• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7
8. 8
8
9. 9
9
10. 10
10
11. 11
11
12. 12
12
13. 13
13
14. 14
14
• Level: GCSE
• Subject: Maths
• Word count: 1346

I have been set the task of finding equable shapes and eventually find a possible formula for them. I will intend to do this by first using the rule of an equable shape, where the perimeter and the area have the same numerical value

Extracts from this document...

Introduction

Maths Investigation Introduction - I have been set the task of finding equable shapes and eventually find a possible formula for them. I will intend to do this by first using the rule of an equable shape, where the perimeter and the area have the same numerical value and extending it to regular polygons. E.g. Taking the sides as X Area = Perimeter X� = 4X X� - 4X = 0 X(X-4) = 0 X = 0 X = 4 I intend to set out the results, eventually in a table and possibly in a graph depending on the type of results I get. I hope to see a pattern in the results so I can predict the results in more complicated shapes. I intend to extend the task to surface area and volume and find a formula for them. I am only going to do regular polygons because irregular polygons could get very complicated and I doubt a general formula could be found for them. The first shape I am going to take is an equilateral triangle because it is a regular polygon and it has the least amount of sides a shape can have. The area of a triangle is 1/2 base * height and I am going to use L for the sides E.g. ...read more.

Middle

The final regular shape I will be testing is the decagon, Area = Perimeter 10 * area of triangle = 10L 10 * 1/2 L * 1/2 L Tan18� = 10L 1/2 ' Tan18� = 2 1/2 = 2Tan18� L = 4Tan18� Here I am doing the triangle again, I am going to split it up into 3 triangles and use the same method on this as the other shapes. Area = Perimeter 3 * area of triangle = 3L 3 * 1/2 L * 1/2 L Tan60� = 3L 1/2 ' Tan60� = 2 1/2 = 2Tan60� L = 4Tan60� As you can see the result I got fits a pattern, I am now going to put these results into a table and hopefully produce a general formula for any regular shape with any amount of sides. Number of sides Formula 3 4Tan60� 4 4Tan45� 5 4Tan36� 6 4Tan30� 8 4Tan22.5� 10 4Tan18� A table of results showing how sides relate to the formula As you can probably see, the greater the number of sides the less the angle is. I predict if you take the number of sides as N, then you start with 4tan and divide 180, because that's what all the angles of a triangle add up to and divide it by then you get your answer. ...read more.

Conclusion

Surface area = volume 5 times area of triangle = surface area of face 2 * 5 * 1/2 L * 1/2 L + 5L� = 5 * 1/2 * L * 1/2 L * L Tan45� Tan36� 1/2 L 1/4 L Tan36� + 1 = Tan36� 1/2 + Tan36� = 1/4 L 4(1/2 + Tan36�) = L 2 + 4 Tan36� = L Already a visible pattern has emerged. All the 3D shapes seem to have only one equable answer. Sides (N) Formula 3 2+4tan60 4 2+4tan45 5 2+4tan36 I predict the general case formula for any number of sides for a regular 3D shape will be 2+4Tan (180�) N Conclusion - I have discovered some key formulas for finding equable lengths for regular shapes and regular 3D shapes. There is a link between the 2D and 3D shapes, both 3 sided shapes had tan of 60 and 4 sided shapes had a tan of 45 and so on. If I continued with my investigation I would of liked to have done something with cones and shapes with a slant. The investigation went quite well although I found it quite difficult at the beginning to work out and understand a method that could be used for a shape with any number of sides. Once I had the method of breaking the shapes down into triangles it was fairly easy. My method could have been improved if it was set out neater then it would be more easy to understand. ?? ?? ?? ?? 1 ...read more.

The above preview is unformatted text

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

Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

Related GCSE Fencing Problem essays

1. Investigating different shapes of gutters.

was 30? and the length (x) was 7cm, which is 1/3 of the length (L) 21cm. (L/3). I tried using different values for L, which I put in a table shown below. Length(L) Best Area (cm) Angle Side length(x) 21 63.65 30 7 24 83.14 30 8 27 105.22 30

2. Investigate different shapes of guttering for newly built houses.

A = 30x�-2x� Differentiating with respect to x dA dx = 30 - 4x� If the area is to be a maximum then: dA dx = 0 30 - 4x = 0 4x = 30 x

1. When the area of the base is the same as the area of the ...

There were many trends and patterns in the graphs. The line graph for the 18cm by 18cm squared net tray showed me that what ever the maximum volume is the area of the base equals to it this is shown on the graph because the maximum volume is when 1cm has been cut off and on the graph at

2. Investigating different shapes to see which gives the biggest perimeter

Once I know the height, I can then work out the area of the isosceles triangle. Pythagoras's Theorem: Pythagoras's theorem suggests that a� (height) + b� (base) = c� (side). If we rearrange the equation, we can work out the height if we know the base and the side.

1. GCSE Maths Coursework Growing Shapes

total pentagons so the formula = 2.5n2 - 2.5n + 1 Check No. of Pentagons = 2.5n2 - 2.5n + 1 = (2.5 x 32) - 2.5 x 3 + 1 = 16 Pattern no. (n) Pentagons Added 1 0 2 5 3 10 4 15 5 20 Pentagons Added

2. The coursework problem set to us is to find the shape of a gutter ...

Table to show sides will never equal a semi circle The table below proves that the no matter how many sides a polygon has it will never reach the area of the circle as the circle has no sides. So as the amount of sides go up the area increases

1. The best shape of guttering

360/K17 (J17/K17)/SIN(360/2*K17) 0.5*N17*N17*sinM17*(K17/2) I17*O17 I17 J17 K17+2 J18/(K18/2) 360/K18 (J18/K18)/SIN(360/2*K18) 0.5*N18*N18*sinM18*(K18/2) I18*O18 I18 J18 K18+2 J19/(K19/2) 360/K19 (J19/K19)/SIN(360/2*K19) 0.5*N19*N19*sinM19*(K19/2) I19*O19 I19 J19 K19+2 J20/(K20/2) 360/K20 (J20/K20)/SIN(360/2*K20) 0.5*N20*N20*sinM20*(K20/2) I20*O20 I20 J20 K20+2 J21/(K21/2) 360/K21 (J21/K21)/SIN(360/2*K21) 0.5*N21*N21*sinM21*(K21/2) I21*O21 I21 J21 K21+2 J22/(K22/2) 360/K22 (J22/K22)/SIN(360/2*K22) 0.5*N22*N22*sinM22*(K22/2) I22*O22 I22 J22 K22+2 J23/(K23/2)

2. The Fencing Problem. My aim is to determine which shape will give me ...

An example would be Shape 24, which has a length of 240 and a width of 260, and a shape 26, which would have a length of 260 and width of 240, they would both produce the same area, which is in this case 62400. Triangles Isosceles Triangles Shape No.

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to