BAM!!!
Mr. Tarrou.
In this video we are going to continue our discussion of the Pythagorean Theorem.
We are finding the lengths of missing sides.
In these examples, I am going to do three of them, the last 2 are closely related to
each other...
We are going to do a couple of examples that have some algebra involved where one missing
side is not just a variable but an algebraic expression.
Here we have x-1.
And you know, you just work through a couple of you happen to have some of these questions
maybe say in an Algebra 2 book.
So we have got x-1 as leg helping to make one of the right angles.
We have got 3 again helping to make this 90 degree right angle.
We have a hypotenuse of 3 square root of 5.
So if we are going to find this value of x, and then length of the side of this right
triangle we have to set up the Pythagorean Theorem.
When we know two sides and we are missing the length one side of triangle, we can solve
for the missing length of that triangle using the Pythagorean Theorem.
So we have a^2 or one of the legs squared plus one of the other legs... or the other
leg squared so a^2+b^2.
Just make sure that the two sides that help to make the right angle are together on the
same side of the equation...
So x-1 squared equals and parenthesis 3 square root of 5.
Again you should make sure that you always use parenthesis when doing substitution into
an algebraic equation or expression.
So we have 3 squared which is 9.
Then we have x-1 squared.
Well x-1 squared is going to be let's see here...
Off to the side (x-1)^2 is x-1 times x-1.
We are going to distribute that together.
Make sure that you do that because a lot of students will try and short cut this process
and end up with only two terms when there is supposed to be three.
So x times x is x^2.
x times -1 is negative one x.
Then another -1x.
And then finally -1 times -1 is 1.
So we have x^2-2x+1.
So that is what we have here.
We have 9 plus, and then x^2-2x+1.
Then... let me make sure this is looking alright.
Yep.
Then we have on the other side...
You know, how do you take 3 square root of 5 and square it?
Well, 3 square root of 5...
When you square it.
When you multiply it by itself, you do want to take those coefficients of the radical
which is the 3 and multiply them together because of the square.
So 3 times 3 is 9.
And the square root of 5 times the square root of 5 is the square root of 25, which
is 5.
All of this is touching.
So all of that is being multiplied together.
So don't do 9 plus 5 is 14 or something like that.
This is all multiplied together, so 9 times 5 is equal to 45.
Ok, so we have 9 and 1 make 10.
We need to bring the 45 over to the left hand side with subtraction.
We are not going to...
We are trying to solve for x, but I cannot isolate x because there is not just one x
term.
There are two.
There is this squared term and a single degree term.
So we need to get this set equal to zero and solve with either factoring, completing the
square, or the quadratic formula.
So we are going to subtract both sides by 45 and get x^2-2x-35 equals 0.
And when our leading coefficient is 1, we can kind of just pay attention to the constant.
What factors of 35 subtract to give us a middle terms of -2?
Let's see....
Factors of 35.
There are not that many of them.
How about 7 and 5.
Now it is a negative 35, so they need to have different signs.
So we are going to have x which...
We are going to factor the first term and put those factors in the first two blanks.
We are going to factor the constant and put those factors in the second of the two blanks
in these binomials.
The square root of x^2 is x.
Factors of 35 that subtract to be -2, we already talked about that 7 and 5.
The middle term is negative, so the larger of the two factors needs to be negative.
We are going to have -5, excuse me, -7 and +5.
This equals zero.
Now after you have taken a quadratic, that is the power of 2... or degree of 2, and factor
it... then you set each factor equal to zero.
Because this one clumping, the parenthesis holds all this together...
this is something times something equals zero.
So one of those two factors has to be zero so that the product is equal to zero.
And we are going to make that happen.
They won't be zero at the same time, just so that one of them are.
So x-7 is equal to zero.
x+5 is equal to zero.
Solve each one of those for x and we get x is equal to 7 and x is equal to -5.
Now, in pure algebra there are two answers.
But we are finding the lengths of a right triangle.
There is no such thing as a negative length.
So while the algebra gives us two possible answers and if I gave you this question in
an algebra 1 or algebra 2 class and said solve it there are two answer.
But we need to make sure that we use the solution, or maybe both of them work...
I don't think they will, but we need to use the solution that gives us a reasonable answer
here based on the fact that we are trying to find the sides of a right triangle.
So we cannot use the x value of negative five because -5-1 is -6.
You cannot have a negative length.
Thus the answer that we can use is x=7.
So we are going to take out the x, put in the fact that we just found that it equals
7, and then 7-1 is equal to 6.
So not only do we know that x is equal to 7, but the side of this triangle... the leg
is equal to a length of 7 minus 1 which is 6 units.
I have got two more examples.
I will be right back after erasing by board.
For our second example we have a circle.
We have a radius with a length of 300.
We have a tangent line to this circle.
It is a tangent line because it intersects once.
And a line is perpendicular to the radius so we know it is a tangent line.
And we have a hypotenuse cut up into two different parts.
The only part that is labelled is the part extending outside the circle of 416.577.
This could be some kind of planetary word problem with a satellite.
It could be some kind of wheel or gear with a tangent line that represents a distance
it travelled, or something like that.
I am just giving you the diagram and we are going to figure out what x is equal to.
Now if you look at it and you go, well...
Let's see.
Pythagorean Theorem is only useful when you know two sides of a right triangle and you
want to find the third missing side.
And if you look at this and you initially think I don't have all three sides.
This is the radius of 300, this length is unknown, and I just have the blue segment
of 461.577.
I don't know what this little piece is.
Well, it is a circle right!
So if you go from the center of a circle to a point on the circle, regardless of what
direction you go in you are using the radius.
That is the radius.
And the radius of a circle is content value.
So this isn't really that difficult of a problem.
My next example which looks similar will be more thought provoking.
I could just recognize that other radius is also 300 and thus this is just as straight
forward as any other Pythagorean problem is.
One leg squared plus the other leg squared is equal to the hypotenuse squared.
Now 461.577 is not the hypotenuse, nor is 300.
But when you add them you get the hypotenuse.
That is 761.577. 300 squared is 90,000. 3 times 3 is 9, and a couple zero and a couple
of zeros more when you square it and you get 90,000... plus x^2.
Now I did work this out to be as straight forward as possible here.
761.577 squared is approximately 580,000 units.
Whatever.
580,000.
We subtract both sides by 90,000 and we get x^2 is equal to 490,000.
Square root both sides and we get x is equal to 700, the missing side of the right triangle.
We are going to use basically the same diagram but I am going to move the x to another location,
somewhere along this hypotenuse and see how much that changes out work.
BAM!!!
Waking you up:)
Now with this example I have put the x on and marked it off as just a portion of the
hypotenuse.
So we need to solve out what x is.
We do know two sides of the right triangle, the radius of 35... the leg, and the other
leg of 50... the length of this tangent segment.
And we know the... sort of... the length of they hypotenuse.
Now the length of the hypotenuse is not x.
It is actually x plus this little segment here of 35.
So this hypotenuse is actually x+35.
Now if I look at and go ok...
I have one leg which is 35... so 35^2.
I have another leg which is 50, so 50^2.
This all equals the hypotenuse squared.
Now the hypotenuse is x+35.
And if I just look at that, BOOM, I see the hypotenuse is a binomial squared and start
working it out.
We have 35 squared which is, referring to my notes here, 1225.
50 squared is 2500.
x+35 squared, x+35 times x+35 is going to be x^2+70x...
I guess should distribute this out.
x^2 plus 35x plus 35x plus, and 35 times 35 is... haha.. it is right there 1225.
And that means that is over here +1225.
So x+35 squared is that trinomial of x^2+70x+1225.
Well, I have got 1225 on both sides, so when I subtract that
from both sides of the equation it is going to cancel out.
I need to do the same thing with 2500.
Subtract that over to the other side as well because it is positive too.
So when we do that subtraction and get that 2500 over from the left to the right hand
side, we are going to get 0=x^2+70x-2500.
Now when was the last time we tried to factor a trinomial with a constant that large?
As it turns out this actually does not factor.
So I would just look at this and straight away start doing the Quadratic Formula.
That is going to be x equals -70 plus or minus the square root of 70 squared minus 4 times
'a' times 'c' all over 2 'a'.
When you have a quadratic equation the you cannot factor, or don't wish to try and factor,
you can apply the quadratic formula to solve it instead.
Which again is the opposite of b plus or minus the square root of b^2-4ac over 2a.
When you have your degrees decreasing, we have a second degree, first degree, and the
constant, 'a' is the leading coefficient or the degree of the second degree term, b is
coefficient of the first degree term, and then c is the constant.
When you work this out you are going to get the answers of approximately 26 and -96.
Because x is a length, you cannot use the negative value.
Now the way I have worked this out is really not the most efficient way because we ended
up being forced into using the Quadratic Formula.
Because of how I labelled the hypotenuse as being x+35.
And that is its length.
If however I came through here and temporarily named the entire hypotenuse as value 'y' the
problem algebraically, as far as the process goes to solve it, becomes a lot simpler.
It is going to be one leg squared which is 35 plus the other leg squared which is 50
squared equals the hypotenuse squared which is y squared.
Now you can see I don't have any binomials that I have to squared that then forces me
to work with the quadratic formula.
How this is going to look of course, it is going to be 1225... that is 35^2... plus 2500...
I just don't want to try and do that in my head and make a mistake on these videos.
I do that enough as it is every once and a while.
2500 plus 1225 is 3725.
I can square root both sides.
I get y is equal to approximately, I am rounding this off a bit, approximately 61.
Now y is the entire length of the hypotenuse.
X is only part of it.
X is that entire hypotenuse with 35 taken out.
So I can, now that I know the entire length is 61 units approximately, I can say...
What color do I have it in?
Blue.
I can say that x is 61, and it is again been rounded off...
it is not rounded very much...
but approximately 61 minus this part of the hypotenuse which is 35.
That is another way to get the answer of 26.
Thinking about relabelling the problem a little bit, then working through it.
A little bit of thought ahead of time, gave us what really was algebraically a simpler
solution.
I am Mr. Tarrou.
Those are a few more complicated examples of the Pythagorean Theorem and how you use
it to solve right triangles.
BAM!
Go Do Your Homework.