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有趣的周長與面積問題 (英) : 三個用到周長與面積的例題
- Lets do some example problems here, so we have the perimeter
- of each of the outer triangles is 30. So for example if I took
- The sum of this side, this side, and that side I will get 30 and that is
- true of all these outer triangles, these 5 outer triangles.
- They then tell us that the perimeter of FGHIJ
- So FGHIJ the perimeter of this pentagon right over here is 50
- So if I add up that side plus that side plus that side plus that side, plus that side,
- I get 50. And then they say what is the perimeter of the star?
- So the perimeter of the star is really the outsides. if you take the bases
- away of each of these triangles. So it is this side, let me do this in a new color actually
- So the perimeter of the triangle i will do in orange. It is going to be this
- plus that plus that plus that plus that plus that
- i think you get the idea plus that plus that plus that plus that
- So the perimeter of the of the star so let me call this: perimeter perimeter of the star
- it is going to be equal to the perimeter of the 5 triangles
- is equal to perimeter of 5 outer triangles. Just call them 5 triangles like this
- minus their basis, right, if i take the perimeter of all of these sides
- If i added up the part that should not be part of the perimeter of the star
- would be this part,that part, that part,that part, that part and that part.
- those are not the part, those are not the part of the perimeter of the star
- so should be the perimeter of the 5 triangles minus the links of their bases
- links of their 5 bases. So what is the perimeter of the 5 triangles?
- well, the perimeter of each of them is 30, perimeter of 5 of them is going to be 5 times 30
- which is 150, now we want to subtract out the links of their 5 bases
- now the links of their 5 bases if we add them up is the exact perimeter of this in inner pentagon
- right over here. So this inner pentagon has a perimeter 50, that is the sum of the 5 bases.
- So that right over here is 50, so the perimeter of the star is going to be 150 minus 50, or or 100.
- All we need is to get the perimeter of all triangles, subtracted out these bases
- which was the perimeter of the inner pentagon and we are done.
- Now lets do the next problem.
- What is the area of this this quadrilateral, something that has 4 sides of ABCD?
- And this is a little bit we have not seen a figure quite like this just yet,
- it on the right hand side looks like a rectangle, and on the left hand side looks like a triangle
- and this is actually trapezoid, but we can actually as you could imagine
- the way we figure out the area of several triangles splitting it up into pieces we can recognise.
- And the most obvious thing to do here is started A and just drop a rock
- drop an altitude right over here, and so this line right over here is going to hit
- at 90 degrees and we could call this point E.
- And what is interesting here is we can split this up into something we recognize
- a rectangle and a right triangle. But you might say how do, how do we figure out what these you know
- we have this side and that side, so we can figure out the area
- of this rectangle pretty straight forwardly. But how would we, how would we figure out the area of this
- triangle? Well if this side is 6 then that means that this that EC is also going to be 6.
- If AB is 6, notice we have a rectangle right over her, opposite side of a rectangle are equal.
- So if AB equals 6, implies that EC is equal to 6, EC is equal to 6, so EC is equal to 6
- and if EC is equal to 6 then that tells us that DE is going to be 3.
- DE is going to be 3, this distance right over here is going to be 3.
- And we know that because if this is 6, this has to be something that we add to 6 to get 9,
- 9 was the length of this entire, of the entire base of this figure right over here.
- 9 was this entire distance so 9 minus 6 gives us 3, and now we have all the information that we need
- to figure out the area. The area of this part right over here of this rectangle is
- just going to be 6 times 7, so is going to be equal to 42 plus the area of this triangle right over here.
- Plus the area of this triangle right over here, and that is one half base times height one half.
- The base over her is 3, one half times 3 and the height over here is once again going to be 7
- this is a rectangle, opposite sides are equal, so if this is 7, this is also going to be 7
- one half times 3 times 7, so it is going be 42, lets see.
- 3 times 7 is 21, 21 divided by 2 is 10.5, 10.5 so this is going to be equal to 52.5
- 52.5 is the area of this entire figure.
- Lets do one more.
- So here I have a bizarre looking, a bizarre looking shape, and we need to figure out its perimeter.
- And it it first seems very daunting because they have only given us this side and this side
- and they have only given us this side right over here.
- And one thing that we are allowed to assume in this and you don't always have to make
- you can't always make that assumption and I just didn't draw it here
- I had time because it would had really crowded out this this diagram.
- Is it all of the angles in this diagrams are right angles,so i could have drawn a right angle here
- a right angle here, a right angle there, right angle there, but as you can see
- it kind of makes things a little bit, it makes things a little bit messy.
- But how do we figure out the perimeter if we don't know these little distances,
- if we don't know these little distances here. And the secret here is to kind of shift the sides
- because all we want to care about is the sum of the sides of the sides.
- So what I will do is a little exercise in shifting the sides. So this side over here I am going to shift
- it and put it right up there, then this side right over here, this length right over here
- I am going to shift and put it right over there. Then let me keep using different colors, and then this
- side right over here I am going to shift it and put it right up here. Then finally Iam going to have
- this side right over here, I can shift it and put it right over there and I think you see
- what is going on right now. Now all of these sides combined are going to be the same as this side
- kind of building, even you know this thing was not a rectangle,its its perimeter is going to be a little bit interesting.
- All we have to think about is this 2 right over here, now lets think about all of these sides that is
- going up and down. So this side i can shift it all the way to the right and go right over here.
- Let me make it clear all inside goes all the way to the end, right that it is the exact same all insde.
- Now this white side I can shift all the way to the right over there, then this green side I can shift
- right over there and then I have, and then I can shift, and then i can shift this.
- Actually let me not shift that green side yet, let me just leave that green side
- so I have not, I have not done anything yet, let me be clear I have not done anything yet with that and
- that I have not shift them over and let me take this side right over here and shift it over.
- So let me take this entire thing and shift over there and shift it over there.
- So before I count these two pieces right over here and we know that each have length 2
- this 90 degrees angle, so this has link to and this has link to.
- Before I count these two pieces, I shifted everything else so I was able to form a rectangle.
- So at least counting everything else I have 7 plus 6, so lets see 7 plus 6
- all of these combined are also going to be 7, plus 7, and all of these characters combined are all also going to be 6,
- plus 6, and then finally I have this 2, right here that I have not counted before, this 2,
- plus this 2, plus this 2. And then we have our perimeter, so what is this giving us,
- 7 plus 6 is 13, plus 7 is 20, plus 6 is 26, plus 4 more is equal to 30.
- And we are done.