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jbarber
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?
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\[A_{washer}=\pi R^2-\pi r^2=\pi\left(R^2-r^2\right)\]
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.