#### Is there a connection between the size (surface area) of a leaf and the between the trunk to the road?

Surface Area of leaves Research question: Is there a connection between the size (surface area) of a leaf and the between the trunk to the road? Aim: To find a connection between the size(surface area) of the leaf and the distance between the trunk and the road. Hypothesis: Yes, I believe that the surface area of the leaves will increase the further away from the road they are. I believe that there are many reasons all due to different pollutions, caused by the cars driving on the road. I'm sure that it affects the leaves in other ways such as premature leaf drop, delayed maturity, plant growth, reproduction....but also the size will get affected. Cars are responsible for a tremendous amount of air pollution and wasted energy that affect humans and our environment. Acid rain, which is caused by air pollution, poisons our water as well as plants. The smoke and fumes from burning fossil fuels rise into the atmosphere and combine with the moisture in the air to form acids rain. The main chemicals here are sulphur dioxide and nitrogen oxides. The tree's roots absorb water from the ground, as a life source and when the acid rain, rains around that tree its life source is poisoned. The acid rain also harm the leaves as fog, acid fog, which the leaves will bath in, and that will make their protective waxy coating can, wear away. Which could lead to water loss, which makes the

#### Symmetry in Nature

Khan Salinder Snowflake . A snowflake is an example of rotational symmetry. When you rotate it 60 degrees you will find that the snowflake will still look the same as it did before it was rotated. Or you can say that it has six lines of symmetry. It can be folded in half in six different ways and both halves look the same. Snowflakes can have either hexagonal or triangular symmetry although the hexagonal snowflake is most common. Beehive A beehive has translational symmetry meaning that it has a repeating pattern of hexagons. Individual cells of a honeycomb have rotational symmetry they can be rotated one sixth of a turn and still look like the same as before the rotation. A honey comb is built slanted so that honey doesn't fall over. The hexagonal shape of a cell gives strong construction and also uses less building material. Seashell A seashell has reflectional or bilateral symmetry. A seashell only has one line of symmetry. It can be split in half so that one side is like a mirror reflection of the other side. The lines on a seashell are arranged in such a way that you see perfect symmetry. Animal Most animals are symmetrical in at least one way. For animals, symmetry is related to fitness. Symmetrical horses can run faster than non-symmetrical horses. There are two types of animals; radiata and bilateria. Radiata has radial symmetry. Bilateria

#### The aim of this coursework is to investigate which shape gives the largest enclosed area for a fixed perimeter of 1000m. In the coursework I will be investigate different shapes with different number of sides to see which encloses the largest area.

Jugdeesh Singh Maths Coursework 2003 Mrs Phull 0 Blue Aim The aim of this coursework is to investigate which shape gives the largest enclosed area for a fixed perimeter of 1000m. In the coursework I will be investigate different shapes with different number of sides to see which encloses the largest area. Three sided shape Triangle The only three sided shapes are triangle. There area depends on the length of each side. For the triangle I will be investigating which triangle with has the largest area by changing the lengths of each side and eventually getting the triangle with the largest area. Prediction: I predict that the triangle with equal sides will have the largest area. This is the equilateral triangle. First I will start by changing the length of the base. To calculate the area of the triangle I will be using the following formula /2 Base x Perpendicular Height E.g. /2 Base x Perpendicular Height (1/2 x 10m) x 15m = 5m x 15m = 75 m2 For some triangles the perpendicular height is not given, therefore we have to work the height out our selves. We will do this by applying Pythagoras' Theorem, which is. The square on the hypotenuse is equal to the sum of the squares on the other two sides. a2 + b2 = c2 E.g. At this point we do not know the perpendicular height so by using Pythagoras' Theorem we can work out the height, but first we must divide the

#### t shape t toal

T-Total Part 1 The aim of the investigation is to find out the relationship between the t-number and t-total. The t-number is the number in the t-shape, which is at the base of the T. The t-total is the sum of all numbers inside the t-shape. I will start my investigation by looking at t-shapes on a 9 by 9 grid. To solve the problem of finding the relationship between the t-number and t-total I will look at the information algebraically. I will firstly assign a letter to the t-number of the shape, this letter will be T. I will then express the rest of the numbers in the t-shape with the letter assigned to the t-number in the t-shape. Therefore it will give me a standard expression to apply to all the t-shapes. Where the expression is equal to the t-total. Examples ` The expression works for all t-shapes in a 9 by 9 grid. I will now simplify the expression into a simple formula. T + (T-9) + (T-17) + (T-18) + (T-19) = T-total 5T - 63 = T-total I will see if this new formula still works. 5 ? 20 - 63 = T- total 00 - 63 =T - total 37 = 37 5 ? 21 - 63 = T- total 05 - 63 = T - total 42 = 42 Part 2 I will now as part of my investigation use different grid sizes, transformations of the t-shape and investigate the relationship between both. Then I will see how the t-number and the t-total relate to the new factors. The smallest grid size can only be a 3 by 3 grid

#### An experiment to find out if seeing the eyes of a well known persons face is a factor of face recognition

An experiment to find out if seeing the eyes of a well known (celebrity's) face is a factor of face recognition. Abstract; The aim of this experiment is to find out if participants can recognise the faces of well known celebrities if the celebrities' eyes are blacked out and if the eyes of a person's face are a major factor of face recognition. I predict that the participants will find it easier to recognise the celebrities' faces in the condition where the eyes are not blacked out more than when the eyes are blacked out. I used a lab experiment and independent group design. The target population from which my participants were used were the 20 students in my AS psychology class (opportunity sample). I randomly separated the 20 participants into 2 groups of 10 for each of the 2 conditions. I then delivered a brief and handed out the pictures of the celebrities face down in front of the participants. I then timed the participants for 2 and a half minutes while they turned over the sheets and wrote their answers upon it. Number of participants Condition 1 Celebrities with eyes ( mean score out of 20) Number of participants Condition 2 Celebrities with blacked out eyes (mean score out of 20) 0 7/20 = 85% 0 3/20 = 65% In consideration of my results I reject my null hypothesis and accept my hypothesis that the participants will find it easier to recognise the

#### 1,000m of fence - what is the maximum area I can enclose?

,000m of fence - what is the maximum area I can enclose? Introduction: In this project, I am trying to find out the maximum area of an enclosure I could make with 1000m of fencing. I am going to attempt to tackle the problem by using different shapes, Including Quadrilaterals, Pentagons, Hexagons, Octogons, Isogons and circles. I already know that the mathematical solution (circles) may not be useful for the farmer because circles do not tessellate. The plots need to tessellate to make use of as much land as is possible - I have shown examples of tessellation below: Historically, farmland in Britain was divided into hexagonal plots to use all available space. Since proper roads were built, this has no longer been possible, as space is at a minimum and roads usually run at the sides of fields. Hypothesis: My hypothesis is the more sides a shape has the more area it will contain and I believe the mathematical solution will be a circle because circles have infinite sides. Logical Ideas for shapes: * Triangle * Quadrilateral * Pentagon * Hexagon - There is a pattern here, so I will jump to: - * Decagon (10 sides) * Isogons (20 sides) * n number of sides How do I know if a Formula Measures Area or Volume? /2 b h measurement x measurement l b w measurement x measurement x measurement Quadrilateral: As with triangles, there are different types of

#### In this work I investigated different shaped tubes, which can be made from rectangular sheets of card, of given width and length. I used Microsoft Excel to collate data.

Tubes Part 1 Introduction: In this work I investigated different shaped tubes, which can be made from rectangular sheets of card, of given width and length. I used Microsoft Excel to collate data. (These sets of data are available in a set of appendixes at the back) Method: Using Microsoft Excel and the dimensions of the card that were given, I was able to collate the data for the volume of different shaped tubes. I used different formulas to try and test out what the maximum volumes were for each given shape. I used 24cm for the length and then 32cm for the length on the same shape to see the comparison. Throughout this investigation the length of the card is going to be represented by L and the width by W. Tube investigated To find the volumes of the different shaped tubes, I worked out the cross sectional area and then multiplied that by the length of the tube. General Formulas for volumes of prisms Cylinder: cross sectional area = ?r2 Volume = ?r2 x length Cuboid with rectangular end face: cross sectional area = width x length Volume = cross sectional area x length of prism Equilateral and Isosceles Triangle face: cross sectional area = 0.5 x base x height Volume = cross sectional area x length of prism Hexagon: cross sectional area = Generalised Formula for working out the volume of any polygon Results: Refer to appendix 1 for the data collected for

#### Borders - Fencing problem.

GCSE Coursework Borders Here are the first four examples of the shapes in this sequence: The sequence begins with a single white square, which is then surrounded by black squares to form the second shape. Each new cross is then formed by completely surrounding the previous cross with a border of black squares. In each new cross, the previous cross can be seen as the area of white squares in the centre. In this investigation I will try to get algebraic formulas from the sequence, each expressing one property in terms of another (e.g. defining the area as the diameter squared). These formulas can then be checked and, proven, using a variety of maths skills. Defining N The variable that I will be using to refer to terms in this investigation will be n. As there could be confusion over this matter I thought that it is important to state that the first term in the sequence which is just one white cross will have a value of n which is equal to zero. This means the n'th term for any shape in the sequence will be the number of squares that extend from the centre square out to the edge. e.g.) After looking at this diagram I worked out that the width of this shape would be: 2n + 1. This also works for all of the other shapes in the sequence here is a table to prove it. n 2n + 1 width (counted) 0 3 3 2 5 5 3 7 7 4 9 9 5 1 1 Perimeter If I count the number of

#### Shapes Investigation I will try to find the relationship between the perimeter (in cm), dots enclosed and the amount of shapes (i.e. triangles etc.) used to make a shape.

GCSE Maths Coursework - Shapes Investigation Summary I am doing an investigation to look at shapes made up of other shapes (starting with triangles, then going on squares and hexagons. I will try to find the relationship between the perimeter (in cm), dots enclosed and the amount of shapes (i.e. triangles etc.) used to make a shape. From this, I will try to find a formula linking P (perimeter), D (dots enclosed) and T (number of triangles used to make a shape). Later on in this investigation T will be substituted for Q (squares) and H (hexagons) used to make a shape. Other letters used in my formulas and equations are X (T, Q or H), and Y (the number of sides a shape has). I have decided not to use S for squares, as it is possible it could be mistaken for 5, when put into a formula. After this, I will try to find a formula that links the number of shapes, P and D that will work with any tessellating shape - my 'universal' formula. I anticipate that for this to work I will have to include that number of sides of the shapes I use in my formula. Method I will first draw out all possible shapes using, for example, 16 triangles, avoiding drawing those shapes with the same properties of T, P and D, as this is pointless (i.e. those arranged in the same way but say, on their side. I will attach these drawings to the front of each section. From this, I will make a list of all

#### Find a formula to enable the perimeter to be found for any odd Pythagorean triple.

Perimeter: I want to find a formula to enable me to find the perimeter for any odd Pythagorean triple. I know that perimeter is a + b + c so I know the formula for the shortest, middle and longest side so I can assume that if I substitute the formulas I know, I can presume I will find the formulas for the perimeter. The perimeter = a + b + c. Therefore I took my formula for 'a' (2n + 1), my formula for 'b' (2n² + 2n) and my formula for 'c' (2n² + 2n + 1). I then did the following: a + b + c (2n + 1) + (2n² + 2n) + (2n² + 2n + 1) = 2n + 1 + 2n + 2n + 1 + 4n² = 4n² + 6n + 1 + 1 = 4n² + 6n + 2 = formula for perimeter I will further prove this formula by using sequences. 3,4,5 5,12,13 7,24,25 9,40,41 11,60,61 12 30 56 90 132 +18 +26 +34 +42 +8 +8 +8 Here the first differences aren't the same, but the second differences are. If the second difference is a constant, then the formula for the nth term contains n². The number in front of n² is half the constant difference. Therefore the first part of the formula is 4n². Here I will substitute 4n² into a table to find the rest of the formula. The difference is the difference between perimeter and the 4n²: Term number: 1 2 3