BAM!!! Mr. Tarrou. In this video we are going to do four examples of finding missing lengths,
missing sides, of a right triangle. Now if you are given a right triangle and you know
two sides and you are missing the length of the third side, whether that is a leg like
this example, or the hypotenuse like my first example is going to be, you can solve those
using the Pythagorean Theorem. Now if it is say... if I only know one side of a right
triangle and the other two sides are unknown, then I am going to need to have some information
about one of these acute angles in the corner and do some function call trig functions of
sine, cosine, and tangent. Those are beyond the scope of this lesson, but just to let
you know you can't always use the Pythagorean when you have a right triangle. if you have
less information, or different information, you will have to some other way to solve for
that missing information. But we are going to have two sides and be looking for a third.
So Pythagorean Theorem says we have got a right triangle again and of course all triangles
have three sides. They have two legs, well a right triangle will have two legs. The two
sides that we call legs are the two sides that come together to make the right angle,
or the 90 degree angle. And then the side that is opposite that 90 degree angle that
is not helping to create that 90 degree angle, that is going to be your hypotenuse. It will
be the longest side of your triangle. So if you do your math and, you know... this is
maybe a 3 and a length of 4, if your hypotenuse is not longer than either 3 or 4... it would
be a 3, 4, 5 triangle by the way... we will be able to figure that out by the end of this
video. If your hypotenuse, the side opposite the right angle, is not the longest measurement
then you have made a mistake somewhere. The formula for the Pythagorean Theorem is...
I have seen this written. You will probably will not see it like this, but leg squared
plus leg squared equals hypotenuse squared. In other words just make sure the longest
side of your triangle, the hypotenuse... the side opposite the 90 degrees or the right
angle, is the side that is by itself. What you will see most of the time in textbooks,
or all the time really, is a^2+b^2=c^2. The two sides that have something in common...
the two sides that help to make the right angle, the 90 degree angle, those two sides
are always together in the equation. The hypotenuse again, the longest side, is always the value
that is by itself whether you know it or not. The hypotenuse has to be the value that is
over to the right hand side of your equation by itself. Ok. We are going to do 4 examples
with increasing difficulty as we go. I am going to do another video with 3 more advanced
examples that have a little bit of algebra involved in them. So we have got this right
triangle because of the corner. We have a leg of 5. This is certainly not drawn to scale.
We have a leg measure is drawn with a leg measure of 5 and the other leg is a length
of 12. You should never trust a diagram in Geometry. Most of the time when you take standardized
tests like the SAT, it would say not drawn to scale. If I wanted to attempt to make this
a little bit better, of course 5 is much shorter than 12. So maybe the diagram should look
something like that. Well we know the two sides that make the right angle and we are
looking for the hypotenuse. I think I had a 'c' there before but it does not matter
what the letter is of course... unless you are using the a, b, and c labels that are
in the Pythagorean Theorem because 'a' and 'b' are marked off as the legs... let's put
that 'c' back in there... and 'c' is the hypotenuse. But a variable is just a letter that represents
an unknown numerical value. It really does not matter though whether you call it a, b,
c, d, e, f, g, h... Most of your problems are going to be in terms of x. Again Pythagorean
Theorem is a^2+b^2=c^2, but that is why I am giving you this more generic formula here
where it is just take a leg and square it, add it to another leg squared, and set it
equal to the hypotenuse squared. I am reinforcing... A variable is just a variable. It is what
that variable represents numerically that is important. So 5 squared plus 12 squared,
the two sides that are helping to make the right angle together on the same side of the
equation equals c squared. Now 5^2 is 25. 12^2 is 144. Now 5^2 is 5 times 5 not 5 times
2. That would be 2 times 5 if it were. 5^2 is 5 times 5 which is 25. 12^2 is 12 times
12 which is 144. That is equal to c squared. This comes out to be 169 when you add them.
You get rid of an exponent, a power, of 2 by square rooting both sides of your equation.
You get c is equal to 13. Now when you square root a number there is a positive and a negative
answer. If this were a word problem or an application problem and we were not talking
about actual measurements, then the square root of 169 would be plus or minus.. or positive/negative
13. But you cannot have a negative distance. That would be like you taking a tape measure,
you know that you pull out of a spool, you don't push that tape measure back inside the
spool and have a negative measurement. You only pull it out and you would have that positive
measurement. So the hypotenuse here is 13. And that side opposite the right angle is
longer than my other two sides... the two sides that are helping to make the right angle...
my two legs. So here we have another problem. The variable is on one of the legs. A lot
of students incorrectly will write 7^2... let me go back to the orange chalk here. 7^2+9^2
equals x^2. A lot of students want to keep the variable alone for all of the problems.
No no no. The hypotenuse, the side opposite the 90 degree angle, must be the side that
is left alone. The two sides that help to make the right angle must stay together on
the same side of the equation. So we have x^2, one of the legs, plus the other leg squared
which is 7^2 equals the hypotenuse squared which is 9^2. This is x^2 plus 49 equals 81.
We are going to subtract both sides by 49 and get x^2 is equal to... let's see here.
81-40 is 41, and 41-9 is 32. Ok. Like I said, these are going to be coming in ever increasing
difficulty as we go through here. We are going to square root both sides of the equation.
We get x is equal to the square root of 32. Now some teachers, and sometimes I am guilty
of this too depending on the level of class, we are like ok... just put in the calculator
and get a decimal. You need to be comfortable using square roots, or just radicals in general,
and simplifying them especially if you are in Geometry going into Algebra 2. You will
need to understand how to reduce a radical. So I want to make this smaller and get an
exact value answer and not a decimal. If I can reduce this square root of 32.. I can
reduce it if there is a perfect square in the 32. Now perfect squares... like 2 times
2 is 4. 4 does divide into 32 but so does a bigger. 3, 3 times 3 is 9. 9 is the next
perfect square. I don't know why I am doing 7?! But 9 is the next perfect square. But
9 does not go into 32 evenly. So 4 squared, or 16, is a perfect square and 16 does go
into 32. 32 divided by 16 is 2. So I am factoring this in such a way that one of my factors
is a perfect square. Then the square root of 16 is 4, and thus missing leg has a length
of 4 square root of 2 units. That would be considered an exact value answer because if
you put in the square root of 2 in your calculator you will get a decimal that runs off the screen.
It does not stop at the end of that screen. It actually goes on forever and ever and ever.
So that decimal is a rounded off value and thus will be an approximate answer. This is
the exact solution. I have two more examples. Thank you very much for watching. I hope this
is helping. BAM!!! For our next example we have the variable on the hypotenuse. Students
do generally tend to get these right more often with the variable on the hypotenuse
when they are first learning this process. But I have 4 square root of 3 here, and again
we are not just going to make decimals here. I want to make sure that you understand how
to deal with square roots. So we are going to increase the difficulty a little bit by
having the radical in the problem in the first place. a^2+b^2=c^2 or the two sides that help
the leg... excuse me... the two legs that help to make the right angle stay together
on the same side of the equation. So we have got 4 square root of 3 squared plus 4 squared
equals x squared... the hypotenuse squared. 4 square root of 3 squared is 4 square root
of 3 times 4 square root of 3. You want to put the coefficients together. 4 times 4 which
is going to be 16. And we want to put these square roots together separately. The square
root of 3 times the square root of 3 is the square root of 9. The square root of 9 is
3. When you multiply square roots together and they have exactly the same value on the
inside of the square root symbol it will cancel out like I just said. The square root of 3
times the square root of 3 is the square root of 9, which comes right down back to 3. Now
the 4 and the square root of 3 are touching, the 4 and the square root of 3 are touching,
so when you do the 4 times 4 and get 16... when you do the square root of 3 times the
square root of 3 and get 3, please remember that you are multiplying those and not...
I don't know... whatever else students come up with sometimes. So 16 times 3. 10 times
3 is 30. 6 times 3 is 18. So it is 48. 4 squared is 16. 48 plus 10 is 58. 58 and 6 is 64. We
are going to square root both sides of the equation. We get a final answer of 8. We have
got one more example. Last example. We have got this right triangle. We have got a missing
leg. We have another leg which is 2 and a half. We have got a hypotenuse which is 3
and 2/3. Yes, there are fractions in here. So we are going to leave those in there. We
are going to deal with them. We are going to make them into improper fractions so they
are easier to work with. Find common denominators. All of that good exciting stuff. Well, we
are finding x. So the two legs, the two sides that make the right angle... the two legs
that make the right angle again stay together. So we have 2 and 1/2 squared plus x squared
is equal to the hypotenuse which needs to be squared. You might notice that every time
I do substation I use parenthesis. All of these values are positive so really at this
point it is a habit. But if you do substitution with say algebraic expressions where there
are more than one term, or if you are substituting in negative values, the parenthesis are very
very important. You do need to basically use parenthesis every time you do substitution.
These are mixed fractions and while it is possible to work with them, it is not very
nice. So we are going to make these improper fractions. We are going to do the denominator
times this whole number out here. So 2 times 2 is 4, then add on the numerator. So 2 times
2 is 4, plus 1 is 5. So we have 5/2 squared plus x^2 equals... 3 times 3 is 9, and 9 plus
2... excuse me... 3 times 3 is 9, and 9 plus 2 is 11 over 3 squared. We are going to take
this fraction and square it. That means that we are going to do 5/2 times 5/2. When you
multiply fractions you go straight across. Or when you square a fraction you square both
the top, numerator, and the denominator. So 5 times 5 is 25. And 2 times 2 is 4, or 2^2
is 4... now plus x^2 is equal to 11^2 which is 11 times 11 which is 121. 3 squared which
is 9. We need to take this 25/4 and move it over to the other side. So we are going to
subtract this side by 25/4. I am going to cancel this out and erase it in just a second.
I am going to bring that minus 25/4 over here. I am leaving some room because of course we
need common denominators to add or subtract fractions. The lowest common multiple between
9 and 4? I am going to take the... either number. Whatever. 9, 18... 4 doesn't go into
9. 4 doesn't go into 18. 9, 18, 27. 4 does not going into 27. 36. So the lowest common
multiple between these numbers is 36. So to get the 9 to be equal to 36 we are going to
multiply it by 4. To get the 4 to equal 36, we are going to multiply it by 9. It seems
like we are just sort of swapping the denominators and multiplying them to the numerator and
denominator. You actually can do that. The reason we are having to do that here is because
4 and 9 do not have any common factors in them. So that is, you know, all we are left
with. The numbers are going to have to be a little bit big. So I am going to get this
out of the way. This is going to cancel out. I need the room. So we have x squared is equal
to... 121 times 4 is 484... minus... I am going to write one big fraction because I
already have my common denominator of 36. 25 times 9. Now I can try to do that, but
it is a quarter. So if you have 4 quarters that is a dollar or 100 cents. If you have
8 quarters that is 2 dollars or 200 cents plus an addition quarter. So 225. 25 times
9 is 225. All of this over 36. And 484 minus 225. Well 484 - 200 is 284, 284 -20 is 264,
264-5... 264 minus 5 should be 259. Let me double check. Yes 259. So x^2 is equal to
259/36. We are going to square root both sides. And we are going to get x is equal to the
square root of 259 over... now 36. The square root of 36 is 6. As I am finishing this up,
I realized that I didn't look to see if there was any perfect squares in 259. Let me pause
this. I am going to take 259 and divide by 4, 9, 16, 25, 36, 49.... run through the perfect
squares and see if there are any in that value and thus have that be reducible. Let me double
check. I will be with you in just a second. Maybe you should do the same. Well. I checked.
That is the simplest this answer gets. There are no perfect squares in 259. And so whatever.
That is the answer. If your teacher is going to be happy with decimals, it is ok just type
all this in there and get a decimal form. But I am going to leave that in decimal form
and call it a day. I am going to one more video. It is again Pythagorean Theorem, with
a couple of questions or examples that involve a little bit of algebra. Just in case you
are in Algebra 2 and you are seeing some of these Pythagorean Theorem problems where the
sides are like x-4 and not just a simple variable like this with the x. So if you need that
just keep watching, and if not... I am Mr. Tarrou. BAM! Go do your homework:)