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USING STRUCTURE A semicircle has endpoints $(-2,-5)$ and $(2,8) .$ Find the arc length of the semicircle.

7.85

Geometry

Chapter 11

Circumference, Area, and Volume

Section 1

Circumference and Arc Length

Area and Perimeter

Surface Area

Volume

Circles

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University of Michigan - Ann Arbor

University of Nottingham

Lectures

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In $28-37, \theta$ is the …

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GEOMETRY A circle has a ra…

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Arc length Find the length…

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So first we need to find where the center of the circle is. We know that the cheap points that are on the semicircle has a diameter of starting at negative to five and to eight. So the center would be the middle of this diameter of the mid point. So we can use our midpoint formula where we add the exits and tonight by two and add the wives and divide by two. And this becomes our center. This, if we do the calculations, is zero and 6.5. So let's put that into our formula for our circle. Radius squared is equal to set of X is gonna put zero minus two. And instead of why, I'm gonna put 6.5 and then minus eight. So I'm plugging in this point into this equation squared. So if I do the calculations, I can square root both sides and everything goes underneath. The radical negative two squared becomes 46.5 minus eight is negative. 1.5 y squared. I get 2.25 and then when I square root 6.25 get 2.5. So that's my radius. So then using that information about another, she hear my art length is gonna be 100 80 over 360 which is half. And then I want to multiply that times my circumference two times pi times radius to 25. When I do all that, I get about seven points 85

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