BAM!!! Mr. Tarrou In this video I want to continue doing more examples of finding segment
areas in circles. In the previous video where I introduced sector area and segment area,
when I got to these problems of finding segment area I did very very long process of finding
the segment area and it was long because..well...I wanted to review some knowledge that we learned
in previous lessons and our class. Working with isosceles triangles, dropping down altitudes,
and I just used the basic formula for finding the area of a triangle which is one-half the
base times the height. I am not going to find the height of this triangle, because what
I am going to do in these two examples is use some area of triangles that depending
on what class you are in you might have seen these in PreCalculus. I use them quite a bit
if that is why you are watching this video and you just have a question about some random
problems. If you are in a Geometry class, your Geometry book should have by now introduced
a couple of ways...various ways...of finding area of triangles. It is not just area is
equal to one-half base times height. Because, if you watched my lesson introducing segment
area, while that was a lot of good review in finding that...finishing that problem,
it was long and cumbersome. So we have two more convenient ways of finding the area of
a triangle. We have the area of a triangle when you know two sides and an included angle.
That means that you know two sides of a triangle and the angle that is in-between those. Well,
the area of that triangle is going to be found more easily by simply doing one-half a times
b, now the lower case letters in my textbooks represent the side... so these would be two
sides of my triangle, and that is times the sine of C. That is going to be the included
angle. So, in words it would be actually easier to understand. The area of a triangle is equal
to one-half multiplied by two sides of the give triangle that you know times the sine
of the included angle, the angle that these two sides help to make. So, the area of this
triangle would be one-half times four times seven times the sine of 27 degrees. Make sure
that your calculator is degree mode and there is your answer. That saves me for question
like this, like I did in the previous lesson, going through all of the work of finding not
only the height of the triangle, but also the base. In this triangle I don't know either
one. Now, another way of finding the area of a triangle, that we are going to use on
the second example in my two example video here, is something called Heron's Formula.
I should have written it up here. h e r o n, Heron's...well apostrophe s, Heron's Formula.
The area of a triangle when you know the three sides of the triangle, you don't know any
of the angles and of course you don't know the height because then you could just do
area is equal to one-half the base times the height. The first thing that you are going
to do is find this variable called s. Now with Heron's formula s is equal to one-half
of the perimeter. You will take these three sides, add them up, and divide them by two.
Once you know what s is equal to, half of the perimeter, the area formula...these go
together...you have to do both. The area formula is equal to the square root of s times s minus
a, times s minus b, times s minus c. Again these two formulas are very very nice because
they really speed up the process of finding the area of the triangle if you are not given
some of the nicest information. So let's see what it looks like here. I have a circle with
a radius of seven and I have got this arc here of 260 degrees. I want to find the area
of this segment. Now, all area problems in Geometry pretty much, I am sure there is some
example that we can come up with, but pretty much all of your area problems where you have
a bunch of shapes together and just a portion is shaded is going to be the same...find the
area of the large region and subtract out the area of the smaller region that is not
shaded in. We are going to find for this, we are going to find the area of the sector
right here. Finding the area of the sector is the measure of your arc, in this case that
is going to be arc AB, over 360 degrees...and again all that is just part over whole. It
is just what percent of the circle do you want to find the area of. It is the measure
of the arc over 360 times, and I said what percent of the area of the circle do you want,
so it is what percent of pi times r squared. This minus the area of the triangle is the
segment. Well here I am going to skip all of the work that i did in my introductory
video. I want to find the area of this triangle. Well one side is seven. The other side is
seven because these are both radius or radii of the circle. This angle inside that is included
between these two sides, well 360 degrees is what you rotate inside of a circle and
260 are already taken up, so 360 minus 260 is a hundred. So, looking at filling the formula
out we have one-half a b times the sine of C. This could also be 1/2 b c time the sine
of A. Whatever two lower case letters are here, the other letter is at the end of that
formula. Your textbook may have given you three versions of that formula. Let's plug
in the numbers. I will need to refer to my notes because this will come out of the calculator.
We have the part of our circle that we are concerned with are the 100 degrees out of
the 360 times, I don't think I did any of this in exact form...no I didn't. So it is
going to be 100 divided by 360 times 3.14 times our radius which is seven squared minus
one-half our sides that include our angle and they are both seven (I could write seven
squared but keeping it exactly how the formula is written) times the sine of 100. That is
it. You probably wonder why I did such a long example in the previous lesson. I got review
60-30 triangles, finding heights and bases, and all kinds of good stuff. We found altitudes
of isosceles triangle by bisecting the vertex and the opposite side with a perpendicular
bisector...blah, blah, blah. If we just want the answer, 100 divided by 360 is .278. We
have 3.14 of course times forty-nine minus (I showed more steps than I expected
that I would have). One-half, oops, point five times seven times seven times the sine
of 100 comes out to be 24.13. This mess when you get it typed into your calculator... If
you have a two line scientific calculator where it is very easy to enter fractions,
really honestly you could type all this in at once and get the answer like that. I am
just showing you some steps though if you need it. 42.77 minus 24.13, and that comes
out to be 18.64. Now, don't forget you are finding area and not length so it is 18.64
units squared. Last example. Here we have a circle with a radius of five and a chord
length of nine. A chord is a segment in a circle whose endpoints are on the circle itself.
Again this is going to be an isosceles triangle because these are both radiuses and their
lengths therefore are going to be equal. I want to find the area of this segment. Well,
we are going to well.... If this side is five and this side is nine, we of course automatically
know the other side is five and we have a triangle where we know all three sides. So,
the measure of our arc? We have a little bit of a problem here don't we. Ok, now far as
the measure of the arc goes we still need to a little bit of scratch work. Now there
is a couple of ways to find the measure of that arc and I am looking at this central
angle that we are going to call theta right here. We can do the Law of Cosine if you know
that. The Law of Cosine says that a side squared is equal to the other two sides squared and
added together minus two times those two sides times the cosine of theta. But, my students
have not learned the Law of Cosine so I am not going to do that. What I am going to do
is pull this triangle out. I am only going to pull half of it out because at this point,
except for these fancy formulas and I have really derived how these come about because
we have not talked about the Law of Sine and Law of Cosine in my Geometry class like I
have in my PreCalc class, and I am just going to look at this right triangle here. Again
I still need to pull from some math skill that I used in previous lesson where I did
my long winded example. This is an isosceles triangle. When you drop and altitude in an
isosceles triangle, it becomes an angle bisector of the vertex and it bisects the base. So,
this nine is going to get bisected or cut in half and this is going to become 4.5. Of
course we know this side here of our isosceles triangle, which is now going to become the...how
could I forget this... the hypotenuse is equal to five. We are going to find h. Actually,
I don't need to find that. I am not using the old method. I want to find out what this
angle is up here. Let's just call this capital A. I want to find the measure of capital A.
From angle A, I know the opposite side and I know the hypotenuse. If you remember SOHCAHTOA,
sine is opposite over hypotenuse, CAH cosine is adjacent over hypotenuse, and TOA tangent
of an angle is equal to the opposite side over the adjacent. I have the opposite and
I have the hypotenuse, so I am going to be using sine. The sine of A is equal to opposite
over hypotenuse which is 4.5 over 5. Of course, this is going to come out of our calculators
so let me refer to my notes. 4.5 divided by 5 is going to be .9. We are going to get the
sine function away from the angle by doing the inverse sine function. The inverse sine
function, remember with a regular trig function you put in an angle measure and from the trig
function you get the sides of the triangle. With the inverse trig function, this .9 is
of course as you just saw me do the ratio of the sides, and you get out an angle measure.
The inverse sine of .9 is approximately 64.16 degrees. That means in my drawing here theta
is not equal to 64.16. Remember that I cut this isosceles triangle in half with an altitude
so I could use some right triangle trigonometry. So, that is half the angle theta. Theta is
two of those measurements. Theta is two times 64.16, so it is going to be 128.32. Let me
just double check here, excellent. Now I can get back to this shorter way of doing this
problem. It is shorter I promise you. The measure of the arc, the central angle is 128.32
degrees so therefore the arc is as well. This is times 3.14 times the radius which is five
squared minus.... Ok, Heron's formula. The triangle is five and five is ten, ten and
nine is of course nineteen. So s is equal to one half of the overall perimeter of the
triangle of nineteen. That means that the answer will be what, nine point five. We have
the square root of 9.5 times 9.5 minus a just pick a side 5, times 9.5 minus five again
because we have two fives, times 9.5 minus nine. When you get all that in your calculator
correctly we get 128.32 over 360 times pi times twenty-five, that comes out to be 27.98.
This comes out to be, if you type it in correctly, 9.81. Our segment area comes out to be 18.17
square units. So, I hope this helps you with your homework and maybe some of the harder
problems at the end of your assignment if you are just in Geometry. Or, maybe just any
other random question in your math class. I am Mr. Tarrou. Thank you for watching. BAM!
Go Do Your Homework!