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 surface area 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 everywhere. Calculate the volume of the metal. (ix) The difference between outside and inside surface of a cylindrical metallic pipe 14cm long is 44cm2 . If the pipe is made of 99 cm3 . Find the outside and inner radii of the pipe.

Volume and surface area.
1. A hollow cylindrical pipe is 21 dm long. Its outer and inner diameters are 10cm and 6cm respectively. Find the volume of copper used in making the pipe. 2. The height of a right circular cylinder is 10.5m. Three times the sum of the areas of its two circular faces is twice the area of the curved surface. Find the volume of the...

...Mathematics
Volume of Solids
Formulae for Volume of Solids
Cube | Cuboid | Triangular Prism | Cylinder | Cone | Pyramid | Sphere | AnyPrism |
s3 | lwh | ½bhl | Πr2h | 1/3πr2h | 1/3Ah | 4/3πr3 | Ah |
A = area of the base of the figure
s = length of a side of the figure
l = length of the figure
w = width of the figure
h = height of the figure
π = 22/7 or 3.14
1. Compute the volume of a cube with side 7cm.Volume of cube: s3
s = 7cm
s3 = (7cm x 7cm x 7cm) = 343cm3
2. Compute the volume of a cuboid (also known as a rectangular prism) with the dimensions 4cm by 13cm by 9cm.
Volume of a cuboid: lwh
l = 4cm
w = 13cm
h = 9cm
lwh = (4cm x 13cm x 9cm) = 468cm3
3. Compute the volume of a triangular prism with a base length of 60cm, a base width of 8cm, and a height of 10cm.
Volume of a triangular prism: ½bhl
½b = (8cm x 1/2) = 4cm
h = 10cm
l = 60cm
½bhl = (4cm x 10cm x 60cm) = 2400cm3
4. Compute the volume of a cylinder which is 2m tall and has a radius 75cm. Convert this litres.
Volume of a cylinder: πr2h
π = 3.1415
r2 = (75cm)2 = 375 cm2
h = 2m = 200cm
πr2h = 235612. 5 cm3
cm L = 1cm 0.001 L
235612.5 cm3/ 1000 = 235.6125 L
5. Compute the volume of a cone with a radius of 200cm and a height of 0.75m.
Volume of a cone: 1/3πr2h
1/3π = 1.047
r2 =...

...Mathematics Class X (Term–II)
13
SURFACEAREAS AND VOLUMES
A. SUMMATIVE ASSESSMENT
(c) Length of diagonal =
TH
G
(a) Lateral surfacearea = 4l2 (b) Total surfacearea = 6l2 (c) Length of diagonal = 3 l 3. Cylinder : For a cylinder of radius r and height h, we have : (a) Area of curved surface = 2πrh
BR
O
ER
2. Cube : For a cube of edge l, we have :
O
YA L
TEXTBOOK’S EXERCISE 13.1 Unless stated otherwise, take =
22 . 7
Q.1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surfacearea of the resulting cuboid. [2011 (T-II)]
1
S
l 2 b2 h2
5. Sphere : For a sphere of radius r, we have : Surfacearea = 4πr2 6. Hemisphere (solid) : For a hemisphere of radius r we have : (a) Curved surfacearea = 2πr2 (b) Total surfacearea = 3πr2
PR
(a) Lateral surfacearea = 2h(l + b) (b) Total surfacearea = 2(lb + bh + lh)
(d) Total surfacearea of hollow cylinder = 2πh(R + r) + 2π(R2 – r2) 4. Cone : For a cone of height h, radius r and slant height l, we have : (a) Curved surfacearea = πrl = r h 2 r 2 (b) Total surface...

...Introduction
Cells have to interact with their environment, chemicals and water and in order to do so they must be able to move across the cell membrane and the cell. The movements within a cell are called Diffusion. When molecules move across a cell membrane it is known as Osmosis. Diffusion is the process by which molecules of a substance move from areas of higher concentration of that substance to areas of lower concentration. Diffusion can be the transfer of anything anywhere. However, that is not true for osmosis. Osmosis is diffusion, but a specific type of diffusion. Osmosis is only the diffusion of water molecules through a selectively permeable membrane and a concentration gradient. Only certain molecules can cross the membrane to either go in our out of the cell.
If outside the cell there is a lower concentration of molecules than inside the cell, it is called hypotonic and water will move from outside the cell into the cell. If the solution outside the cell has a larger concentration of dissolved materials the solution is hypertonic, so the water would move from the cell into the solution.
Aim
The aim of this experiment was to verify the concept of Osmosis and Diffusion with a semi-permeable membrane (dialysis tubing), it will be exposed to different environments and concentration gradients.
Hypothesis
I thought that that dialysis tubing would end up weighing more as there would be less water and more molecules...

...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...

...corresponding base. If the altitude were increased by
4cm and the base decreased by 2cm, the area of the triangle would remain the same. Find the base and altitude of the triangle.
2. Some toffees are bought at the rate of 11 for Rs10 and same numbers are bought at the
rate of 9 for Rs 10. If the whole lot is sold at one rupee per toffee, find the gain or loss percent.
3. Chandu purchased a watch at 20% discount on its marked price but sold it at marked price.
Find the gain percent of Chandu on this transaction.
4. A motorboat covers a certain distance downstream in a river in five hours. It covers the same distance upstream in five hours and half. The speed of water is 1.5 km/hr. Find the speed of the boat in still water.
5. Factorize:
(i) 2x2+y2+8z3-22xy-42yz+8xz
(ii) x6 – 3x4y2 +3x2y4 –y6
6. Evaluate: (367/2 –369/2)/ 365/2
7. Divide 34x-22x3-12x4-10x2-75 by (3x+7) and check your answer.
8. The digit in the tens place of a number is three times that in the ones place. If the digits are reversed, the new number will be 36 less than the original number. Find the number.
9. A well is dug 20m deep and it has a diameter 7m. The earth, which is so dug out, is spread out on a rectangular plot 22m long and 14m broad. What is the height of the platform so formed?
10. The total surfacearea of a hollow cylinder open at both ends is 4620sqcm, area of the base ring is...

...SurfaceArea Formulas
In general, the surfacearea 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
|SurfaceArea of a Cube = 6 a 2 |
[pic](a is the length of the side of each edge of the cube)
In words, the surfacearea 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 surfacearea of a cube is 6 times one of the sides squared.
|SurfaceArea of a Rectangular Prism = 2ab + 2bc + 2ac |
[pic](a, b, and c are the lengths of the 3 sides)
In words, the surfacearea 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...

...SurfaceareaSurfacearea is the measure of how much exposed area a solid object has, expressed in square units. Mathematical description of the surfacearea is considerably more involved than the definition of arc length of a curve. For polyhedra (objects with flat polygonal faces) the surfacearea is the sum of the areas of its faces. Smoothsurfaces, such as a sphere, are assigned surfacearea using their representation as parametric surfaces. This definition of the surfacearea is based on methods of infinitesimal calculus and involves partial derivatives and double integration.
General definition of surfacearea was sought by Henri Lebesgue and Hermann Minkowski at the turn of the twentieth century. Their work led to the development of geometric measure theory which studies various notions of surfacearea for irregular objects of any dimension. An important example is the Minkowski content of a surface.
Definition of surfacearea
While areas of many simple surfaces have been known since antiquity, a rigorous mathematical definition of area requires a lot of care. Surfacearea...

...Background information:
1) What is diffusion?
Movement of a particular type of molecule from an area of high concentration to an area of low concentration.
2) How is diffusion used by living cells?
Living cells bring in food, water and oxygen, and excrete wastes through the process of diffusion
3) List two body systems in vertebrates that are dependent on diffusion
Digestive system and respirational system
4) What is meant by the term metabolism
the chemical processes that occur within a living organism in order to maintain life.
5) Why cant a single cell grow very large?
If it grows too large it will not be able to take sufficient food and oxygen or eliminate waste molecules fast enough.
6) Why must large organisms be multicellular?
To overcome the problem of small cell size. Metabolism is the total of all the chemical processes that take place in the body. These chemical processes convert the food you eat into the energy and materials needed for all life processes.
Inputs and Outputs
Anabolism and Catabolism
• Metabolic processes can be regarded as either anabolic or catabolic.
Catabolism
• Catabolic reactions are destructive metabolic processes during which complex substances are broken down into simpler ones.
• Catabolic processes release energy.
• Cellular respiration is a good example of a catabolic process. Below is a summary of the cellular respiration process.
glucose + oxygen —> water + carbon dioxide...