The Pythagorean Theorem: Extensions and Applications
Summary
TLDRThis video explores various applications of the Pythagorean theorem, starting with finding the distance between points on an XY grid, such as those used in computer graphics. It explains how to calculate this distance using the difference in coordinates and introduces the concept of delta (Δ) for differences. The video then extends the theorem to three dimensions, illustrating how to find distances across 3D objects using a similar approach. Finally, it demonstrates how to derive the equation of a circle on a plane, showing practical uses of the Pythagorean theorem in algebra and geometry.
Takeaways
- 📏 The video discusses extensions and applications of the Pythagorean theorem, focusing on finding distances between points on a graph.
- 📉 The distance formula can be applied to find the distance between two points on an XY grid by creating a right triangle and using the Pythagorean theorem.
- 🔢 To find the horizontal and vertical distances between points, subtract the respective X and Y coordinates and use these differences in the distance formula.
- 🔄 The formula for the distance between two points on a plane is derived by squaring the differences in X and Y values, adding them, and taking the square root.
- 🌐 The video extends the concept to three dimensions, explaining how to find the diagonal distance across a three-dimensional object using a similar approach.
- 🏠 In three dimensions, the distance formula incorporates the length, width, and height of the object, leading to a generalized Pythagorean theorem for 3D space.
- 🔍 The process of finding the diagonal in 3D involves first finding the distance across the floor (a 2D problem) and then using this distance in a new right triangle that includes the height.
- 📐 The video also explains how to find the equation of a circle on a plane, using the Pythagorean theorem to relate the radius and the coordinates of points on the circle.
- 🔵 The equation of a circle is derived from the relationship between the radius and the coordinates of any point on the circle, written as x^2 + y^2 = r^2.
- 🎨 Applications in graphics and geometry software are highlighted, showing how programs like GeoGebra use these equations to plot circles and other shapes on a coordinate plane.
Q & A
What is the primary application of the Pythagorean theorem discussed in the video?
-The primary application discussed is finding the distance between two points on an XY grid using the Pythagorean theorem.
How is the distance between two points on a graph calculated?
-The distance is calculated by finding the horizontal and vertical distances between the points, squaring them, adding the squares, and then taking the square root of the sum.
What are Delta X and Delta Y in the context of the distance formula?
-Delta X represents the difference in the X coordinates of the two points, and Delta Y represents the difference in the Y coordinates.
How does the video extend the Pythagorean theorem to three dimensions?
-The video extends the Pythagorean theorem to three dimensions by demonstrating how to calculate the diagonal distance across a three-dimensional object using the lengths, widths, and heights of the sides.
How is the diagonal distance across a three-dimensional space calculated?
-The diagonal distance is calculated by taking the square root of the sum of the squares of the length, width, and height.
What is the equation for a circle derived in the video?
-The equation for a circle is \( x^2 + y^2 = r^2 \), where r is the radius of the circle.
How is the concept of a right triangle used to find the equation of a circle?
-The concept is used by noting that the radius forms the hypotenuse of a right triangle, with the legs being the X and Y coordinates, thus leading to the equation \( r^2 = x^2 + y^2 \).
What is the significance of using coordinate numbers in calculating distances on a graph?
-Using coordinate numbers allows for precise distance calculations, which is crucial when dealing with non-integer or decimal coordinates where simply counting squares is not feasible.
How does the video suggest visualizing three-dimensional space for applying the extended Pythagorean theorem?
-The video suggests visualizing the room you're in as a box-like structure and imagining the diagonal distance between a top corner and an opposite bottom corner.
What practical applications of the distance formula are mentioned in the video?
-Practical applications mentioned include plotting graphics on a computer screen, determining distances in various fields like graphic design, and defining geometrical shapes like circles.
Outlines
📏 Applying the Pythagorean Theorem to Graph Distances
This paragraph introduces the use of the Pythagorean theorem to find distances between points on a two-dimensional graph, such as an XY grid found on television or computer screens. It explains the concept by visualizing the distance as the hypotenuse of a right triangle formed by the horizontal and vertical distances between two points. The process involves identifying the points by their coordinates, calculating the differences in the x (horizontal) and y (vertical) values, squaring these differences, summing them, and then taking the square root of the result to find the distance. The paragraph also introduces the notation Δx for the difference in x-coordinates and Δy for the difference in y-coordinates, emphasizing the general applicability of this method even when dealing with decimal numbers.
📐 Extending the Pythagorean Theorem to Three Dimensions
The second paragraph extends the concept of the Pythagorean theorem from two to three dimensions, illustrating how to find the distance across space within a room, from one corner where a wall meets the ceiling to the opposite corner on the floor. It likens this to finding the diagonal of a rectangle but in three dimensions, involving the length (L), width (W), and height (H) of the room. The process involves first calculating the distance across the floor (X) using the Pythagorean theorem for a two-dimensional rectangle (L^2 + W^2), and then using this result as one leg of a new right triangle to find the total diagonal distance (D) using the formula D = √(X^2 + H^2). This paragraph also provides an example calculation for a room with given dimensions, demonstrating the practical application of this three-dimensional extension of the Pythagorean theorem.
📘 Finding the Equation of a Circle Using Distance
The third paragraph discusses the application of distance concepts to derive the equation of a circle. It explains that any point on a circle is at a constant distance (the radius, R) from the center of the circle. By considering a right triangle formed by the radius and the x and y coordinates of a point on the circle, the relationship R^2 = x^2 + y^2 is established, which is the equation of the circle. The paragraph also demonstrates the use of GeoGebra software to visualize and manipulate the equation of a circle, showing how changing the radius value alters the size of the circle on the graph, and how the algebraic equation directly corresponds to the geometric representation.
🎨 Practical Applications of Distance Formulas in Graphics
The final paragraph ties together the concepts discussed in the previous sections, emphasizing their practical applications in various fields, particularly in graphic design and computer graphics. It highlights that understanding the mathematical principles behind these applications allows for better use of graphic manipulation tools and can be valuable for professionals in the field. The paragraph concludes by noting the importance of the Pythagorean theorem and its extensions in defining not only straight lines but also curves, such as circles, on graphs.
Mindmap
Keywords
💡Pythagorean Theorem
💡Distance Formula
💡Coordinates
💡Right Triangle
💡Delta (Δ)
💡Square Root
💡Hypotenuse
💡Three-Dimensional Extension
💡Equation of a Circle
💡Graph Plotting
Highlights
The video discusses extensions and applications of the Pythagorean theorem.
It explains how to find the distance between points on an XY grid, like on a television or computer screen.
The distance formula is derived by creating a right triangle with horizontal and vertical distances as legs.
The horizontal and vertical distances between points are calculated using coordinate differences.
The distance formula is represented as the square root of the sum of the squares of the differences in X and Y coordinates.
The concept of Delta X and Delta Y is introduced to represent the differences in X and Y coordinates.
A practical application of the distance formula is demonstrated with an example using counting squares on a graph.
Decimal coordinates are also considered for calculating distances, emphasizing the formula's utility beyond whole numbers.
The video extends the Pythagorean theorem to three dimensions, showing how to find the diagonal of a room.
A two-step process is introduced to calculate distances in three-dimensional space, involving the diagonal of a rectangle on the floor.
The three-dimensional distance formula is presented as the square root of the sum of the squares of length, width, and height.
An example calculation is shown to find the diagonal of a three-dimensional object using the extended Pythagorean theorem.
The video demonstrates finding the equation of a circle using the Pythagorean theorem, with the circle's radius as the hypotenuse.
Geogebra is used to visually represent the equation of a circle and its relationship to the Pythagorean theorem.
The equation x^2 + y^2 = R^2 is shown to describe any circle, where R is the radius.
The video highlights the importance of understanding algebraic concepts for practical applications in graphic design and art.
The video concludes by emphasizing the versatility of the Pythagorean theorem in various mathematical and real-world contexts.
Transcripts
okay this video is going to deal with
several extensions and applications of
the Pythagorean
theorem and one that I'd like to start
off with is how to find distance between
points on a graph on XY grid like this
one place by the way where XY grids
occur is on a television screen or a
computer screen
and you're trying to um uh plot graphics
and so forth if you have a program that
plots the graphs it identifies the
points by numbers how far over and how
far up from some coordinate system like
this and one of the things you might
want to know is how far is it between
two
points um there's a lot of applications
of this but let's just see in general
how would we find that
distance well first let's draw that
distance in and if you look at it a
little bit it looks like we have a
triangle
here if I go along the coordinate grid
looks like we have a horizontal and a
vertical line connecting this to make a
right triangle and the distance we're
looking for is the hypotenuse so if we
could find the horizontal distance
between these points and the vertical
distance between these points we could
take those numbers square them add them
together and take the square root and we
got the distance okay
well first let's identify the points by
their numbers notice this one is over
two and up three so I write two
three this one is over 1 2 3 4 5 6 7 and
up 1 2 3 4 five so this is
7 five all
right now we could just look at the
graph and say oh from here we have to
step over here 1 2 3 4 5 squares and up
two but uh what if these were decimal
numbers and so forth sometimes we don't
want to Simply count squares on a graph
we want to compute the distance based on
these coordinate numbers themselves and
so let's see what these numbers
mean first of all notice that this first
number the two is how far over
horizontally this point is so to get to
this point we go over two and up
three this point this number seven is
how
far horizontally we have to go to get to
this distance over here and so if I want
to find the horizontal distance between
the points it's like taking 7 minus 2
and that gives me
five okay so I take the difference so I
take 7 - 2 and that's it's going to give
me the horizontal distance notice the
same thing applies vertically the three
here is how far up I have to go uh from
the x- axis to get up to this
level and five is how far I have to go
up from the x-axis to get to this level
and so how far different are they how
far apart are they
vertically well I Take 5 - 3 and that
gives us 2 2 so I take the 7 - 2 and
then I take my 5 -
3 if I Square each of these and add them
together and take the square root that's
going to give me the distance between
the
points okay so the way we write this
there's a abbreviation we sometimes use
the difference in the X values we
sometimes write as simply Delta X that's
the Greek letter for Delta
and it has a D sound in Greek and so it
stands for the difference of the X
values this x value is 7 this one is two
so Delta X or the difference of the X's
is five take Delta X and square it then
I take Delta y this y value is five this
one is three the difference of the Y
values is 2 so I take my Delta Y and
square it add them together and take the
square root and this is the distance
formula for a two-dimensional
surface okay let's just go ahead and
finish it out if I do it from the
numbers or if I do it from Counting
squares either one 7 - 2 is 5 is
25 5 - 3 is 2^ 2ar is 4 add them
together I get 29 and so the distance
between the points is a square < TK of
29 at this stage we're just going to use
a
calculator okay and so if I just take 29
and hit the square root by the way for
the rest of this um the
exercises um in this uh unit uh we'll go
ahead and use a calculator um unless
we're specifically practicing the square
root technique from the last video okay
and so there you go four uh
5.38 or you might round it up to 5.39
okay that's the distance formula in two
Dimensions let's see how we can
generalize what we know about the
Pythagorean theorem to finding a
distance across space so a diagonal uh
across a three-dimensional object look
at look around yourself at the room
you're sitting in assuming you're in a
standard sort of box-like room and
imagine
wanting to know the distance from the
top corner where one of the walls meets
the ceiling and the bottom corner here
where the opposite wall or the opposite
uh vertical line like this meets the
floor and we want to take uh let's use a
different color here we want to find
this
distance okay so if you look around
physically in the room you're sitting in
and try to um imagine in your
circumstance what I'm trying to draw
here that'll make it more real to you so
if I have a two-dimensional situation I
could find the diagonal of a rectangle
and uh by finding um the hypotenuse of a
right triangle notice that if I split a
rectangle like this I have a right
triangle whose legs are simply L and W
the length and the width of the
rectangle okay so in this case the
distance
here is simply the square root of the
sum of the squares of the two legs l^2 +
w^2 if I had numbers I would Square them
add them together and take the square
root as we've been doing
before now what do we do for a
three-dimensional
situation well we have to build somehow
on what we already know we can't just
jump to conclusions here okay so let's
go back and see is this distance we're
looking for part of a right triangle
whose uh legs we can find all right and
notice that if I go this length along
think of that as a hypotenuse of a
triangle and think of the the height
along this edge here as one leg then the
distance across the floor would be the
other leg so look at this red triangle
here that's uh simply a right triangle
and that has two legs if I could find
both legs this one I know is H if I can
find this length I can then uh put them
together and find uh this distance D
that I'm looking for
here okay well we're going to have to
somehow figure out this distance across
the floor and if you look at it isn't
that a lot like finding the distance
across a rectangle because the floor
itself looks like a rectangle laid down
flat
and so if I think of this triangle let's
make that green and in fact um let's
just shade it in this triangle
here the legs of the triangle are L and
W just like they were here so the
distance across the floor let's call
that distance
X as this uh becomes a leg of the
vertical uh triangle so we have a
two-step process start with L and W and
I can say x is equal to the square < TK
of l^2 +
w^2 this is good we'll just use green to
refer to this
triangle and um uh what we're then going
to want to do is find use that X as a
leg of the red triangle so let's switch
to red
now and for this red triangle we want uh
D which is the hypotenuse is the square
< TK of x^2 +
h^2
okay so what is
X2 well here is
X if I take X
squar that's going to be uh if I take
the square root of something and I turn
around and square it it's like undoing
the square root so that's simply l s
+
w^2 and so I can substitute
this for my x^2 let's see what we get D
equals the square < TK of X2 which is
this plus
h^2 now that is rather remarkable look
what we have we have a new formula that
looks a a lot like the Pythagorean
theorem the diagonal across a rectangle
in two Dimensions is simply the square
root of l^ 2 +
w^2 the diagonal three-dimensionally
across a three-dimensional room is the
square < TK of l^2 + w^2 +
h^2 so it's like an
extension okay so if we have
a uh say a rectangle like this and if
the width in this direction is say three
and the distance in this way is five and
the distance this way is
two let's pull up a
calculator and we can just do it all
right here I take 3 squared so I'll take
3^
squared
plus 5^ squar 5^ s let's say equals
plus 2
squ
equals 38 and then take the square root
so I take the length the width and the
height Square each of them add them
together and take the square root so the
diagonal
here is going to be
6.16 in whatever units we're measuring
and if these are feet that'll be 6.16
feet
okay so this is a useful extension of
the Pythagorean theorem we now have
something that looks a lot like the
Pythagorean theorem but it's a a
three-dimensional
version okay in Algebra 1 uh you learned
how to find equations of lines and
that's about as far as you got you
really didn't find equations of lots of
different things on the plane but from
what we've learned so far it's very easy
to find the equation of a circle so here
is a grid
and here's a circle whose radius is four
so let's say radius equals 4 and so the
question is can I find an equation that
describes this circle and it's
remarkably simple to do that let's see
what we're talking about pick any point
on the circle label it
XY okay in fact XY could stand for any
point in the plane but I want to say
what is the condition so that the XY
point that I picked is actually on the
circle and so you have to say what do
you mean when you talk about a circle if
I drew a circle with a compass I'd put
the point of the compass here and then
stretch it out to a certain distance as
a radius and then swing it around what
I'm physically doing with the compass is
locating on the plane all the points
that are a given distance this radius
from this point so I'm saying this
distance is going to be my radius and
any point on the plane that is this
distance from the center point is going
to be on the
circle well let's try it look at this hm
that's a diagonal line and notice if I
drop a perpendicular here and go over
there I have a right triangle and in
fact the XY Point here tells how far
over and how far up I have to go to get
from the origin up to here so this side
of the triangle is actually X and this
side of the triangle is actually Y and
then I can write down the relation among
these three variables by noticing X and
Y are the legs of a right triangle and R
is the hypotenuse so R 2 is equal to
x^2 +
y^2 in fact what we have here is an
equation that describes the circle any
point for which the X and Y coordinates
of the point satisfy this relationship
will be on the
circle okay let's try out this equation
of a circle in
geogebra so down here I'm going to input
x^2 Plus y^2
equals and then it's going to be R 2 so
let's pick a radius what if I wanted
this circle to have a radius of 2 so I
want it to be this big okay well uh x^2
+ y^2 = R2 so I'm going to type 2^2 or
simply four and let's see what we get
there we are we have a circle of the
given radius and notice over here in the
algebra view that's why they call it Geo
jebra it's algebra and geometry combined
it says x^2 + y^2 = 4 and that's the
equation for this circle if I turn off
this circle it disappears right you can
click this little um ball out in front
of the C and so there you have it
okay if I change this to be
X2 +
y^2 equals and some other amount like
let's say 1 which is 1^ squar we expect
to get a circle that's in here and sure
enough what if I want for equation with
radius of three then I'm going to say
x2+
y^2
equals uh I have remember I have to say
3^ squar so it' be
9 and there we go so you can make a
Target so when we specify these things
algebraically we now know how to specify
a circle what if using geogebra I simply
put this uh uh Center Point
here and then pull it out and go out to
say the
four okay this time I specified the
circle geometrically by using the
geometry tools here
but look what it put over here on the
algebra view of the program it says x^2
+ y^2 = 16 so we didn't enter the
equation it used the equation to
actually uh compute the circle and
figure out which pixels on the screen
it's going to turn blue here or whatever
this color is and it's going to be
determined by x^2 + y^2 ALS r^2 where R
is 4 okay so in the background uh
programs like Photoshop and other kinds
of uh graphic manipulation programs are
doing this kind of thing all the time
and if you're a a graphic artist or U
say a a commercial artist so forth you
would you need to have some appreciation
for what's going on behind the scenes to
be able to use the tools well a lot of
times you don't have to use the algebra
directly but uh somebody's doing that
and those people I'm sure getting paid
very well okay there you have it uh
equations not only of straight lines but
we've moved on to equations of other
objects in this case circles because
circles are defined in terms of the
distance from a center point out to the
circle and we know how to find distances
between points on the
plane so we found uh three different
extensions or if you want to call them
applications of the Pythagorean theorem
here we have the distance
formula we have an extension of the
Pythagorean theorem into three
dimensions and now we found how to use
distance as a way of defining um curves
that we can put on graphs in this case a
circle okay
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