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A machinist creates a washer by drilling a hole through the center of a circular piece of metal. If the piece of metal has a radius of x + 10 and the hole has a radius of x + 6, what is the area of the washer?

Mathematics
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ok so you have two circles. The area of the washer is the area of the piece of metal minus the area of the hole.\[A_{washer}=A_{metal}-A_{hole}\]
both are circles and the area of a circle with radius \(r\) is \(A=\pi r^2\)
so let \(r\) be the radius of the hole, and \(R\) be the radius of the metal. Then you have\[A_{washer}=\pi R^2-\pi r^2=\pi\left(R^2-r^2\right)\]

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Other answers:

Ok so from here there are a few ways you can go. I prefer just to plug in the values, expand it out, and simplify\[A_{washer}=\pi\left((x^2+20x+100)-(x^2+12x+36)\right)\]\[A_{washer}=\pi\left(20x+100-12x-36\right)\]\[A_{washer}=\pi\left(8x+100-36\right)\]\[A_{washer}=\pi\left(8x+64\right)\]
\[A_{washer}=8\pi(x+8)\]
this is of course assuming that the washer is 2D.

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