As we were approaching the problem of how to design a gutter which would carry the most water from the roof with a 12 inch wide piece of material, we came up with a thesis; the length of the roof can be any number according to the roof sections, the maximum volume of the gutter will then depend on the maximum area out of the shapes' side (flank) with respect to the 12in-wide-material piece. Thus the areas of four different shapes were found. By process of elimination, the optimal area for each shape was found by deriving the equations of area as well as basic geometry. The shapes used for this project were a semi-circle, triangle, square, and rectangle. For all four shapes we maximized the area rather than the volume because the length of the gutter system does not play an important factor.

For the semi-circle, it was easy to assume that the circumference was 24 inches, considering that half of the circumference needed to be equal to 12 inches, the width of the manufactured piece of material. The equation for circumference is C=2πr. By plugging in 24 for C, we were able to find the radius, r, to be 12/π. Since finding the optimum area is needed, we could plug in the radius into the equation A=πr², the area of a circle. By substitution, we found the area to be 144/π or 45.8 in². However, this is the area of the full circle, so half of this number is equal to the semi-circle's area. Thus, 22.9 in² is the maximum amount of water a semi-circle can hold.

If we assume the shape of the gutter to be an isosceles triangle, then the angle between the two isosceles sides results in the maximum value of the area of the triangle. It makes sense for both sides, AB and AC, to be of equal length, so each side is 6 in. The diagram of the triangle is drawn below. AB + AC = 12 (12in-wide-piece)

AB = AC (isosceles triangle)
AB = 12/2 = 6in, AC = 6in also.
The...

...Name: Ernest Ng
Class: 4G (23)
Date: 2-7-06
Mathematics ACE: Ellipse Areas
Before we embark on solving the problem, let us first explore the definition of ellipse.
[pic]
An ellipse is a curve that is the locus of all points in the plane the sum of whose distances [pic] and [pic] from two fixed points [pic] and [pic] (the foci) separated by a distance of [pic] is a given positive constant [pic]
[pic]
While [pic] is called the major axis, [pic] is the semi major axis, which is exactly half the distance across the ellipse. Similarly, the corresponding parameter [pic] is known as the semi minor axis.
Parallels drawn from the formula for the area of circle ([pic]) and formula for the area of an ellipse (A = [pic]ab)
Formula for the area of a circle:[pic] where [pic] is the area, and [pic] is the radius.
In the case of a circle, radius a represents the semi major axis while radius b represents the semi minor axis. One can thus find the area of the circle through the formula A = [pic]ab, where a is equal to b. Hence circle, in actual fact, is a unique case of ellipse.
Proving that the area of an ellipse is πab
Procedures to take (Theory)
1. We have to let an ellipse lie along the x-axis and find the equation of the ellipse curve.
2. Upon finding the equation of the ellipse curve, we have to change the subject of the...

...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...

... SECTION A (40 Marks)
Attempt all questions from this Section
Question 1.
a) What number must be subtracted from 2x3 – 5x2 + 5x so that the resulting polynomial has a factor 2x – 3 ? [3]
b) D, E, F are mid points of the sides BC, CA and AB respectively of a Δ ABC. Find the ratio of the areas of Δ DEF and Δ ABC. [3]
c) A man borrowed a sum of money and agrees to pay off by paying Rs 3150 at the end of the first year and Rs 4410 at the end of the second year. If the rate of compound interest is 5% per annum, find the sum borrowed. [4]
Question 2.
a) The y-axis is a line of symmetry for the figure ACBD where A, B have co-ordinates (3, 6), (– 3, 4) respectively. (i) Find the co-ordinates of C and D. (ii) Name the figure ACBD and find its area. [3]
b) PAQ is a tangent at A to the circumcircle of Δ ABC such that PAQ is parallel to BC, prove that ABC is an isosceles triangle. [3]
c) A rectangular piece of paper 30 cm long and 21 cm wide is taken. Find the area of the biggest circle that can be cut out from this paper. Also find the area of the paper left after cutting out the circle. [Take π = 22/7] [4]
Question 3.
a) Construct a 2 × 2 matrix whose elements aij are given by aij = i + j. [3]
b) The point P (– 4, – 5) on reflection in y-axis is mapped on P’. The point P’ on reflection...

...
(6.03)
61) What is the area of a square that has a side length of:
a) 9 units
b) 11 units
Area of a square: = (Side length)2
a) A = (9)2 b) A = (11)2
= 81 u2 = 121 u2
62) A rectangle has a length of 8 in and a width of 5 in. Find the area.
Area of a rectangle: = (L)(W)
A = (L)(W)
=(8 in)(5 in)
= 40 in2
63) Find the length of a rectangle if the width is 8 cm and the perimeter 40 cm.
Perimeter: = 2(L + W)
40 = 2(L + 8)
2 2
20 = L + 8
- 8 - 8
12 = L
L = 12 cm
64) The diagonals of a square measure 12 cm. Find the area of the square. (Show two methods)
[pic]
Method 1: Method 2:
Using Pythagorean Theorem Introducing Diagonal Formula
a2 + a2 = 122 Area = (d1)(d2)
2(a2) = 144 2
2 2 = (12)(12) 6
a2 = 72 2 1
Area = (a)(a) Area = 72 cm2
= a2
Area = 72 cm2
65) Prove the following: If the...

...Area of some rectilinear figures
Area of a Triangle : 1/2 x Base x Heights
Area of Equilateral Triangle : Sqrt(3)/4 x (Side)2
Area of a Triangle - Hero's formula :
Sqrt{s (s - a) (s - b) (s - c)} where 2s = a + b + c.
Area of Rectangle = Length x Breadth
Area of a Square = (Side)2
Area of Four Walls = 2 x height (Length + breadth)
Area of Rhombus = 1/2 x (Products of Diagonals)
Area of Trapezium = 1/2 x (Sum of Parallel Sides) x (Distance between Parallel lines)
Area of Parallelogram = Base x Altitude
Somestandards units of Area
100mm2 = 1 cm2, 100 cm2, = 1 dm2
100dm2 = 1 m2, 100 m2, = 1 are
100arcs = 1 hactare
100 hactare = 1 km2
Area related to Circles:
Circumference of a Circle or Perimeter of a Circle: The Distance arround the circle or the length of a circle is called its circumference or perimeter. |
| Circumference (Perimeter) of a circle = d or 2r, where, r is the radius of the circle and = 22/7 |
Circle: If r be the radius of a circle, and its center is O. |
| Area of a Circle = r2 |
| Area of a Semi-Circle = r2 |
| Area of a Quadrant = r2 |
Perimeter of a Semi Circle: If r be the radius of a circle, and its center is O. |
| Perimeter of a Semi Circle : r + 2r |
Area of a Ring: If r be the...

...Area 51
Area 51 is a nickname for a military base that is located in the southern portion of Nevada in the western United States (83 miles north-northwest of downtown Las Vegas). Situated at its center, on the southern shore of Groom Lake, is a large secretive military airfield. The base's primary purpose is to support development and testing of experimental aircraft and weapons systems.
Its secretive nature and undoubted connection to classified aircraft research, together with reports of unusual phenomena, have led Area 51 to become a focus of modern UFO and conspiracy theories. Some of the activities mentioned in such theories at Area 51 include:
- The storage, examination, and reverse engineering of crashed alien spacecraft (including material supposedly recovered at Roswell), the study of their occupants (living and dead), and the manufacture of aircraft based on alien technology.
- Meetings or joint undertakings with extraterrestrials.
- The development of exotic energy weapons (for SDI applications or otherwise) or means of weather control.
- The development of time travel and teleportation technology.
- The development of unusual and exotic propulsion systems related to the Aurora Program.
- Activities related to a supposed shadowy one world government or the Majestic 12 organization.
Theories
Misdirection
Some people believe that if everyone focuses on Area 51 the...

...Force vs. Area
Connor Blackmon
Chemistry I H, 1st Period
Mrs. Kris Clements
October 18, 2012
Problem
Will a balloon pop if it is places on a bed of nails and pressure is applied?
Hypothesis
If a balloon is placed on a bed of nails and a force is applied, then the balloon will burst.
Variables
Independent variable- Force applied to the balloon and number of nails
Dependent variable- Does the balloon burst?
Materials
14 inch by 14 in by .75 in plywood board x2
196 nails
4 rods (14 inches tall)
Ruler
Pen
Drill
10 latex and 10 rubber balloons
Weights (1 lb, 5 lbs, 10 lbs; multiple of each weight)
Procedures:
Assembling the Board:
Using a pen and ruler, every one inch make a mark on one of the boards, these marks should be parallel to each other
use a drill to place a nail at each one of the points made on the board, all nails will be used
On the four corners drill a hole for a 14 inch rod facing the same way as the nails
Using the drill again, make four holes in the corners of the other plywood board for the rods to slide through
Experiment Procedures:
Inflate the rubber balloons to 11 inches in diameter, all balloons should be plus or minus .2 of an inch in diameter
Place rubber balloon on the middle of the bed of nails
Slide plywood board trough the rods to sandwich to balloon
Record if the balloon pops or not and weight applied to balloon including the weight of the plywood...

...Scheduled Areas and Tribal Areas : Scheduled Tribes live in contiguous areas unlike other communities. It is, therefore, much simpler to have an area-approach for development activities and also regulatory provisions to protect their interests. In order to protect the interests of Scheduled Tribes with regard to land alienation and other social factors, provisions of ‘‘Fifth Schedule" and ‘‘Sixth Schedule"have been enshrined in the Constitution.
The Fifth Schedule under Article 244(1) of Constitution defines ‘‘Scheduled Areas" as such areas as the President may by Order declare to be Scheduled Areas after consultation with the Governor of the State. The Sixth Schedule under Article 244 (2) of the Constitution relates to those areas in the States of Assam, Mehalaya, Tripura and Mizoram which are declared as ‘‘Tribal Areas" and provides for District Councils and/or Regional Councils for such Areas. These Councils have been conferred with wide ranging legislative, judicial and executive powers. The Fifth Schedule Areas : The criteria for declaring any area as a ‘‘Scheduled Area" under the Fifth Schedule are : (a) Preponderance of tribal population, (b) Compactness and reasonable size of the area, (c) A viable administrative entity such as a district, block or taluk, and (d) Economic...