Surface Area Formulas
In general, the surface area is the sum of all the areas of all the shapes that cover the surface of the object. Cube | Rectangular Prism | Prism | Sphere | Cylinder | Units

Note: "ab" means "a" multiplied by "b". "a2" means "a squared", which is the same as "a" times "a".

Be careful!! Units count. Use the same units for all measurements. Examples

|Surface Area of a Cube = 6 a 2 |

[pic](a is the length of the side of each edge of the cube)
In words, the surface area of a cube is the area of the six squares that cover it. The area of one of them is a*a, or a 2 . Since these are all the same, you can multiply one of them by six, so the surface area of a cube is 6 times one of the sides squared.

|Surface Area of a Rectangular Prism = 2ab + 2bc + 2ac |

[pic](a, b, and c are the lengths of the 3 sides)
In words, the surface area of a rectangular prism is the are of the six rectangles that cover it. But we don't have to figure out all six because we know that the top and bottom are the same, the front and back are the same, and the left and right sides are the same. The area of the top and bottom (side lengths a and c) = a*c. Since there are two of them, you get 2ac. The front and back have side lengths of b and c. The area of one of them is b*c, and there are two of them, so the surface area of those two is 2bc. The left and right side have side lengths of a and b, so the surface area of one of them is a*b. Again, there are two of them, so their combined surface area is 2ab.

|Surface Area of Any Prism |

[pic] (b is the shape of the ends)
Surface Area = Lateral area + Area of two ends
(Lateral area) = (perimeter of shape b) * L
Surface Area =...

...Planning
Aim
To determine how the surfacearea of the tablets affects the rate of the reaction. To determine which form of tablets gives the biggest surfacearea resulting in the fastest reaction rate.
Investigation question:
What is the relationship between the total surfacearea of the tablets and the rate of the reaction?
Hypothesis:
The rate of reaction will be the fastest when the tablets crushed into powder as there is a bigger total surfacearea resulting in more effective collisions between particles.
Variables:
Independent variable: Different forms of tablets.
Dependant variable: Time the syringe took to stop moving as the tablets dissolve.
Fixed variables:
*External temperature
*volume of HCl
*Temperature –all 3 final runs were done on the same day so whether was not an issue and did not affect the results
*Use of catalyst – a catalyst was not used in any of the experiments
* Use the same person to observe the reaction because different people have different eyesight
Background information relating to the experiment
In this experiment we are looking at one effect that influences the rate of reaction , namely total surfacearea. The reaction rate (rate of reaction) or speed of reaction for a reactant or product in a particular reaction is defined as how fast or slow a reaction takes place....

...Surfacearea / Volume ratio Experiment
Introduction:
The surfacearea to volume ratio in living organisms is very important. Nutrients and oxygen need to diffuse through the cell membrane and into the cells. Most cells are no longer than 1mm in diameter because small cells enable nutrients and oxygen to diffuse into the cell quickly and allow waste to diffuse out of the cell quickly. If the cells were any bigger than this then it would take too long for the nutrients and oxygen to diffuse into the cell so the cell would probably not survive. Single celled organisms can survive as they have a large enough surfacearea to allow all the oxygen and nutrients they need to diffuse through. Larger multi celled organisms need specialist organs to respire such as lungs or gills.
Apparatus Needed For the Experiments:
1. Beakers
2. Gelatin blocks mixed containing universal indicator
3. 0.1M Hydrochloric acid
4. Stop Watch
5. Scalpel
6. Tile
7. Safety glasses
Method:
1. A block of gelatin which has been dyed with universal indicator should be cut into blocks of the following sizes (mm).
5 x 5 x 5
10 x 10 x 10
15 x 15 x 15
20 x 20 x 20
10 x 10 x 2
10 x 10 x 10 (Triangle)
10 x 15 x 5
20 x 5 x 5
The triangle is of the following dimensions.
The rest of the blocks are just plain cubes or rectangular blocks.
Universal indicator is a neutral indicator....

...INTERNATIONAL BACCALAUREATE
BIOLOGY INTERNAL ASSESMENT
DATA COLLECTION AND PROCESSING
CONCLUSION AND EVALUATION
RELATIONSHIP BETWEEN SURFACEAREA AND THE VOLUME OF A CELL, HOW IT AFFECTS THE RATE OF DIFFUSION
Nazirul Ibrahim
04.10.2011
DATA COLLECTION AND PROCESSING
We have determined the following variables:
* dV- the time for the color indicator phenolphthalein to reach the center of the cube (diffusion)
* iV- the size of the cube
* cV- the amount of Sodium Hydroxide the perfect dimensions of the cube, how the cube is fully submerged in the solution, the temperature of the cube and the constant surrounding temperature, the concentration of the NaOH solution
During this experiment of investigating the relationship between the surfacearea and the volume of a cell, the following was determined:
* The amount of time (sec) taken for the pink color to reach the center of the cube
* How ratio of the cube (the volume and surfacearea) affects the time of diffusion
* The ratio of the
With the results obtained accurately, calculations were made to find out the following:
* The mean of each group’s data
* The standard deviation of each group’s data
As each trial consist for 5 different sizes of the agar cubes (1cm, 1.5cm, 2cm, 2.5cm and 3cm) and that the trials are repeated 5 times, we expect that results will be as follows:...

...SurfaceArea to Volume Ratio and the Relation to the Rate of Diffusion
Aim and Background
This is an experiment to examine how the SurfaceArea / Volume Ratio affects the rate of diffusion and how this relates to the size and shape of living organisms.
The surfacearea to volume ratio in living organisms is very important. Nutrients and oxygen need to diffuse through the cell membrane and into the cells. Most cells are no longer than 1mm in diameter because small cells enable nutrients and oxygen to diffuse into the cell quickly and allow waste to diffuse out of the cell quickly. If the cells were any bigger than this then it would take too long for the nutrients and oxygen to diffuse into the cell so the cell would probably not survive.
Single celled organisms can survive as they have a large enough surfacearea to allow all the oxygen and nutrients they need to diffuse through. Larger multi-celled organisms need organs to respire such as lungs or gills.
Method
The reason I chose to do this particular experiment is because I found it very interesting and also because the aim, method, results- basically the whole experiment would be easily understood by the average person who knew nothing about SurfaceArea/Volume Ratio. The variable being tested in this experiment is the rate of diffusion in relation to the size of the gelatin...

...AREA
(i) The area of a rhombus is equal to the area of a triangle whose base and the corresponding altitude are 24.8 cm and 16.5 cm respectively. If one of the diagonal of the rhombus is 22 cm, find the length of the other diagonal.
(ii) The floor of a rectangular hall has a perimeter 250m. If the cost of paining the four walls at the rate of Rs 10 per m2 is Rs 1500. Find the height of the hall.
(iii) A room is half as long again as it is broad. The cost of carpeting the room at Rs 3.25 per m2 is Rs 175.50 and the cost of papering the walls at Rs 1.40 per m2 is Rs 240.80. If 1 door and 2 windows occupy 8m2, find the dimensions of the room.
(iv) A river 2m deep and 45m wide is flowing at the rate of 3 km per hour. Find the volume of water that runs into the sea per minute.
(v) A closed cylinder has diameter 8cm and height 10cm. Find its total surfacearea and volume.
(vi) The volume of a metallic cylinder pipe is 748cm3 . Its length is 14 cm and external diameter 18cm. Find its thickness.
(vii) A cylindrical bucket, 28cm in diameter 72cm high is full of water. The water is emptied into a rectangular tank, 66cm long and 28cm wide. Find the height of the water level in the tank.
(viii) A cylindrical tube, open at both ends, is made of metal. The internal diameter of the tube is 10.4cm and its length is 25cm. The thickness of the metal is 8mm...

...investigation is to prove the effect of increasing size on the efficiency of diffusion. Diffusion is the process that cells use to obtain oxygen, water and food. Also, how they lose waste substances, for example, urea and carbon dioxide. Basically, Diffusion is when particles move from an area of high concentration to an area of low concentration. The surfacearea to volume ratio of the cell is an important factor in diffusion. It is the effect of this factor that will be investigated in this practical.
Materials used in the practical consist of blocks of agar jelly containing indicator, a sharp knife, used to cut the agar jelly into the recommended sizes, diluted sulphuric acid (300mL) , ruler, to measure the sizes of the cubes, absorbant paper towel, three 250mL beakers, and a measuring cylinder.
To perform this experiment, the following procedures will need to be follow precisely:
Obtain three blocks of agar jelly. Carefully cut three cubes, as close as possible to 1 cm, 2 cm, and 3 cm in size. Try to be exact, as this will help in the calculations later. Record the exact dimensions of the cube, in a table (see back of booklet).
Calculate the volume, the surfacearea, and surfacearea/volume ratio for each cube. These will be known as the ‘outer’ cube dimensions.
Measure 100 mL of sulphuric acid into each of the three beakers.
CAUTION:...

...Kimberly M Dollar
000234333
EFT4: Math: Task 5: SurfaceArea of Cubes
Introducing SurfaceArea
For a fifth or sixth grade class to understand the concept surfacearea in relation to a cube they need to understand what a cube is first. They will learn that a cube is a special type of rectangular solid. The length, width, and height of a cube are exactly the same. After explaining what a cube is they will need to understand what it means to find the surfacearea. The surfacearea is not the same as finding the volume of a cube. The surfacearea is the area on the outside of a three-dimensional shape, like the cube. The surfacearea of a cube is six times the surfacearea of one side of the cube. There are six sides to one cube, after learning this about a cube the appropriate formula to find the surfacearea is:
Surfacearea of a cube=6s^2 (The “6” represents the number of sides; “s” represents one side of a cube; “^2” represents taking one side and timing it by itself; the end result gives the surfacearea of a cube).
Prerequisite Skills
The necessary prerequisite skills required to determine the surfacearea...

...have to be taken in and to also be removed. This is where the surfacearea to volume ratio comes into place; the reason why this ratio is so important is because the surfacearea of a cell essentially affects the rate of the transferring of useful substances (through diffusion and osmosis etc.) in and out of the organism. On the other hand, the total volume of the organism also affects the rate of the making of material inside the cell and the ability to hold all of the substances. Whilst organisms are slowly growing and developing day by day, the volume of the organism increases, but not to the same extent as the surfacearea; this is because the organisms’ surfacearea increases at a much slower rate than its volume. Through research and experiments, it is apparent that as the organism grows, its surfacearea to volume ratio slowly decreases, the table to the right also proves that this theory is true; meaning that it would become increasingly difficult for the organism to obtain the required nutrients and also expelling the wastes produced by metabolism. In the end, it becomes impossible for diffusion to occur efficiently, where the cells becomes too large so they would divide through the process of mitosis. For example, the alveoli in our lungs have a relatively high large surfacearea to volume ratio,...