To better understand certain problems involving rockets
it is necessary to use some mathematical ideas from
the study of triangles.
Let us begin with some definitions and terminology which we will use on this slide.
We start with a right triangle. A right triangle is a
three sided figure with one angle equal to 90 degrees. A 90 degree angle is
called a right angle and that is where the right triangle gets its name.
We define the side of the triangle opposite from the right angle to
be the hypotenuse, h. It is the longest side of the three sides
of the right triangle. The word "hypotenuse" comes from two Greek words
meaning "to stretch", since this is the longest side.
We are going to label the other two sides a and b.
The Pythagorean Theorem is a statement relating the lengths
of the sides of any right triangle.
The theorem states that:
For any right triangle, the square of the hypotenuse
is equal to the sum of the squares of the other two sides.
Mathematically, this is written:
h^2 = a^2 + b^2
The theorem has been known in many cultures, by many names, for many years.
Pythagoras, for whom the theorem is named, lived in ancient Greece, 2500 years ago.
It is believed that he learned the theorem during his studies in Egypt. The
Egyptians probably knew of the relationship for a thousand years before
Pythagoras. The Egyptians knew of this relationship for a triangle with sides in the
ratio of "3 - 4 - 5".
5^2 = 3^2 + 4^2
25 = 9 + 16
Pythagoras generalized the result to any right triangle. There are many different
algebraic and geometric proofs of the theorem. Most of these begin with a
construction of squares on a sketch of a basic right triangle. On the figure at
the top of this page, we show squares drawn on the three sides of the triangle.
A square is the special case of a rectangle in which all the sides are equal
in length. The
areaA of a rectangle is the product of the sides.
So for a square with a side equal to a, the area is given by:
A = a * a = a^2
So the Pythagorean theorem states the area h^2 of the square drawn on the
hypotenuse is equal to the area a^2 of the square drawn on side a
plus the area b^2 of the square drawn on side b.
We begin with a right triangle on which we have constructed squares on the two sides, one
red and one blue. We are going to break up the pieces of these two squares and move them into
the grey square area on the hypotenuse. We won't lose any material during the
operation. So if we can exactly fill up the square on the hypotenuse,
we have shown that the areas are equal. You work through the construction by clicking on the
button labeled "Next". You can go "Back" and repeat a section, or go all the way back tothe beginning by clicking on "Reset".
What is it doing? The first step rotates the triangle down onto the blue square.
cuts the blue square into three pieces, two triangles and a red rectangle.
The two triangles are exactly the same size as the original triangle.
The "bottom" of the original triangle exactly fits
the vertical side of the square, because the sides of a square are equal.
The red rectangle has its vertical sides equal to the base of the original triangle,
and its horizontal sides equal to the difference between the "bottom" side and the
"vertical" side of the original triangle.
Using the terminology from the figure at the top of this page, the dimensions
of the red rectangle are:
vertical length = b
horizontal length = b - a
The next step is to move the red rectangle
over adjacent to the red square. The rectangle sticks out the top of the red square
and the two triangles remain in the blue square. The next step is to move one of the
blue triangles vertically into the hypotenuse square. It fits exactly along the side of
hypotenuse square because the sides of a square are equal. The next step is to move the
other blue triangle into the hypotenuse square. (We are half way there!) The next step
is to slide the form of the original triangle to the left into the red region. The triangle cuts
the red region into three pieces, two triangles and a small yellow square.
The original triangle fits exactly into this region because of two reasons;
the vertical sides are identical, and the horizontal side of the red region is equal to
the length of the red square plus the horizontal length of the red rectangle which we
moved. The horizontal length of the red region is:
horizontal length = a + (b - a) = b
The horizontal length of the red region is exactly the length of the horizontal side
of the original triangle.
The yellow square has dimensions b - a on each side.
The next step is to move one of the red triangles into the
hypotenuse square. Again it's a perfect fit.
The next step is to move the final red triangle into
the hypotenuse square. Now if we look at the grey square that remains in the
hypotenuse square, we see that its dimensions are b - a; the long side
of the triangle minus the short side. The final step is to move the yellow square into
this hole. It's a perfect fit and we have used all the material from the original red
and blue squares.