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Maths Investigation: Drainage Channels.

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Introduction

Maths Investigation: Drainage Channels. We have been given a length of metal that is 60cm wide. What shape should the metal be bent into so it will hold the greatest volume of water? 60cm X cm X cm Conditions: 1. You must have a symmetrical cross section. 2. The channel must be widest at the top. Semicircle The circumference of the whole circle would be 120cm but the circumference of the semicircle would be 60cm. 120 = 2?R 60 = ?R R = 19.1cm Area = ?R� A = ??x 19.1� A = ? x 3364.76 A = 1145.92 1145.92 is the area of the whole circle so the area of the semicircle is half of that. ...read more.

Middle

224.88cm� Example 2 Sin10� x X/30 X = 0.174 x 30 X = 5.21 Cos10� x h/30 H = 0.98 x 30 H = 29.54 Area = 5.21 x 29.54 = 154.0597cm� As you can see this is a long drawn out process, so if we were to create a table it would save having to draw all the triangles out. This is on the next page Rectangles X X 60 - 2X Example 1 5cm 5cm 50cm Area = 50 x 5 Area = 250cm� Graph of results on next page. Trapeziums. 20cm 20cm 20cm 20cm 20cm Opp xcm Adj Hyp H 50� 20cm Sin50� = x/20 X = 20sin50 Cos50� = h/20 H = 20cos50� Instead of doing this drawn out process I found a formula that worked, it is Area = (a+b/2) ...read more.

Conclusion

X 20 X S Z H H Z S B X H Z S SinZ = x/s X = s x sinz� Cosz = h/s H = S x cosz Area = b + (b + 2s sinz�)/2 = Ans Area = Ans x 20cosz� There are graphs on the next few pages to represent these results. (To enter these results into a spreadsheet I needed to know about radians which is how a spreadsheet works.) 2 x 3.142 = 360� 1� = 3.142/180� N Sides 180/n 60/n 36� Side = 60/n Angle = 180/n 180/n 90/n A H O 30/n TOA CAH SOH (Tan 90/n = 30/n) /h h = (30/n) / tan 90/n Area = 30/n x (30/n) /tan 90/n = 30/n x 30/n tan 90/n Formula for Area is = 900/ (n x tan 90/n) ...read more.

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