Basic Algebra
By
Ian Beardsley
Formulas Derived from the Parallelogram
Remarks. Squares and rectangles are parallelograms 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:
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:
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 = bh
A =
1
2
bh
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:
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:
(More derived from the parallelogram)
Divide rectangles into four quadrants, and show that
A.
B.
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.
A =
1
2
rc
A = π r
2
(x + a)(x + b) = x
2
+ (a + b)x + ab
(x + a)(x + a) = x
2
+ 2a x + a
2
There are four interesting squares to complete.
1) The area of a rectangle is 100. The length is equal to 5 more than the width multiplied
by 3. Calculate the width and the length.
2) Solve the general expression for a quadratic equation,
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 . Find t.
Show that for a right triangle where a is the hypotenuse, b and c are legs.
It can be done by inscribing a 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 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.
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 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.
a x
2
+ bx + c = 0
x = vt +
1
2
at
2
a
2
= b
2
+ c
2
r
2
= x
2
+ y
2
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 alternate 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 alternate 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.
2. Given two angles and a side of a triangle, the other two sides can be found.
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.
c
2
= a
2
+ b
2
− 2abcos(C )
a
sin(A)
=
b
sin(B)
=
c
sin(C )
