## Friday, November 30, 2012

### Proving Two Parallel Lines - Circles

Category: Plane Geometry

"Published in Newark, California, USA"

If two circles are tangent externally and a line is drawn through a point of contact and terminated by the circles. Prove that the radii drawn to its extremities are parallel. Photo by Math Principles in Everyday Life
Solution:

Consider the given figure Photo by Math Principles in Everyday Life

Proof:

1. Statement: ∠1 ≅ ∠2

Reason: Vertical angles are congruent.

2. Statement: OP OA and O'P O'B

Reason: All points in a circle are equidistant from its center.

3. Statement: AB is drawn through point P.

Reason: Given item.

4. Statement: ΔOAP and ΔO'PB are isosceles triangles.

Reason: An inscribed triangle in a circle which consist of a center of a circle and the two end points of a chord is always an isosceles triangle.

5. Statement: ∠1 ≅ ∠3 and ∠2 ≅ ∠4

Reason: The two opposite angles of an isosceles triangle are congruent.

6. Statement: ∠1 ≅ ∠2 ≅ ∠3 ≅ ∠4

Reason: Transitive property of congruence.

7. Statement: ∠AOP = 180º - (∠1 + ∠3)
∠PO'B = 180º - (∠2 + ∠4)

Reason: The sum of the interior angles of a triangle is 180º.

8. Statement: ∠AOP ≅ ∠PO'B

Reason: By computation at #7, if ∠1 ≅ ∠2 ≅ ∠3 ≅ ∠4, then ∠AOP ≅ ∠PO'B.

9. Statement: OA ║ O'B

Reason: If a transveral line (OO') passed the two alternating interior angles (∠AOP and ∠PO'B) that are congruent, then it follows that the two lines (OA and O'B) which are adjacent to the alternating interior angles are parallel.

## Thursday, November 29, 2012

### Money - Investment Problem

Category: Algebra

"Published in Newark, California, USA"

Mr. Manalang has 3 children who are entering college. He was offered to pay 75% of the tuition fee for the second and 50% for the third if he enrolls his 3 children in the same school. If the total tuition fee amounts to ₱ 4,500.00, how much is the regular tuition fee?

Solution:

The above problem statement involves the money problem. Let's analyze the whole statements carefully as follows.

Let x be the amount of tuition fee for Mr. Manalang's 1st child.

Let 0.75 x be the amount of tuition fee for Mr. Manalang's 2nd child.

Let 0.50 x be the amount of tuition fee for Mr. Manalang's 3rd child.

If you add all the tuition fees of Mr. Manalang's children, the total amount is ₱ 4,500.00. The working equation for this problem will be

x + 0.75 x + 0.50 x = 4,500

2.25 x = 4,500

x = 2,000

Therefore, the regular tuition fee is ₱ 2,000.00. This is also the tuition fee for Mr. Manalang's 1st child.

Note: The monetary sign, ₱, is Philippine Pesos. The given word problem was in 1964.

## Wednesday, November 28, 2012

Category: Algebra

"Published in Newark, California, USA"

Given the quadratic equation in general form:

Prove that the Quadratic Formula has a formula of

Solution:

There are three ways in solving the quadratic equation. First,  you can solve the quadratic equation by factoring. Second, if you cannot factor the quadratic equation, you can solve it by completing the square. Third, you can solve the quadratic equation by using the Quadratic Formula. Right now, we will derive the formula for Quadratic Formula. Let's consider the quadratic equation in general form

Divide both sides of the equation by the coefficient of x2 which is a.

Transpose the third term to the right side of the equation

Apply the completing the square method to the above equation

Get the LCD at the right side of the equation

Take the square root on both sides of the equation

Therefore,

## Tuesday, November 27, 2012

### Homogeneous Functions - Arbitrary Constant

Category: Differential Equations, Integral Calculus

"Published in Newark, California, USA"

Find the particular solution for

when

Solution:

If you examine the given equation, it is a differential equation because it has dy and dx in the equation. The type of equation is Homogeneous because the functions and variables cannot be separated by Separation of Variables. There's a method to solve the Homogeneous Functions. Consider the given equation

Let y = vx

dy = vdx + xdv

Substitute y and dy to the above equation, we have

The above equation can now be separated by Separation of Variables. Arrange the equation according to their variables

Integrate both sides of the equation

Take the inverse natural logarithm on both sides of the equation

but y = vx and v = y/x

To solve for C, substitute the following:

to the above equation, we have

Therefore,

## Monday, November 26, 2012

### Dividing Rational Fractions

Category: Algebra

"Published in Suisun City, California, USA"

Perform the indicated operations and simplify

Solution:

This is a division of a rational fraction with another rational fraction. As a rule in Mathematics that we need to get the reciprocal of the divisor first and then perform the multiplication as follows

Factor all the polynomials in the numerator and denominator. Do this by trial and error so that the middle term of a polynomial is matched.

Simplify the above equation

## Sunday, November 25, 2012

### Solving Parallelogram Equation

Category: Analytic Geometry

"Published in Suisun City, California, USA"

Given a parallelogram with vertices A(- 2, - 1), B(4, 2), C(7, 7) and D(1, 4).

a. Find the equation of a parallelogram as a function of x.

b. Prove that the given parallelogram is real a parallelogram.

Solution:

The first that we have to do is to plot the vertices of a parallelogram and draw the figure as well. Photo by Math Principles in Everyday Life

a. To get the equation of a parallelogram as a function of x, we need to get the equations of the sides of the parallelogram using the two point form.

For AB, substitute the values of points A and B

For BC, substitute the values of points B and C

For CD, substitute the values of points C and D

For DA, substitute the values of points D and A

Therefore, the equation of a parallelogram as a function of x is

b. Since the slopes of the opposite sides of a parallelogram are equal, then the given parallelogram is real a parallelogram.

mAB = mCD = ½

mBC = mDA = 5/3