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Archive for the ‘Formulas’ Category

Friendly Formula: the Distance Formula


A few days ago I posted a “Friendly Formula” for the Midpoint Formula.

Today I am presenting a Friendly Formula for the Distance Formula, an important formula in Algebra 1 courses.

Friendly Formulas make algebra
less intimidating!


First I’m going to present the Friendly Formula for the Distance Formula and demonstrate how to use it. Then I’ll explain why it makes sense.

Buckle your seatbelts ’cause here it is: the distance between any two points on the coordinate plane is simply the SQUARE ROOT of …
(the x-distance squared) plus (the y-distance squared).

And here’s an example of how easy it can be to use this formula.
Suppose you want the distance between the points (2, 5) and (4, 9).

First figure out how the distance between the x-coordinates, 2 and 4.
Well, 4 – 2 = 2, so the x-distance = 2.
Now square that x-distance: 2 squared = 4

Next find the distance between the y-coordinates, 5 and 9:
Well, 9 – 5 = 4, so the y-distance = 4.
Now square that y-distance: 4 squared = 16

Next add the two squared values you just got: 4 + 16 = 20

Finally take the square root of that sum: square root of 20 = root 20.

That final value, root 20, is the distance between the two points.

Now we get to the question of WHY this Friendly Formula makes sense. I will explain that in my next post.

HINT: The Distance Formula is based on the Pythagorean Theorem. See if you can spot the connection.

EXTRA HINT: Make a coordinate plane. Plot the two points I used in this example, and construct a right triangle in which the line connecting these two points is the hypotenuse. If you can figure this out, the “Aha!” moment is a glorious event!

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Category:

"Friendly Formulas", Algebra, Coordinate Plane, distance, Formulas, Formulas, Uncategorized

Invisible Misunderstandings: Square roots of 2 and 3


Would you say that the square root of two is an important number in math? Hmmm … and would you agree that the square root of three, while perhaps not quite so important, is still a quantity whose value students should be able to estimate?

Why not, right? After all, these numbers play key roles in the 30-60-90 and 45-45-90 “special triangles.” And therefore they both appear a lot in geometry, and a great deal in trig. And on top of that, root two, widely believed to be the first irrational number discovered, shows up in a wide range of other math contexts as well.

First letter of a text about the square root o...

square root of 2 w/ "parent" triangle

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Category:

Algebra, Area, Formulas, Geometry, Logic of Geometry, Math Facts, Math in General, Math Instruction Techniques, Mental Math, Proofs, Proportions, Radicals, Tutoring, Uncategorized

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Who Invented the Coordinate Plane?


A fly …

Who would think that a mere fly could play a major role in the history of human thought?

But when it comes to the development of Algebra, that’s the story. I’ll explain how this works just a bit later in this blog. But it is all related to what is happening now in algebra classes all around the world.

For it’s spring, that time of year again when we get out the graph paper and the ruler. Kids are working on the Cartesian coordinate plane.

One about I like about the coordinate plane is that there’s an interesting story about how it was discovered, or should I say, invented. [Hard to know the right word for an intellectual Invention like the coordinate plane.]
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Category:

Algebra, Coordinate Plane, Formulas, Geometry, Great Mathematicians, Logic of Geometry, Making Math Fun, Math in General, Math Instruction Techniques, Math stories, Uncategorized, Using STORIES, Using stories in math

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Answer, Fun Math Problem


Answer to problem about the circular and square pegs and holes.

The “fit” for each situation is the following ratio:
(Area of Inner Figure) ÷ (Area of Outer Figure)

For the square peg in a round hole —
Call the radius of the circle r.
Then the diagonal of square “peg” = 2r
Notice that by slicing the square along its diagonal,
we get a 45-45-90 triangle, with the diagonal being
the hypotenuse and the sides being the two equal legs.
Using the proportions in a 45-45-90 triangle,
side of square peg = r times the square root of 2
Multiplying this side of the square by itself gives
us the area of the square, which comes out as:
2 times the radius squared

This being the case,
Area of square is: 2 times radius squared, and
Area of circle is: Pi times radius squared, and so …

Cancelling the value of the radius squared, we get:
Ratio of (Area of square) to (Area of circle) is:
2÷Pi = 0.6366

For the round peg in a square hole —
Call radius of the circle r.
And since the diameter of the circle is the same length as
the side of the square, the side of the square = 2r
Multiplying the side of the square by itself to get the
area of the square, we find that the area of the square
is given by: 4 times radius squared.

This being the case,
Area of circle is: Pi times radius squared
Area of square is: 4 times radius squared, and so …

Ratio of (Area of circle) to (Area of square) is therefore:
Pi ÷ 4 = 0.7854

Of the two ratios, the ratio of the circular peg in a square hole
is greater than that of the square peg in a circular hole.

Therefore we can say that the circular peg in a square hole
provides a better fit than a square peg in a circular hole.

And that is the answer!

Category:

Area, Challenge Problem, Formulas, Formulas, FUN PROBLEM, Geometry, Making Math Fun, Math in General, Practice Problems, Uncategorized

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FUN MATH PROBLEM — Circling the Square & Vice-Versa


From time to time I will post interesting math problems.

Feel free to try these problems. Share them with friends and colleagues. Use them however you see fit!

I will post the answer to the problems two days later, after people have had time to respond.

To post your response, simply send an email to me @ info@SingingTurtle.com
and make your Subject: Fun Problem.

The problem: Which provides the fuller fit? Putting a circular peg in a square hole, or putting a square peg in a circular hole? To get credit, show all work, and justify your answer by expressing each “fit” as a percent.

A few term-clarifications, to help you do this correctly:

a) By “fit,” I mean the ratio of the smaller shape to the larger shape, expressed as a percent. For
example, if a ratio is 4 to 5, that would represent a “fit” of 80 percent.

b) For the circular peg in the square hole, assume that the diameter of the circle equals the side of the
square. For the square peg in a circular hole, assume that the diameter of the circle equals the diagonal of the square.

c) By “fuller fit,” I mean the larger of the two ratios.

Have fun!

Category:

Area, Formulas, Formulas, FUN PROBLEM, Geometry, Making Math Fun, Math in General, Math Instruction Techniques, Practice Problems, Probability, Proofs, Proportions, Puzzles, Ratios, Uncategorized

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Answer to Problem of the Week – 10/18/2010


Problem of the Week – Answer

Here again is the problem:

Katja and Anthony are on a sightseeing trip in the western United States. Beginning where they land in Santa Fe, NM, they drive 80 miles east to see the historic wild west town of Las Vegas, NM. Then they travel 50 miles north to visit the Kiowa National Grasslands . Next they drive 140 miles west to visit Chaco Canyon National Historical Park. Finally they journey 130 miles south to visit El Malpais National Monument. When they reach El Malpais, how many miles are they from their starting point in Santa Fe?

And here is the answer, sent in by Eric Trujillo, a computer engineer based in Salem, OR.

“After taking trip, the pair are 100 miles away from their SF starting point. I calculated by drawing a diagram and then using the Pythagorean theorem. Sides triangle were 60 and 80 miles, so hypotenuse = 100 miles, using 3-4-5 right triangle relationship.”

Thank you for that reply, Eric.

Category:

Algebra, Coordinate Plane, Formulas, Pythagorean Theorem

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Problem of the Week – 10/18/2010


[Note:  I really am not getting a check from the New Mexico Tourism Department for this post, though I wouldn’t mind if they sent me one!]

Katja and Anthony are on a sightseeing trip in the western United States. Beginning where they land in Santa Fe, NM, they drive 80 miles east to see the historic wild west town of Las Vegas, NM. Then they travel 50 miles north to visit the Kiowa National Grasslands . Next they drive 140 miles west to visit Chaco Canyon National Historical Park. Finally they journey 130 miles south to visit El Malpais National Monument. When they reach El Malpais, how many miles are they from their starting point in Santa Fe?

Please explain how you found your answer, and send answers either as comments to this post, or as emails w/ subject POTW, sent to josh@SingingTurtle.com    I will not post your comments unless and until I determine that it is correct. And then, only on the day when I send out the answer on my blog.

 

Chaco's smaller kivas numbered around 100, eac...

Kivas in Chaco Canyon, New Mexico

 

Category:

Algebra, Brain Teasers, Challenge Problem, Coordinate Plane, Formulas, Geometry, Making Math Fun, Problem of the Week

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ANSWER TO FRIDAY’S PROBLEM:


ANSWER TO FRIDAY’S PROBLEM:

The problem, once again:

For any polygon, a “diagonal” is defined as a line segment that runs from one vertex  to another, running  through the polygon’s interior. Find a formula that determines the number of diagonals in any convex polygon with n sides. Once you have the formula, use it to figure out the number of diagonals in a convex polygon with 1,000 sides (don’t try this by hand! — that’s why algebra was invented).

The winning answer was provided by Chris Mark. The formula, for a convex polygon with n sides, is this:  Number of Diagonals =  [n(n– 3)]/2. For n = 1000, the number of diagonals = 498,500.

The reasoning behind the formula. A polygon has as many vertices as sides. So a polygon with n sides also has n vertices. Now, consider any vertex of the n-gon. From that vertex the number of diagonals that can be drawn is (n – 3). That is because we cannot draw a diagonal to 3 vertices:  the vertex chosen, and the two adjacent vertices. So the expression (n – 3) = the number of diagonals that can be drawn from any vertex. We multiply (n – 3) by n to obtain the total number of diagonals that can be drawn from all n vertices. But if we simply multiply n by (n – 3), we’d be counting each diagonal twice. To eliminate that problem, we divide the product, [ n(n – 3)], by 2, and that provides the correct formula:

Diagonals  =  [n(n – 3)]/2

Applying this formula to a convex polygon with 1,000 sides, we see that the number of diagonals =  498,500.

In addition to providing the answer, Chris pointed out that the problem need not be restricted to regular polygons, as it was when posted. This formula works for all convex polygons, regular or not.

Thanks, Chris. And thanks to everyone who submitted answers.

Category:

Algebra, Formulas, Formulas, Geometry, Logic of Geometry, Problem of the Month, Proofs

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