# fencing problem part 2/8

Extracts from this document...

Introduction

This is a graph of the areas of isosceles triangles. It contains more details than the table above and there are more results shown.

My results show that the triangle with the largest area for a perimeter is the equilateral triangle. This also follows my previous observation but with a slight change. This triangle has no difference between all three sides, not just two sides. As there is only one form of equilateral triangle for each perimeter, I will now work out the area of an equilateral triangle with a perimeter of a thousand metres.

To find out the area of this equilateral triangle, I will use Heron’s Formula. The formula is

Where S stands for semiperimeter.

(Half of the total perimeter).

48112.52m2

Middle

Brahmagupta's formula is .

Brahmagupta’s formula can only be used in cyclic quadrilaterals.

Some examples of cyclic quadrilaterals are below.

Many cyclic quadrilaterals are rectangles, trapeziums or squares, there are also many cyclic quadrilaterals which are not any of the three shapes above but I will not be including them in my investigation because this would be very time consuming as there area an infinite number of these quadrilaterals. I will now investigate the areas of various trapeziums.

I will use a perimeter of 1000m and a fixed base of 200m. I will then set a length for the line parallel to the base, and this length will be changed. The other two

Conclusion

60000.00

200

225

287.5

287.5

61035.98

200

250

275

275

61618.79

200

275

262.5

262.5

61704.31

200

300

250

250

61237.24

200

325

237.5

237.5

60146.31

200

350

225

225

58336.31

200

375

212.5

212.5

55674.14

200

400

200

200

51961.52

200

425

187.5

187.5

46875.00

200

450

175

175

39804.21

My results show me that the trapezium with the least difference between three unfixed sides has the largest area. I predict that the trapezium with the least difference in the three unfixed lengths would have a bigger area.

I will now work out the area of the trapezium which has no difference between the three unfixed sides. I will work out the length of the three unfixed sides by dividing 800m (1000m – base) by three. The answer is 266.6667m. Using Brahmagupta’s Formula, I have worked out that the area of the trapezium is 61734.20m2. As I predicted, this trapezium has more area than all the other trapeziums of base 200m.

This student written piece of work is one of many that can be found in our GCSE Fencing Problem section.

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