 m × 357 m × 294 m is 47160.14 m2

Results

I have now completed my task under this topic of my coursework. Four different types of triangles have been investigated. I have shown the results below in a tabulated form:

 

I have created a graph, which represents the data that has been shown above:

By looking at the graph above it is obviously noticed that the equilateral triangle possesses the most area out of all the other triangles where as a right angles triangle has the least area.  

Join now!

Quadrilaterals

During the second part of my coursework I shall be investigating different types of quadrilaterals. The following are the five main quadrilaterals that I shall be scrutinizing:

  • Square: this is a regular quadrilateral. It has equal sides and equal angles.
  • Rectangle: this is the second quadrilateral that I shall be investigating. It has two pairs of equal sides with all equal angles. I shall be investigating many kinds of rectangles.
  • Trapezium: this is a quadrilateral with a single ...

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Fencing problem.
Document Details
  • Word Count
    7410
  • Page Count
    42
  • Level
    GCSE
  • Subject
    Maths
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