[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:01.85,0:00:06.04,Default,,0000,0000,0000,,The theorem of Pythagoras is a\Nwell known theorem.
Dialogue: 0,0:00:06.60,0:00:11.19,Default,,0000,0000,0000,,It's also a very old one. Not\Nonly does it bear the name of
Dialogue: 0,0:00:11.19,0:00:14.80,Default,,0000,0000,0000,,Pythagoras in ancient Greek,\Nbut it was also known to the
Dialogue: 0,0:00:14.80,0:00:17.10,Default,,0000,0000,0000,,ancient Babylonians and to the\Nancient Egyptians.
Dialogue: 0,0:00:18.23,0:00:22.43,Default,,0000,0000,0000,,Most school boys school girls\Nlearn of it as a square plus B
Dialogue: 0,0:00:22.43,0:00:25.66,Default,,0000,0000,0000,,squared equals C squared, but\Nthe actual statement of the
Dialogue: 0,0:00:25.66,0:00:27.60,Default,,0000,0000,0000,,theorem is more to do with
Dialogue: 0,0:00:27.60,0:00:31.99,Default,,0000,0000,0000,,areas. So let's have a look at\Nthe statement of the theorem.
Dialogue: 0,0:00:33.11,0:00:37.72,Default,,0000,0000,0000,,The theorem talks about\Na right angle triangle.
Dialogue: 0,0:00:39.27,0:00:43.22,Default,,0000,0000,0000,,So then we have a right angle\Ntriangle and we show that with
Dialogue: 0,0:00:43.22,0:00:46.87,Default,,0000,0000,0000,,this little mark here, so we\Nhave a square in that corner.
Dialogue: 0,0:00:47.70,0:00:52.66,Default,,0000,0000,0000,,What the theorem says is that\Nthe square on the hypotenuse.
Dialogue: 0,0:00:52.66,0:00:57.17,Default,,0000,0000,0000,,Now that means we have to\Nidentify this side the
Dialogue: 0,0:00:57.17,0:01:02.13,Default,,0000,0000,0000,,hypotenuse, and it's always the\Nlongest side in the right angle
Dialogue: 0,0:01:02.13,0:01:07.54,Default,,0000,0000,0000,,triangle and it's always the one\Nthat is opposite to the right
Dialogue: 0,0:01:07.54,0:01:12.50,Default,,0000,0000,0000,,angle. So it's this side here\Nwhich bears the name the
Dialogue: 0,0:01:12.50,0:01:18.98,Default,,0000,0000,0000,,hypotenuse. So the theorem then\Nsays the square on the
Dialogue: 0,0:01:18.98,0:01:26.20,Default,,0000,0000,0000,,hypotenuse, so if they draw a\Nsquare on this hypotenuse, so
Dialogue: 0,0:01:26.20,0:01:29.48,Default,,0000,0000,0000,,there we've got our square.
Dialogue: 0,0:01:30.21,0:01:35.57,Default,,0000,0000,0000,,And it says that the area of\Nthat Square is equal to the sum
Dialogue: 0,0:01:35.57,0:01:36.72,Default,,0000,0000,0000,,of the areas.
Dialogue: 0,0:01:37.30,0:01:42.68,Default,,0000,0000,0000,,Of the squares on the\Ntwo shorter sides.
Dialogue: 0,0:01:43.82,0:01:47.13,Default,,0000,0000,0000,,So these are all squares.
Dialogue: 0,0:01:47.13,0:01:52.54,Default,,0000,0000,0000,,All the same, all got right\Nangles in them.
Dialogue: 0,0:01:53.12,0:01:55.77,Default,,0000,0000,0000,,This is a square as well.
Dialogue: 0,0:01:56.29,0:02:02.25,Default,,0000,0000,0000,,All sides are equal. Got a\Nright angle. What this says
Dialogue: 0,0:02:02.25,0:02:09.84,Default,,0000,0000,0000,,is that this area A plus this\Narea B is equal to that area.
Dialogue: 0,0:02:09.84,0:02:12.01,Default,,0000,0000,0000,,C A+B equals C.
Dialogue: 0,0:02:13.14,0:02:20.21,Default,,0000,0000,0000,,OK. Let's have a look at\Na special case and see if we can
Dialogue: 0,0:02:20.21,0:02:22.25,Default,,0000,0000,0000,,see how that actually works.
Dialogue: 0,0:02:22.26,0:02:30.11,Default,,0000,0000,0000,,Got here a\Nsingle red triangle.
Dialogue: 0,0:02:33.46,0:02:35.83,Default,,0000,0000,0000,,It's a very special triangle.
Dialogue: 0,0:02:37.34,0:02:41.56,Default,,0000,0000,0000,,This side along the bottom is\None unit long and that side
Dialogue: 0,0:02:41.56,0:02:46.49,Default,,0000,0000,0000,,along there is 2 units long, so\NI'm just going to fix it. Little
Dialogue: 0,0:02:46.49,0:02:47.90,Default,,0000,0000,0000,,bit of Blu tack.
Dialogue: 0,0:02:48.78,0:02:54.51,Default,,0000,0000,0000,,On there I'm just going to fix\Nit so that in fact I can draw
Dialogue: 0,0:02:54.51,0:03:01.70,Default,,0000,0000,0000,,round it. So that\Nwe've drawn around that
Dialogue: 0,0:03:01.70,0:03:09.58,Default,,0000,0000,0000,,triangle, now I want\Nto fix some squares
Dialogue: 0,0:03:09.58,0:03:13.53,Default,,0000,0000,0000,,on the sides of
Dialogue: 0,0:03:13.53,0:03:20.61,Default,,0000,0000,0000,,this triangle. So there's\Nthe square on that side, but the
Dialogue: 0,0:03:20.61,0:03:27.32,Default,,0000,0000,0000,,square on this side is going to\Nlook in fact just a tiny bit
Dialogue: 0,0:03:27.32,0:03:32.59,Default,,0000,0000,0000,,different. It's going to be an\Nassembly of triangles, each one
Dialogue: 0,0:03:32.59,0:03:37.38,Default,,0000,0000,0000,,identical to this original\Ntriangle, so there's half of it.
Dialogue: 0,0:03:37.97,0:03:40.32,Default,,0000,0000,0000,,Now I'm going to put the other
Dialogue: 0,0:03:40.32,0:03:43.37,Default,,0000,0000,0000,,half on. Here.
Dialogue: 0,0:03:44.11,0:03:48.05,Default,,0000,0000,0000,,Lastly, debate.
Dialogue: 0,0:03:49.38,0:03:53.66,Default,,0000,0000,0000,,I will make it up with\Nthis bit.
Dialogue: 0,0:03:56.26,0:04:00.68,Default,,0000,0000,0000,,OK, now remember what the\Ntheorem says. It says that if we
Dialogue: 0,0:04:00.68,0:04:05.83,Default,,0000,0000,0000,,take the area of this square and\Nadd it to the area of that
Dialogue: 0,0:04:05.83,0:04:09.88,Default,,0000,0000,0000,,square, we should get the area\Nof this square on the
Dialogue: 0,0:04:09.88,0:04:11.72,Default,,0000,0000,0000,,hypotenuse. Let's see how right
Dialogue: 0,0:04:11.72,0:04:17.64,Default,,0000,0000,0000,,that is. Each one of these red\Ntriangles that I've got here and
Dialogue: 0,0:04:17.64,0:04:21.83,Default,,0000,0000,0000,,that I'm putting on is exactly\Nthe same as these triangles
Dialogue: 0,0:04:21.83,0:04:24.88,Default,,0000,0000,0000,,here, so I can put one on there.
Dialogue: 0,0:04:28.00,0:04:33.96,Default,,0000,0000,0000,,That one\Non there.
Dialogue: 0,0:04:33.96,0:04:36.95,Default,,0000,0000,0000,,Now put
Dialogue: 0,0:04:36.95,0:04:39.93,Default,,0000,0000,0000,,the square.
Dialogue: 0,0:04:40.24,0:04:45.08,Default,,0000,0000,0000,,In there.\NThat
Dialogue: 0,0:04:45.08,0:04:50.52,Default,,0000,0000,0000,,one\Nin
Dialogue: 0,0:04:50.52,0:04:57.31,Default,,0000,0000,0000,,there.\NAnd then finally put
Dialogue: 0,0:04:57.31,0:05:00.56,Default,,0000,0000,0000,,this one in there.
Dialogue: 0,0:05:01.19,0:05:06.39,Default,,0000,0000,0000,,What we can see there is that\Nquite clearly the area of this
Dialogue: 0,0:05:06.39,0:05:11.99,Default,,0000,0000,0000,,square is made up of the area of\Nthat square plus the area of
Dialogue: 0,0:05:11.99,0:05:17.24,Default,,0000,0000,0000,,that square. OK, if this theorem\Nis true, and certainly it seems
Dialogue: 0,0:05:17.24,0:05:22.11,Default,,0000,0000,0000,,to be true for this, and it has\Nbeen proved that it's true, then
Dialogue: 0,0:05:22.11,0:05:24.20,Default,,0000,0000,0000,,how do we get the traditional?
Dialogue: 0,0:05:24.97,0:05:30.96,Default,,0000,0000,0000,,Schoolboy schoolgirl theorem. A\Nsquared plus B squared equals C
Dialogue: 0,0:05:30.96,0:05:37.93,Default,,0000,0000,0000,,squared. So draw our\Nright angle triangle.
Dialogue: 0,0:05:39.76,0:05:43.38,Default,,0000,0000,0000,,Will put on the
Dialogue: 0,0:05:43.38,0:05:46.24,Default,,0000,0000,0000,,squares. On the sides.
Dialogue: 0,0:05:46.81,0:05:50.99,Default,,0000,0000,0000,,And\Nwe
Dialogue: 0,0:05:50.99,0:05:55.17,Default,,0000,0000,0000,,label\Nthese
Dialogue: 0,0:05:55.17,0:05:57.26,Default,,0000,0000,0000,,squares
Dialogue: 0,0:05:57.26,0:06:03.37,Default,,0000,0000,0000,,A&B&C. We label\Nthe lengths of the sides that
Dialogue: 0,0:06:03.37,0:06:05.71,Default,,0000,0000,0000,,lay. It'll be a little C.
Dialogue: 0,0:06:06.43,0:06:11.38,Default,,0000,0000,0000,,Then the theorem tells us that\Nif we add together the areas of
Dialogue: 0,0:06:11.38,0:06:16.34,Default,,0000,0000,0000,,the squares on the two shorter\Nsides, the result is the area of
Dialogue: 0,0:06:16.34,0:06:20.53,Default,,0000,0000,0000,,the square on the longest side\Nof the triangle the hypotenuse.
Dialogue: 0,0:06:20.53,0:06:25.48,Default,,0000,0000,0000,,So A plus B is equal to see, but\Nthis is a square.
Dialogue: 0,0:06:26.20,0:06:32.25,Default,,0000,0000,0000,,And so it's area is A squared\Nand this is also a square, so
Dialogue: 0,0:06:32.25,0:06:34.41,Default,,0000,0000,0000,,its area is B squared.
Dialogue: 0,0:06:34.94,0:06:41.87,Default,,0000,0000,0000,,This is a square and its area is\NC squared and so we have the
Dialogue: 0,0:06:41.87,0:06:45.52,Default,,0000,0000,0000,,traditional result. Always when\Nanswering questions on
Dialogue: 0,0:06:45.52,0:06:49.28,Default,,0000,0000,0000,,Pythagoras Theorem, make sure\Nthat you read the question.
Dialogue: 0,0:06:49.28,0:06:53.45,Default,,0000,0000,0000,,Don't just write down this\Nresult Willy nilly. Make sure
Dialogue: 0,0:06:53.45,0:06:56.78,Default,,0000,0000,0000,,you've actually read the\Nquestion and that you're
Dialogue: 0,0:06:56.78,0:07:01.79,Default,,0000,0000,0000,,thinking about which sides are\Nwhich. Let's have a look at one
Dialogue: 0,0:07:01.79,0:07:03.04,Default,,0000,0000,0000,,or two examples.
Dialogue: 0,0:07:03.58,0:07:10.33,Default,,0000,0000,0000,,First of all, let's take\Na triangle whose two shorter
Dialogue: 0,0:07:10.33,0:07:16.40,Default,,0000,0000,0000,,sides are of length three\Nand of length 4.
Dialogue: 0,0:07:16.92,0:07:22.67,Default,,0000,0000,0000,,And the question is, how long is\Nthe longest side the hypotenuse?
Dialogue: 0,0:07:23.33,0:07:27.62,Default,,0000,0000,0000,,So we can label our sides, let's\Ncall this one a.
Dialogue: 0,0:07:28.22,0:07:33.69,Default,,0000,0000,0000,,Call this one B and then the one\Nwe want to find is see.
Dialogue: 0,0:07:34.52,0:07:42.31,Default,,0000,0000,0000,,So using the theorem, we know\Nthat A squared plus B squared.
Dialogue: 0,0:07:42.31,0:07:43.81,Default,,0000,0000,0000,,Is C squared?
Dialogue: 0,0:07:44.52,0:07:51.63,Default,,0000,0000,0000,,A is 3 three squared\Nplus B is 4, four
Dialogue: 0,0:07:51.63,0:07:54.47,Default,,0000,0000,0000,,squared equals C squared.
Dialogue: 0,0:07:55.30,0:08:01.66,Default,,0000,0000,0000,,3 squared is 9 and 4 squared. Is\N16 equals C squared, and so 25
Dialogue: 0,0:08:01.66,0:08:08.02,Default,,0000,0000,0000,,is equal to C squared, and so we\Nnow need to take the square root
Dialogue: 0,0:08:08.02,0:08:12.26,Default,,0000,0000,0000,,of both sides, and so five is\Nequal to see.
Dialogue: 0,0:08:13.16,0:08:18.80,Default,,0000,0000,0000,,That's the length of the\Nhypotenuse. The longest side of
Dialogue: 0,0:08:18.80,0:08:24.96,Default,,0000,0000,0000,,the triangle. Let's take another\Nexample. This time one where we
Dialogue: 0,0:08:24.96,0:08:30.94,Default,,0000,0000,0000,,know the hypotenuse. Let's say\Nit's of length 13 and we know
Dialogue: 0,0:08:30.94,0:08:37.41,Default,,0000,0000,0000,,one of the shorter sides. Let's\Nsay that's of length 5, but we
Dialogue: 0,0:08:37.41,0:08:40.90,Default,,0000,0000,0000,,want to know is what's the other
Dialogue: 0,0:08:40.90,0:08:48.07,Default,,0000,0000,0000,,shorter side. Let's label these.\NIf I call this One X, let's say,
Dialogue: 0,0:08:48.07,0:08:51.48,Default,,0000,0000,0000,,and this one, why I'm call this
Dialogue: 0,0:08:51.48,0:08:56.94,Default,,0000,0000,0000,,one zed. Because I've label them\Ndifferently, doesn't make the
Dialogue: 0,0:08:56.94,0:09:01.63,Default,,0000,0000,0000,,relationship any different. It\Nstill X squared, plus Y squared
Dialogue: 0,0:09:01.63,0:09:07.26,Default,,0000,0000,0000,,equals zed squared. The sum of\Nthe squares on the two shorter
Dialogue: 0,0:09:07.26,0:09:11.95,Default,,0000,0000,0000,,sides is equal to the square on\Nthe hypotenuse. Let's
Dialogue: 0,0:09:11.95,0:09:17.58,Default,,0000,0000,0000,,substituting. Now X is 5, so\Nthat's 5 squared plus Y squared.
Dialogue: 0,0:09:17.58,0:09:23.21,Default,,0000,0000,0000,,That's what we're trying to find\Nis equal to zed squared and
Dialogue: 0,0:09:23.21,0:09:26.21,Default,,0000,0000,0000,,said. Is 13, so that's 13
Dialogue: 0,0:09:26.21,0:09:32.52,Default,,0000,0000,0000,,squared. So 25\Nplus Y squared
Dialogue: 0,0:09:32.52,0:09:35.65,Default,,0000,0000,0000,,is equal to
Dialogue: 0,0:09:35.65,0:09:42.39,Default,,0000,0000,0000,,169. And so\NY squared must be
Dialogue: 0,0:09:42.39,0:09:44.80,Default,,0000,0000,0000,,equal to 144.
Dialogue: 0,0:09:45.57,0:09:51.46,Default,,0000,0000,0000,,And now if we just turn over the\Npage and take that last line
Dialogue: 0,0:09:51.46,0:09:56.94,Default,,0000,0000,0000,,again, Y squared is equal to\N144. We need to take the square
Dialogue: 0,0:09:56.94,0:10:02.41,Default,,0000,0000,0000,,root of both sides. So why must\Nbe the square root of 144?
Dialogue: 0,0:10:02.51,0:10:06.18,Default,,0000,0000,0000,,12 Will do.
Dialogue: 0,0:10:06.70,0:10:13.38,Default,,0000,0000,0000,,One more. Let's take our\Nright angle triangle here, like
Dialogue: 0,0:10:13.38,0:10:20.41,Default,,0000,0000,0000,,this. And let's say this is\N17 and this is 15.
Dialogue: 0,0:10:20.93,0:10:27.76,Default,,0000,0000,0000,,This one is the one we want to\Nknow now. It's still a right
Dialogue: 0,0:10:27.76,0:10:32.64,Default,,0000,0000,0000,,angle triangle, the hypotenuse\Nis still the side that is
Dialogue: 0,0:10:32.64,0:10:38.01,Default,,0000,0000,0000,,opposite to the right angle.\NStill, the longest side in the
Dialogue: 0,0:10:38.01,0:10:40.45,Default,,0000,0000,0000,,triangle. Let's label our sides.
Dialogue: 0,0:10:41.34,0:10:46.64,Default,,0000,0000,0000,,And again the labelings\Ndifferent but the relationship
Dialogue: 0,0:10:46.64,0:10:54.60,Default,,0000,0000,0000,,remains the same P squared plus\NQ Squared is equal to R-squared.
Dialogue: 0,0:10:54.60,0:10:57.92,Default,,0000,0000,0000,,So let's substitute in our
Dialogue: 0,0:10:57.92,0:11:01.37,Default,,0000,0000,0000,,numbers. P squared that's
Dialogue: 0,0:11:01.37,0:11:08.32,Default,,0000,0000,0000,,15 squared. Q is what\Nwe're trying to find sales plus
Dialogue: 0,0:11:08.32,0:11:15.14,Default,,0000,0000,0000,,Q squared is equal to R-squared\Nand R is 17. So we have
Dialogue: 0,0:11:15.14,0:11:18.29,Default,,0000,0000,0000,,17 squared. Now we need some
Dialogue: 0,0:11:18.29,0:11:21.20,Default,,0000,0000,0000,,numbers. Q
Dialogue: 0,0:11:21.20,0:11:29.11,Default,,0000,0000,0000,,squared equals.\N15 squared, that's
Dialogue: 0,0:11:29.11,0:11:36.18,Default,,0000,0000,0000,,225. And 17 squared\Nis 289. You can work these two
Dialogue: 0,0:11:36.18,0:11:41.57,Default,,0000,0000,0000,,out on a Calculator or you can\Nuse long multiplication. Doesn't
Dialogue: 0,0:11:41.57,0:11:44.51,Default,,0000,0000,0000,,matter which. Now let's find Q
Dialogue: 0,0:11:44.51,0:11:50.65,Default,,0000,0000,0000,,squared. That's equal to 299\Ntakeaway 225, which is 64. And
Dialogue: 0,0:11:50.65,0:11:56.25,Default,,0000,0000,0000,,so taking the square roots again\NQ is equal to 8.
Dialogue: 0,0:11:57.33,0:12:03.03,Default,,0000,0000,0000,,In each of these three, the\Nanswers of being exact, they've
Dialogue: 0,0:12:03.03,0:12:05.10,Default,,0000,0000,0000,,been whole number values.
Dialogue: 0,0:12:05.93,0:12:11.00,Default,,0000,0000,0000,,And these whole number triples\Nor pythagorean triples as we
Dialogue: 0,0:12:11.00,0:12:13.91,Default,,0000,0000,0000,,call them. Occur quite
Dialogue: 0,0:12:13.91,0:12:18.09,Default,,0000,0000,0000,,frequently. But you probably\Nwon't be lucky when you do
Dialogue: 0,0:12:18.09,0:12:21.14,Default,,0000,0000,0000,,questions you in getting an\Nexact answer, you almost
Dialogue: 0,0:12:21.14,0:12:24.87,Default,,0000,0000,0000,,certainly have to use a\NCalculator for many of them, and
Dialogue: 0,0:12:24.87,0:12:29.28,Default,,0000,0000,0000,,you have to decide to shorten\Nthe answer to a given number of
Dialogue: 0,0:12:29.28,0:12:32.67,Default,,0000,0000,0000,,decimal places or a given number\Nof significant figures. I've
Dialogue: 0,0:12:32.67,0:12:36.74,Default,,0000,0000,0000,,used whole numbers 'cause it's\Neasier for me to work with them
Dialogue: 0,0:12:36.74,0:12:40.80,Default,,0000,0000,0000,,on this page, but do remember\Nthey won't always be exact. In
Dialogue: 0,0:12:40.80,0:12:44.87,Default,,0000,0000,0000,,fact, more than likely they\Nwon't be exact and you will have
Dialogue: 0,0:12:44.87,0:12:46.23,Default,,0000,0000,0000,,to work them out.
Dialogue: 0,0:12:46.27,0:12:47.63,Default,,0000,0000,0000,,Using a Calculator.
Dialogue: 0,0:12:48.35,0:12:52.68,Default,,0000,0000,0000,,Let's have a look at another\Napplication of Pythagoras.
Dialogue: 0,0:12:53.22,0:12:59.72,Default,,0000,0000,0000,,Supposing we know the dimensions\Nof a box, let's say we know that
Dialogue: 0,0:12:59.72,0:13:01.72,Default,,0000,0000,0000,,it's 3 by 4.
Dialogue: 0,0:13:02.39,0:13:05.45,Default,,0000,0000,0000,,By 12 so that means\Nwe know that it's.
Dialogue: 0,0:13:06.69,0:13:07.69,Default,,0000,0000,0000,,Three wide.
Dialogue: 0,0:13:08.84,0:13:12.26,Default,,0000,0000,0000,,For long. And.
Dialogue: 0,0:13:13.54,0:13:14.50,Default,,0000,0000,0000,,12 high
Dialogue: 0,0:13:15.66,0:13:18.66,Default,,0000,0000,0000,,Let's complete our box.
Dialogue: 0,0:13:19.64,0:13:27.08,Default,,0000,0000,0000,,So it's 3 by 4\Nby 12 high. Let's make
Dialogue: 0,0:13:27.08,0:13:34.52,Default,,0000,0000,0000,,it look a little bit\N3 dimensional by putting in.
Dialogue: 0,0:13:35.34,0:13:38.43,Default,,0000,0000,0000,,In the dotted line, the bits\Nthat we can't see.
Dialogue: 0,0:13:39.53,0:13:44.81,Default,,0000,0000,0000,,Question. How big is the\Ndiagonal of the box? The
Dialogue: 0,0:13:44.81,0:13:49.63,Default,,0000,0000,0000,,diagonal not that runs across a\Nface but runs from one corner
Dialogue: 0,0:13:49.63,0:13:53.65,Default,,0000,0000,0000,,across to another corner. How\Ncan we work this out?
Dialogue: 0,0:13:54.51,0:14:01.18,Default,,0000,0000,0000,,Which is a cuboid. Each of these\Nsides is a rectangle. All the
Dialogue: 0,0:14:01.18,0:14:06.31,Default,,0000,0000,0000,,joints are at right angles.\NLet's put this diagonal in.
Dialogue: 0,0:14:07.19,0:14:12.97,Default,,0000,0000,0000,,So there it goes spanning the\Nbox well across here, down to
Dialogue: 0,0:14:12.97,0:14:17.79,Default,,0000,0000,0000,,here and across here we've got\Nanother right angle triangle.
Dialogue: 0,0:14:18.96,0:14:24.31,Default,,0000,0000,0000,,So we know that if we knew that,\Nwe could find that. But this
Dialogue: 0,0:14:24.31,0:14:29.27,Default,,0000,0000,0000,,gives us a right angle triangle\Ndown here. And we do know these
Dialogue: 0,0:14:29.27,0:14:33.86,Default,,0000,0000,0000,,two sides, so let's see if we\Ncan work that out. Let's.
Dialogue: 0,0:14:34.40,0:14:38.100,Default,,0000,0000,0000,,Make clear we've got a right\Nangle triangle here and one
Dialogue: 0,0:14:38.100,0:14:46.88,Default,,0000,0000,0000,,here. So first of all, how\Nlong is that? How long is X
Dialogue: 0,0:14:46.88,0:14:51.90,Default,,0000,0000,0000,,the diagonal of this base\Nrectangle here? Well, pythagoras
Dialogue: 0,0:14:51.90,0:14:58.59,Default,,0000,0000,0000,,tells us that 3 squared +4\Nsquared is equal to X squared,
Dialogue: 0,0:14:58.59,0:15:01.38,Default,,0000,0000,0000,,so that's 9 + 16.
Dialogue: 0,0:15:02.13,0:15:08.55,Default,,0000,0000,0000,,Is 25 and that must mean that\NX is equal to 5.
Dialogue: 0,0:15:09.29,0:15:15.74,Default,,0000,0000,0000,,So we've now got X is 5 and\Nwe can use Pythagoras again in
Dialogue: 0,0:15:15.74,0:15:22.79,Default,,0000,0000,0000,,this triangle. 5 squared plus\N12 squared is equal to the
Dialogue: 0,0:15:22.79,0:15:29.10,Default,,0000,0000,0000,,length of the diagonal squared.\NLet's call that Y equals Y
Dialogue: 0,0:15:29.10,0:15:35.99,Default,,0000,0000,0000,,squared, so we have 25 +\N144 is equal to Y squared,
Dialogue: 0,0:15:35.99,0:15:42.88,Default,,0000,0000,0000,,so that's 169 is equal to\NY squared. So why must be
Dialogue: 0,0:15:42.88,0:15:46.32,Default,,0000,0000,0000,,the square root of 169, which
Dialogue: 0,0:15:46.32,0:15:51.79,Default,,0000,0000,0000,,is 13? So we can use the\Ntheorem of Pythagoras in three
Dialogue: 0,0:15:51.79,0:15:55.97,Default,,0000,0000,0000,,dimensions. We can use it to\Nsolve problems that are set up
Dialogue: 0,0:15:55.97,0:15:57.36,Default,,0000,0000,0000,,in 3 dimensional objects.
Dialogue: 0,0:15:58.06,0:16:01.52,Default,,0000,0000,0000,,What about making use of it in
Dialogue: 0,0:16:01.52,0:16:04.98,Default,,0000,0000,0000,,other ways? What about
Dialogue: 0,0:16:04.98,0:16:10.97,Default,,0000,0000,0000,,cartesian geometry? Cartesian\Ngeometry is geometry that set
Dialogue: 0,0:16:10.97,0:16:17.30,Default,,0000,0000,0000,,out on a plane that's got\NCartesian coordinates 11 the
Dialogue: 0,0:16:17.30,0:16:19.83,Default,,0000,0000,0000,,.22, and so on.
Dialogue: 0,0:16:20.51,0:16:24.74,Default,,0000,0000,0000,,One of the other things about\NPythagoras theorem is this.
Dialogue: 0,0:16:25.61,0:16:32.36,Default,,0000,0000,0000,,It tells us if we\Nhave a particular triangle, it
Dialogue: 0,0:16:32.36,0:16:37.08,Default,,0000,0000,0000,,tells us the relationship\Nbetween the sides.
Dialogue: 0,0:16:38.02,0:16:41.63,Default,,0000,0000,0000,,And it works in reverse if.
Dialogue: 0,0:16:42.52,0:16:44.02,Default,,0000,0000,0000,,It's.
Dialogue: 0,0:16:44.95,0:16:52.72,Default,,0000,0000,0000,,The area.\NOf the
Dialogue: 0,0:16:52.72,0:16:59.96,Default,,0000,0000,0000,,square. On\Nthe longest
Dialogue: 0,0:16:59.96,0:17:03.73,Default,,0000,0000,0000,,side. Is
Dialogue: 0,0:17:03.73,0:17:05.54,Default,,0000,0000,0000,,equal.
Dialogue: 0,0:17:06.05,0:17:11.72,Default,,0000,0000,0000,,Tool.\NArea of
Dialogue: 0,0:17:11.72,0:17:14.63,Default,,0000,0000,0000,,the squares.
Dialogue: 0,0:17:15.87,0:17:19.54,Default,,0000,0000,0000,,On the two
Dialogue: 0,0:17:19.54,0:17:24.43,Default,,0000,0000,0000,,shorter sides.\NOf
Dialogue: 0,0:17:24.43,0:17:26.88,Default,,0000,0000,0000,,a
Dialogue: 0,0:17:26.88,0:17:29.32,Default,,0000,0000,0000,,triangle.
Dialogue: 0,0:17:29.32,0:17:36.36,Default,,0000,0000,0000,,Then\NThe triangle
Dialogue: 0,0:17:36.36,0:17:39.85,Default,,0000,0000,0000,,is right
Dialogue: 0,0:17:39.85,0:17:45.97,Default,,0000,0000,0000,,angled. So this is turned\Nthe theorem round. Instead of
Dialogue: 0,0:17:45.97,0:17:51.10,Default,,0000,0000,0000,,saying if the triangles right\Nangled, then this is So what
Dialogue: 0,0:17:51.10,0:17:56.69,Default,,0000,0000,0000,,it's saying is if this is so,\Nthen the triangle is right
Dialogue: 0,0:17:56.69,0:18:02.28,Default,,0000,0000,0000,,angled and we can make use of\Nthis in areas like Cartesian
Dialogue: 0,0:18:02.28,0:18:07.41,Default,,0000,0000,0000,,coordinate geometry to find out\Nwhether or not a triangle is
Dialogue: 0,0:18:07.41,0:18:09.74,Default,,0000,0000,0000,,right angled or not so.
Dialogue: 0,0:18:10.26,0:18:14.00,Default,,0000,0000,0000,,Let's take the three
Dialogue: 0,0:18:14.00,0:18:16.76,Default,,0000,0000,0000,,points. 34
Dialogue: 0,0:18:17.39,0:18:23.42,Default,,0000,0000,0000,,26\NOne note and let's see
Dialogue: 0,0:18:23.42,0:18:26.71,Default,,0000,0000,0000,,if they do give us a\Nright angle triangle.
Dialogue: 0,0:18:28.23,0:18:33.47,Default,,0000,0000,0000,,So first of all, let's plot\Nthese points. First of all, will
Dialogue: 0,0:18:33.47,0:18:35.66,Default,,0000,0000,0000,,mark off on the axes.
Dialogue: 0,0:18:36.89,0:18:39.36,Default,,0000,0000,0000,,Roughly equal spaces.
Dialogue: 0,0:18:40.89,0:18:45.27,Default,,0000,0000,0000,,The\Nsame
Dialogue: 0,0:18:45.27,0:18:47.46,Default,,0000,0000,0000,,markings.
Dialogue: 0,0:18:48.66,0:18:54.74,Default,,0000,0000,0000,,On the Y axis, again\Nroughly equal spaces.
Dialogue: 0,0:18:55.27,0:18:59.77,Default,,0000,0000,0000,,And now we're plot the points.\NThe point one note will be here.
Dialogue: 0,0:18:59.77,0:19:01.84,Default,,0000,0000,0000,,Will call that the point a.
Dialogue: 0,0:19:02.80,0:19:08.68,Default,,0000,0000,0000,,The Point B will be the .26 that\Nwill be roughly there.
Dialogue: 0,0:19:09.27,0:19:16.23,Default,,0000,0000,0000,,And the Point C will be the .3\Nfour and so that will be three
Dialogue: 0,0:19:16.23,0:19:19.01,Default,,0000,0000,0000,,and four that will be roughly
Dialogue: 0,0:19:19.01,0:19:21.88,Default,,0000,0000,0000,,there. Let's join up.
Dialogue: 0,0:19:22.49,0:19:23.49,Default,,0000,0000,0000,,These points.
Dialogue: 0,0:19:25.79,0:19:28.85,Default,,0000,0000,0000,,To give us a triangle.
Dialogue: 0,0:19:29.73,0:19:33.45,Default,,0000,0000,0000,,Now it doesn't look right angle,\Nbut we don't really know 'cause.
Dialogue: 0,0:19:33.45,0:19:35.00,Default,,0000,0000,0000,,These were only rough markings.
Dialogue: 0,0:19:35.79,0:19:40.78,Default,,0000,0000,0000,,Is it a right angle triangle or\Nnot? You mustn't believe what
Dialogue: 0,0:19:40.78,0:19:45.36,Default,,0000,0000,0000,,our eyes tell us, especially\Nwhen we can do calculation which
Dialogue: 0,0:19:45.36,0:19:47.85,Default,,0000,0000,0000,,will help us to be exact.
Dialogue: 0,0:19:47.87,0:19:53.15,Default,,0000,0000,0000,,So I need to work out the\Nlengths of the sides of the
Dialogue: 0,0:19:53.15,0:19:55.18,Default,,0000,0000,0000,,triangles, so let me surround.
Dialogue: 0,0:19:57.90,0:19:59.83,Default,,0000,0000,0000,,Is triangle with.
Dialogue: 0,0:20:00.33,0:20:01.89,Default,,0000,0000,0000,,A rectangle.
Dialogue: 0,0:20:06.24,0:20:10.49,Default,,0000,0000,0000,,And because I've surrounded by a\Nrectangle, these are all.
Dialogue: 0,0:20:11.05,0:20:12.46,Default,,0000,0000,0000,,Right angles here.
Dialogue: 0,0:20:13.14,0:20:16.72,Default,,0000,0000,0000,,And I can see now what the\Nlengths of these little bits
Dialogue: 0,0:20:16.72,0:20:22.43,Default,,0000,0000,0000,,are. So let me workout AB\Nsquared to begin with. a B
Dialogue: 0,0:20:22.43,0:20:27.72,Default,,0000,0000,0000,,squared. Now I can do that\Nbecause this side here is of
Dialogue: 0,0:20:27.72,0:20:32.57,Default,,0000,0000,0000,,length 6, so that's 6 squared\Nplus this length squared here,
Dialogue: 0,0:20:32.57,0:20:34.78,Default,,0000,0000,0000,,which is a blank one.
Dialogue: 0,0:20:35.70,0:20:40.50,Default,,0000,0000,0000,,That's 36 plus one is 37, so\NI've used Pythagoras in this
Dialogue: 0,0:20:40.50,0:20:46.50,Default,,0000,0000,0000,,right angle triangle to find out\Nwhat a B is, or at least to find
Dialogue: 0,0:20:46.50,0:20:51.70,Default,,0000,0000,0000,,out what AB squared is. Let me\Ndo that again here, this time
Dialogue: 0,0:20:51.70,0:20:56.90,Default,,0000,0000,0000,,for the side BCBC squared is and\NI'm going to use this right
Dialogue: 0,0:20:56.90,0:21:02.10,Default,,0000,0000,0000,,angle triangle here, and again I\Ncan see that this length here is
Dialogue: 0,0:21:02.10,0:21:06.50,Default,,0000,0000,0000,,one, so that's one squared plus\Nthe square of this length.
Dialogue: 0,0:21:06.54,0:21:08.71,Default,,0000,0000,0000,,So that's 2 squared.
Dialogue: 0,0:21:09.30,0:21:16.11,Default,,0000,0000,0000,,That's 1 + 4 gives\Nmain 5. Finally, let's do
Dialogue: 0,0:21:16.11,0:21:18.15,Default,,0000,0000,0000,,a C squared.
Dialogue: 0,0:21:19.23,0:21:25.80,Default,,0000,0000,0000,,And again, I can see that this\Nlength here is of length 4, so
Dialogue: 0,0:21:25.80,0:21:31.42,Default,,0000,0000,0000,,that's 4 squared plus the square\Nof this length here. That's 2
Dialogue: 0,0:21:31.42,0:21:37.05,Default,,0000,0000,0000,,squared is equal to 16 + 4, and\Nthat gives me 20.
Dialogue: 0,0:21:37.58,0:21:43.61,Default,,0000,0000,0000,,Now, if this is a right angle\Ntriangle, if I take the squares
Dialogue: 0,0:21:43.61,0:21:48.72,Default,,0000,0000,0000,,of the lengths of the two\Nshorter sides, that's a CNBC.
Dialogue: 0,0:21:50.03,0:21:54.32,Default,,0000,0000,0000,,Add the squares of those\Nlens together. I should get
Dialogue: 0,0:21:54.32,0:21:59.04,Default,,0000,0000,0000,,the square of the length of\Nthe third side. If it's
Dialogue: 0,0:21:59.04,0:22:03.76,Default,,0000,0000,0000,,right angle and if I don't\Nthen it's not right angle,
Dialogue: 0,0:22:03.76,0:22:09.34,Default,,0000,0000,0000,,so let's do that AC squared\Nplus BC squared is 5 + 20.
Dialogue: 0,0:22:11.15,0:22:17.52,Default,,0000,0000,0000,,5 for BC squared, 20 for AC\Nsquare, and that gives me 25.
Dialogue: 0,0:22:19.13,0:22:26.73,Default,,0000,0000,0000,,But a B squared is\N37. That does not equal
Dialogue: 0,0:22:26.73,0:22:34.33,Default,,0000,0000,0000,,37 and therefore the triangle\NABC is not right angle.
Dialogue: 0,0:22:34.34,0:22:41.53,Default,,0000,0000,0000,,The fact that Pythagoras theorem\Nis about squares is fairly well
Dialogue: 0,0:22:41.53,0:22:48.07,Default,,0000,0000,0000,,known. What's not as well known,\Nalthough it's fairly obvious
Dialogue: 0,0:22:48.07,0:22:50.69,Default,,0000,0000,0000,,once you've seen it.
Dialogue: 0,0:22:51.36,0:22:57.86,Default,,0000,0000,0000,,Is that if you take any regular\Nfigure or similar figures and
Dialogue: 0,0:22:57.86,0:23:03.28,Default,,0000,0000,0000,,place them on the sides of a\Nright angle triangle?
Dialogue: 0,0:23:03.97,0:23:10.39,Default,,0000,0000,0000,,Then the area of the figure\Non the hypotenuse is equal to
Dialogue: 0,0:23:10.39,0:23:17.34,Default,,0000,0000,0000,,the sum of the areas of the\Nsame figures on the other two
Dialogue: 0,0:23:17.34,0:23:18.42,Default,,0000,0000,0000,,shorter sides.
Dialogue: 0,0:23:19.62,0:23:24.93,Default,,0000,0000,0000,,Let's make it perhaps as\Ndramatic as we can. Supposing
Dialogue: 0,0:23:24.93,0:23:30.77,Default,,0000,0000,0000,,report semi circles on the sides\Nof this right angle triangle.
Dialogue: 0,0:23:31.61,0:23:38.67,Default,,0000,0000,0000,,Is it the case that the area\Nof the semi circles on the two
Dialogue: 0,0:23:38.67,0:23:44.71,Default,,0000,0000,0000,,shorter sides add up to the area\Nof the semicircle on the
Dialogue: 0,0:23:44.71,0:23:49.25,Default,,0000,0000,0000,,hypotenuse? Is that so? Well,\Nlet's label the sides.
Dialogue: 0,0:23:49.89,0:23:53.31,Default,,0000,0000,0000,,Play, it'll be, it'll
Dialogue: 0,0:23:53.31,0:24:00.77,Default,,0000,0000,0000,,see. And workout the area area\Nof a is now the area of a
Dialogue: 0,0:24:00.77,0:24:06.64,Default,,0000,0000,0000,,semicircle is half the area of\Nthe circle and the area of a
Dialogue: 0,0:24:06.64,0:24:12.07,Default,,0000,0000,0000,,circle is pie are squared π\Ntimes the square of the radius.
Dialogue: 0,0:24:12.07,0:24:18.39,Default,,0000,0000,0000,,Now the diameter of the circle\Nis a, so the radius is a over
Dialogue: 0,0:24:18.39,0:24:20.20,Default,,0000,0000,0000,,2, so that's π.
Dialogue: 0,0:24:21.04,0:24:27.42,Default,,0000,0000,0000,,Times a over 2 squared. If we\Nwork that out, that's Pi over 8A
Dialogue: 0,0:24:27.42,0:24:33.35,Default,,0000,0000,0000,,squared, over 2 all squared is a\Nsquared over 4. That's where the
Dialogue: 0,0:24:33.35,0:24:39.74,Default,,0000,0000,0000,,a square comes from and the two\Ntimes by the four gives us the
Dialogue: 0,0:24:39.74,0:24:46.42,Default,,0000,0000,0000,,8. Be what's the area\Nof this semicircle here? Well,
Dialogue: 0,0:24:46.42,0:24:52.70,Default,,0000,0000,0000,,again. The diameter of the\Nsemicircle is B, so the radius
Dialogue: 0,0:24:52.70,0:24:59.90,Default,,0000,0000,0000,,is be over 2, so the area\Nis 1/2 Pi R-squared and are. We
Dialogue: 0,0:24:59.90,0:25:07.09,Default,,0000,0000,0000,,just said was be over 2, so\Nwe square that and we have π
Dialogue: 0,0:25:07.09,0:25:09.66,Default,,0000,0000,0000,,over 8B squared, this one.
Dialogue: 0,0:25:10.59,0:25:18.16,Default,,0000,0000,0000,,Same again, C equals 1/2.\NThai R-squared and R is
Dialogue: 0,0:25:18.16,0:25:25.73,Default,,0000,0000,0000,,1/2 the diameter 1/2 of\NCC over 2 all squared,
Dialogue: 0,0:25:25.73,0:25:29.52,Default,,0000,0000,0000,,and so that's Pi over
Dialogue: 0,0:25:29.52,0:25:36.84,Default,,0000,0000,0000,,8C squared. So let's add\Nthese together A&B and see what
Dialogue: 0,0:25:36.84,0:25:44.62,Default,,0000,0000,0000,,we get a is π over 8A\Nsquared. B is π over 8B squared,
Dialogue: 0,0:25:44.62,0:25:51.83,Default,,0000,0000,0000,,and there's a common factor here\Nof π over 8. We can take
Dialogue: 0,0:25:51.83,0:25:57.38,Default,,0000,0000,0000,,out, leaving us with A squared\Nplus B squared, but.
Dialogue: 0,0:25:58.00,0:26:02.06,Default,,0000,0000,0000,,The theorem of Pythagoras tells\Nus that in a right angle
Dialogue: 0,0:26:02.06,0:26:06.12,Default,,0000,0000,0000,,triangle, a squared plus B\Nsquared is C squared, and so
Dialogue: 0,0:26:06.12,0:26:08.70,Default,,0000,0000,0000,,that's π over 8 times by C
Dialogue: 0,0:26:08.70,0:26:11.90,Default,,0000,0000,0000,,squared. Which is just say.
Dialogue: 0,0:26:12.58,0:26:18.42,Default,,0000,0000,0000,,And so the question that we\Nasked at the top is true. This
Dialogue: 0,0:26:18.42,0:26:22.91,Default,,0000,0000,0000,,does work for semi circles. It\Nwould work for equilateral
Dialogue: 0,0:26:22.91,0:26:27.40,Default,,0000,0000,0000,,triangles. It would work for\Nhexagons. It would work for
Dialogue: 0,0:26:27.40,0:26:29.19,Default,,0000,0000,0000,,similar isosceles triangles and
Dialogue: 0,0:26:29.19,0:26:34.71,Default,,0000,0000,0000,,so on. Pythagoras theorem is\Nquite a wonderful theorem, and
Dialogue: 0,0:26:34.71,0:26:39.64,Default,,0000,0000,0000,,there's quite a lot to be gained\Nfrom exploring pythagorean
Dialogue: 0,0:26:39.64,0:26:46.05,Default,,0000,0000,0000,,triple, such as three 455-1213\Nand the other one that we saw in
Dialogue: 0,0:26:46.05,0:26:48.02,Default,,0000,0000,0000,,this lecture, eight 1570.