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A circle has center π and radius 13 centimeters.
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A line passes through the points π΅, πΆ, and π·, where πΆ and π· are on the circle, π΅ is 25 centimeters from the point π, and πΆπ΅ equals πΆπ·.
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Calculate the length of line segment πΆπ· and the perpendicular distance π₯ between the line and the point π.
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Round your answers to two decimal places.
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Letβs start by sketching an image.
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We have a circle with center π.
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And then we have three points π΅, πΆ, and π·.
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A line passes through all three of these points.
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But only πΆ and π· are on the circle.
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If we have a line that looks like this, we know that πΆ and π· lie on the circle.
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And we know that the distance from πΆ to π· is equal to the distance from π΅ to πΆ.
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For now, letβs label the distance from πΆ to π΅ and the distance from πΆ to π· as lowercase π.
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We know that thereβs a line segment from π΅ to π.
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This distance is 25 centimeters.
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We want to find the perpendicular distance between the segment πΆπ· and point π.
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And weβve been given that that is π₯.
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So now we have that π is equal to πΆπ·, which is equal to πΆπ΅, and that π₯ is the perpendicular distance from π to πΆπ·.
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Weβre trying to find both of these values.
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In order to do this, letβs remember that this circle has a radius of 13 centimeters.
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This means the distance from πΆ to π is equal to the distance from π· to π.
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Both of these values are 13 centimeters.
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This means we can conclude that triangle πΆππ· is an isosceles triangle.
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Weβll label the perpendicular intersection between the line that passes through π and the line πΆπ· be point π΄.
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When we do this, we can say that the line segment ππ΄ bisects the line segment πΆπ· because the height of an isosceles triangle bisects the base.
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Weβve labeled the distance πΆπ· π.
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If we have some point that bisects that, then we can say that segment πΆπ΄ is equal to segment π·π΄.
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And weβll label that as π over two.
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From here, weβll have to consider some right triangle values.
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We have the larger right triangle π΅π΄π.
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This triangle has its longest side, its hypotenuse, as 25 centimeters.
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It has a side of length π₯.
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And then its final side will be made up of π plus π over two, making its third side equal in length to three π over two.
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We also have a smaller right triangle, triangle πΆπ΄π.
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The hypotenuse of this triangle is one of the radiis of this circle πΆπ, and itβs 13 centimeters.
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It shares the side π΄π, which is π₯ in length.
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Its final side has a length of π over two.
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At this point, we want to come up with some equations that will help us solve for π₯ and π.
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To do this, we can use the Pythagorean theorem, which says π squared equals π squared plus π squared, where π is the hypotenuse and π and π are the other two sides in a right triangle.
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Substituting what we know for triangle π΅π΄π, we get 25 squared equals π₯ squared plus three π over two squared.
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And for triangle πΆπ΄π, we have 13 squared equals π₯ squared plus π over two squared.
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If we square all the terms we know, we recognize that each of these equations has an π₯ squared term.
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We can rearrange this first equation by subtracting nine π squared over four from both sides of the equation, giving us π₯ squared is equal to 625 minus nine π squared over four.
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We can take the value that weβve found for π₯ squared and plug it in to our second equation, which now says 169 equals 625 minus nine π squared over four plus π squared over four.
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Combining these like terms, negative nine π squared over four plus π squared over four is negative eight π squared over four.
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And eight over four is two.
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Next, we subtract 625 from both sides of the equation, which gives us negative 456 is equal to negative two π squared.
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Dividing both sides of the equation by negative two, and we have π squared is equal to 228.
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Taking the square root of both sides gives us π equals 15.099 and so on.
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Rounding that to two decimal places gives us 15.10 centimeters.
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Even though we took the square root, because weβre dealing with length, weβre only interested in the positive square root value for π.
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Remember, weβve already written an equation for π₯ in terms of π: π₯ squared equals 625 minus nine π squared over four.
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When solving for π₯, we want to use our nonrounded π-value.
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This maintains our accuracy as weβre only rounding in the final step.
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When we plug this in our calculator, we find that π₯ squared equals 112.
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And the square root of that value is 10.583 continuing.
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Rounded to two decimal places, this becomes 10.58 centimeters.
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So, to summarize, the length of πΆπ· is what weβve labeled on our diagram as π.
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And that means πΆπ· is equal to 15.10 centimeters to two decimal places.
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And the perpendicular distance π₯ to two decimal places is 10.58 centimeters.