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G01-PythThm.mws

High School Modules > Geometry by Gregory A. Moore

The Theorem of Pythagoras

An exploration of this famous and fundamental theorem, and its applications.

[Directions : Execute the Code Resource section first. Although there will be no output immediately, these definitions are used later in this worksheet.]

0. Code

> restart; with(plots):

```Warning, the name changecoords has been redefined
```

> libname:="C:\\Program Files\\Maple 7\\LIB\\Geom\\", libname:

>

1. The Theorem

We all know what a right triangle looks like, right?

> RtTriPlot( [0,8],[6,0],[0,0]);

Or this :

> RtTriPlot( [0,5],[12,0],[0,0]);

The theorem is this : (trumpets please....)

> a^2 + b^2 = c^2;

Although it sometimes takes on this disguise :

> c = sqrt( a^2 + b^2);

2. The Proof

There are over 100 proofs that have been devised over the past two millennia for this fundamental theorem. We will create a famous diagram which can be an aid in proving the validity of the theorem.

> a := 5; b := 7;

> display( polygonplot( [[0,0],[b,0],[0,a]],color = red ),
polygonplot( [[b,0],[a+b,0],[a+b,b]], color = green ),
polygonplot( [[a+b,b],[a+b,a+b],[a,a+b]], color = blue ),
polygonplot( [[0,a],[a,a+b],[0,a+b]], color = violet ),
polygonplot({[[b,0],[a+b,b],[a,a+b],[0,a]]},color = yellow), scaling= constrained);

How does this prove the theorem? Well we compute the area for this diagram in two different ways.

1. Area of Pieces

Each of the four triangles (green, red, blue, purple) has dimensions a, b, and c.

> a:= 'a': b:= 'b':
`Area of each triangle` = (1/2)*a*b;

> `Area of all four triangles` = 4*(1/2)*a*b;

> `Area of Yellow Square in the middle` = c^2;

> `Total Area of Diagram` = %% + %;

2. Total Area

We can also compute the area in a different way. The entire diagram is actually a square. The length of the side of the big square is a + b, since each side of the square consists one small side of the triangle and one larger leg of the triangle.

> `Total Area of Diagram` = (a+b)^2;

Equating the Two Types of Area

Now we equate the two different computations for the area, and solve.

> 2*a*b + c^2 = (a+b)^2;

> expand(%);

> lhs(%)-rhs(%) = 0;

> c^2 = solve(%, c^2);

Voila!...oops....I mean, "QED" (quo est demonstratum...thus it is demonstrated).

3. Finding the Sides of Right Triangles

Given any two sides of a right triangle its possible to find the third side using this formula. The one key issue about using the formula is to substitute into it properly. Just remember that c, the hypotenuse, is the longest side of the triangle, and its square is equal to the sum of the squares of the other two.

> Pyth := c^2 = a^2+b^2;

Example 3.1: A right triangle has legs of side 5 and 12, find the hypotenuse

In this case, we are given the two legs - the smaller two sides, a and b.

> subs( {a = 5, b = 12}, Pyth);

> c = 'sqrt(169)';
expand(%);

> RtTriPlot2(5,12);

That worked nicely to an integer, but usually, there is a square root involved in one way or another.

Example 3.2: A right triangle has legs of side 2 and 3, find the hypotenuse

In this case, we are given the two legs - the smaller two sides, a and b.

> subs( {a = 2, b = 3}, Pyth);

> c = sqrt(13);

> RtTriPlot2(2,3);

Here is the other variation - given the hypotenuse and one leg.

Example 3.3: A right triangle has leg 10 inches long and hypotenuse 15 inches, find the other side.

In this case, we are given the two legs - the smaller two sides, a and b.

> subs( {a = 10, c = 15}, Pyth);

> b^2 = solve(%, b^2);

> b = 'sqrt(125)';
expand(%);

> RtTriPlot2( 10, 5*sqrt(5) );

4. Well Known Right Triangles

There are two very common families of right triangles, which are well worth remembering.

45-45-90 Triangle

When the two legs are equal, the two base angles are equal also - 45 degrees each.

> RtTriPlot2( 1, 1);

> subs( {a = x, b = x}, Pyth);

> c = sqrt( 2*x^2);
c = x*sqrt(2);

> RtTriPlot2(7,7);

The hypotenuse is always
times the side of the either of the two legs on a 45-45-90 triangle.

The other common triangle is this.

30-60-90 Triangle

This kind of triangle can be considered to be half of an equilateral triangle, and so the shorter base is half the hypotenuse. Knowing this much allows us to express the third side (the longer leg) in terms of the other two also.

> RtTriPlot2(1,sqrt(3));

> RtTriPlot2( sqrt(3)/2, 1/2);

> subs( {a = x, c = 2*x}, Pyth);

> b^2 = solve(%, b^2);

This gives us this relationship among the sides :
1 : : 2

> a = x;
b = x*sqrt(3);
c = 2*x;

> Tri369 := {a = x, b = x*sqrt(3), c = 2*x };

Often we see this kind of triangle expressed in this form, where the hypotenuse is 1.
: : 1

> subs(x = 1/2, Tri369);

> RtTriPlot2(7,7*sqrt(3));

Please take a look at the "Pythagorean Triples" worksheet for some interesting related material.

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