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Math Notebook of Ian Beardsley
Formulas Derived from the Parallelagram
Remarks. Squares and rectangles are parallagrams that have four sides the same length, or two sides the same length. We
can determine area by measuring it either in unit triangles or unit squares. Both are fine because they both are equal sided,
equal angled geometries that tessellate. With unit triangles, the areas of the regular polygons that tessellate have whole
number areas. Unit squares are usually chosen to measure area.
Having chosen the unit square with which to measure area, we notice that the area of a rectangle is base times height
because the rows determine the amount of columns and the columns determine the amount of rows. Thus for a rectangle we have:
A=bh
Drawing in the diagonal of a rectangle we create two right triangles, that by symmetry are congruent. Each right triangle
therefore occupies half the area, and from the above formula we conclude that the area of a right triangle is one half base
times height:
A=(1/2)bh
By drawing in the altitude of a triangle, we make two right triangles and applying the above formula we find that it holds
for all triangles in general.
We draw a regular hexagon, or any regular polygon, and draw in all of its radii, thus breaking it up into congruent triangles.
We draw in the apothem of each triangle, and using our formula for the area of triangles we find that its area is one half
apothem times perimeter, where the perimeter is the sum of its sides:
A=(1/2)ap
A circle is a regular polygon with an infinite amount of infitesimal sides. If the sides of a regular polygon are increased
indefinitely, the apothem becomes the radius of a circle, and the perimeter becomes the circumference of a circle. Replace
a with r, the radius, and p with c, the circumference, and we have the formula for the area of a circle:
A=(1/2)rc
We define the ratio of the circumference of a circle to its diameter as pi. That is pi=C/D. Since the diameter is twice
the radius, pi=C/2r. Therefore C=2(pi)r and the equation for the area of a circle becomes:
A=(pi)r^2
Math Notebook of Ian Beardsley
(More derived from the parallelogram)
Divide rectangles into four quadrants, and show that
A. (x+a)(x+b)=(x^2)+(a+b)x+ab
B. (x+a)(x+a)=(x^2)+2ax+(a^2)
A. Gives us a way to factor quadratic expressions.
B. Gives us a way to solve quadratic equations. (Notice that the last term is the square of one half the middle coefficient.)
Remember that a square is a special case of a rectangle.
There are four interesting squares to complete.
1) The area of a rectangle is 100. The length is equal to to 5 more than the width multiplied by 3. Calculate the width
and the length.
2) Solve the general expression for a quadratic equation, a(x^2)+bx+c=0
3) Find the golden ratio, a/b, such that a/b=b/c and a=b+c.
4) The position of a particle is given by x=vt+(1/2)at^2. Find t.
Show that for a right triangle (a^2)=(b^2)+(c^2) where a is the hypotenuse, b and c are legs. It can be done by inscribing
as square in a square such that four right triangles are made.
Use the Pythagorean theorem to show that the equation of a circle centered at the origin is given by r=x^2+y2 where r
is the radius of the circle and x and y the orthogonal coordinates.
Derive the equation of a straight line: y=mx+b by defining the slope of the line as the change in vertical distance per
change in horizontal distance.
Math Notebook of Ian Beardsley
Triangles
All polygons can be broken up into triangles. Because of that we can use triangles to determine the area of any polygon.
Theorems Branch 1
1. If in a right triangle a line is drawn parallel to the base, then the lines on both sides of the line are proportional.
2. From (1) we can prove that: If two triangles are mutually equiangular, they are similar.
3. From (2) we can prove that: If in a right triangle a perpendicular is drawn from the base to the right angle, then
the two triangles on either side of the perpendicular, are similar to one another and to the whole.
4. From (3) we can prove the Pythagorean theorem.
Theorems Branch 2
1. Draw two intersecting lines and show that opposite angles are equal.
2. Draw two parallel lines with one intersecting both. Use the fact that opposite angles are equal to show that opposite
interior angles are equal.
3. Inscribe a triangle in two parallel lines such that its base is part of one of the lines and the apex meets with the
other. Use the fact that opposite interior angles are equal to show that the sum of the angles in a triangle are two right
angles, or 180 degrees.
Theorems Branch 3
1. Any triangle can be solved given two sides and the included angle.
c^2=a^2+b^2-2abcos(C)
2. Given two angles and a side of a triangle, the other two sides can be found.
A/sin(a)=b/sin(B)=c/sin(C)
3.Given two sides and the included angle of a triangle you can find its area, K.
K=(1/2)bc(sin(A))
4.Given three sides of a triangle, the area can be found by using the formulas in (1) and (3).
Question: what do parallelograms and triangles have in common?
Answer: They can both be used to add vectors.
Math Notebook of Ian Beardsley
Trigonometry
When a line bisects another so as to form two equal angles on either side, the angles are called right angles. It is customary
to divide a circle into 360 equal units called degrees, so that a right angle, one fourth of the way around a circle, is 90
degrees. The angle in radians is the intercepted arc of the circle, divided by its radius, from which we see that in the
unit circle 360 degrees is 2(pi)radians, and we can relate degrees to radians as follows:
Degrees/180 degrees=Radians/pi radians
An angle is merely the measure of separation between two lines that meet at a point.
The trigonometric functions are defined as follows:
…cos x=side adjacent/hypotenuse
…sin x=side opposite/hypotenuse
…tan x=side opposite/side adjacent
…csc x=1/sin x
…sec x=1/cos x
…cot x=1/tan x
We consider the square and the triangle, and find with them we can determine the trigonometric function of some important
angles.
Square (draw in the diagonal): cos 45 degrees =1/sqrt(2)=sqrt(2)/2
Equilateral triangle(draw in the altitude): cos 30 degrees=sqrt(3)/2; cos 60 degrees=1/2
Using the above formula for converting degrees to radians and vice versa:
30 degrees=(pi)/6 radians; 60 degrees=(pi)/3 radians.
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