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Master Chord Properties: Key Concepts in Circle Geometry

CC.HSF.TF.A.1

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Converting between degrees and radians

Trigonometric ratios of angles in radians

Radian measure and arc length

CC.HSF.TF.A.2

Unit circle

CC.HSF.TF.A.3

Solving expressions using 45-45-90 special right triangles

Solving expressions using 30-60-90 special right triangles

Find the exact value of trigonometric ratios

CC.HSF.TF.B.5

Graphing transformations of trigonometric functions

Determining trigonometric functions given their graphs

CC.HSF.TF.B.6

Find the difference of squares: (a - b)(a + b) = (a^2 - b^2)

Find the difference of squares

Solving two-step linear equations using addition and subtraction: ax + b = c

Two-step Linear Equations

Two-step Linear Equations LHS

Solve Linear Equations Using Addition and Subtraction – Word Problem

Two-step Linear Equations — Word Problem

CC.HSF.TF.B.7

Evaluating inverse trigonometric functions

CC.HSF.TF.C.8

Pythagorean identities

CC.HSF.TF.C.9

Sum and difference identities

Sum and Difference Identities

Double-angle identities

Cofunction identities

CC.HSG.C.B.5

CC.HSG.CO.A.1

Line symmetry

Angles in a circle

Chord properties

Circles and circumference

Circle properties

Parallel and perpendicular line segments

Identifying Parallel Lines

Identifying Perpendicular Lines

CC.HSG.SRT.C.6

Use sine ratio to calculate angles and sides (Sin = o / h)

Use tangent ratio to calculate angles and sides (Tan = o / a)

CC.HSG.SRT.C.7

Use cosine ratio to calculate angles and sides (Cos = a / h)

CC.HSG.SRT.C.8

Combination of SohCahToa questions

Word problems relating ladder in trigonometry

Word problems relating guy wire in trigonometry

CC.HSG.SRT.D.10

Law of sines

Ambiguous case: SSA triangles

Law of cosines

CC.HSG.SRT.D.11

Applications of the sine law and cosine law

CC.HSG.SRT.D.9

Area of triangles: 1/2 a*b sin(C)

math

Intro

- Hi, welcome to this question right here. So, we know that A to C is four centimeters. So, let me just put that down right here for you. So, there's four centimeters from here to here, okay? And B to D is from here to here, it's 15 centimeters in total. But instead of me doing the 15, I'm going to divide it into two first, okay? Because. . . So, 15 divided by two is 7.5 but, because this is a bisector so, these two sides right here, the two segments here and here, they're both equal. So, the total is 15, if you divide it by two you get 7.5 from here and it's 7.5 here as well, okay? So, we're still good? 
Take a look, they're looking for the diameter of the circle but, we can actually draw our own radius. See, from here to here, A to B, the reason why I actually draw a line right here is because I want to form a right triangle, okay? 
So, there's our right triangle right here. I know this side's a four, this is my right angle 90 degrees. And I know this side's 7.5. This is actually known as your C, which is the radius or the hypotenuse, the C or the hypotenuse, right? 
So, why are we looking for this? Because this is your radius, okay? So, why are we looking for the radius? Well, if you can figure out your radius times by two, what do you have? Diameter, right? So, that's why we need to find the radius first. So to find the C, remember you have the Pythagoras theorem. So, we've got four square plus 7.5 square and just find the C. All you have to do is square root the whole thing, once you figure this out. So, grab your calculator and go four square, plus 7.5 square and don't forget to square root the whole thing. I have 8.5. Now, remember, this is your radius, okay? 
The radius is 8.5, what about diameter? 
Well, why don't we just times by two. So, the diameter for this question is just 8.5 times by two, the answer turns out to be just 17 centimeter. And that is the answer for this question. Thanks for watching.

In a circle, when a line passes through the center of the circle and is perpendicular to a chord, the line will bisect the chord. In other words, the perpendicular bisector of a chord always passes through the center of the circle. 