How do you decompose a polygon?

1 answer:

4 0

<span>How to decompose polygons to find the area. I'm Bon Crowder and we're talking about taking apart polygons so we can look at their areas. So here we have two polygons. We have a trapezoid that's kind of awkward looking and then we have this other kind of crazy polygon. So when you want to find the area it's like trying to figure out how much carpet you need to put on the floor of a very strange room. Here you can remember the area for the trapezoid but if you don't remember that and you don't have Google on hand really quickly, you can go OK well if I chop off that triangle and chop off that triangle then I have a triangle, a rectangle and a triangle. And if you remember that your area is one half base times height for the two triangles and the area of the rectangle is length times width. Then you can find each area and add them together. For this crazy guy we can do this one of two ways. We can either consider the extra triangle here and then we have the area of the whole rectangle and then the area of the triangle and then subtract or we can look at the area of the rectangle which is this one and then two triangles and then add. So there's a couple of different ways to decompose that one. I'm Bon Crowder and that's how you decompose polygons to find the area</span>

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What is the antiderivative of sin^2(x)cos^2(x)?

marin [14]

Answer:

$\frac{x}{8}-\frac{\sin(4x)}{32}+C$

Step-by-step explanation:

[Most of the work here comes from manipulating the trig to make the term (integrand) integrable.]

Recall that we can express the squared trig functions in terms of cos(2x). That is,

$\cos(2x)=2\cos^2x-1 \\ \cos(2x)=1 - 2\sin^2x.$

And so inverting these,

$\cos^2x=\frac{1}{2} (1+\cos2x) \\ \sin^2x=\frac{1}{2} (1-\cos2x)$ .

Multiply them together to obtain an equivalent expression for sin^2(x)cos^2(x) in terms of cos(2x).

$\sin^2x \cdot \cos^2x =\frac{1}{2} (1-\cos2x) \cdot \frac{1}{2} (1+\cos2x) = \frac{1}{4}(1-\cos^2(2x)).$

Notice we have cos^2(2x) in the integrand now. We've made it worse! Let's try plugging back in to the first identity for cos^2(2x).

$\cos(2x)=2\cos^2x-1 \Rightarrow \cos(4x)=2\cos^2(2x)-1 \Rightarrow \cos^2(2x) = \frac{1}{2}(1+\cos(4x))$

So then,

$\sin^2x \cdot \cos^2x = \frac{1}{4}(1-\cos^2(2x)) = \frac{1}{4}(1-\frac{1}{2}(1+\cos(4x))) = \frac{1}{4}(1-\frac{1}{2}-\frac{1}{2}\cos(4x))=\frac{1}{8}(1-\cos(4x)).$