Math problem, What Math Problem?

Sometimes we feel sorry for our kids, especially those that never had a chance to have a tutor. We have technology and we have a will. We can be read in Bangladesh, and we can be seen in Santa Fe, Argentina.

This monster called math can be insidious and a pain in the neck. We have said it before, Math is a subject similar to learning another language. Syntax and Semantics are laid down from the start and maybe that's when our first graders start thinking the worse from math.

Here are some examples with graphics, and how we can solve them.

The graphic at the right shows a compound shape. The problem is given just as shown on Graph 1.

There are different ways to approach a problem... like in life itself. On Graph 2, we have broken down the compound shape to two well known rectangular shapes. Like in a business of our own, we find the width and length of the sides of those rectangles.

A= A1 + A2

A=b.h Area= base x height

A1=22x12 A2=7X8

After simplification we can gather enough information from the graphs:

A= 22X12 +7X8 Where A is the total area

A=320 m

The area of that compound shape is 320 meters. We could've taken a different approach, but for now, this was the easiest one.

On this Graph 3, your teacher wants you to find the value of 'm.'

Weighing the geometrical implications and the numbers that you were already given, you can solve this 'teaser' by using ratios or proportions.

If you have 2/3= 6/9=24/36

Then, your ratios are keeping the equation at bay.

In the case shown in Graph 3

5/4=10/m

m=8

So, using proportions we found the value of IJ, REPRESENTED BY:

'm= 8'

Sounds like the biggest monster of all, right? We can make it easier by going to a regular kind of chat:

Suppose that they give you this humongous 'Pizza pie' shaped like in Graph 4.

If you are able to break them down in triangles shapes like in Graph 5, you will be given 180 bucks for each triangle that you can get out of Graph 4.

So, a trapezoid or Quadrilateral shaped as in Graph4, is composed of four angles on those four corners.

Let's call the missing angle : x

adding those four angles should make 180+180=360 degrees.

X + 116 + 113 + 34 = 360 degrees

Solving this simple equation will give us the missing angle:

X =360-34-113-116

X=360-263

X=97, THE MISSING ANGLE VALUE IS 97 DEGREES.

GIVEN

Arc AB=60° Arc CD=100°

angle 1 = .5((100 - 60))

angle 1 = 20^o

Not that you want to be a rocket scientist, but your smart ass teacher will come back to hunt you with this Theorem by 8th Grade. So take heed!

On Graph 6, we have ANGLE 1 formed by two secants that intersect on P, and happen to lay their telescopic 'legs' on C , D, A and B.

Given ARC AB and ARC CD

The theorem states that the measure of Angle1 is :

angle 1 = 0.5((arc CD) - (arc AB))