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Master Chord Properties: Key Concepts in Circle Geometry

NL.SO.1202A1

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Word problems relating ladder in trigonometry

Word problems relating guy wire in trigonometry

NL.SO.1202G2

Solving expressions using 45-45-90 special right triangles

Solving expressions using 30-60-90 special right triangles

Pythagorean theorem

Pythagorean Theorem

Using the pythagorean relationship

Applications of pythagorean theorem

NL.SO.1202G3

Similar triangles

Similar polygons

Isosceles and equilateral triangles

NL.SO.1202G4

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

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

Combination of SohCahToa questions

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

NL.SO.1202G5

Parallel and perpendicular lines in linear functions

Angles in a circle

Chord properties

Tangent properties

Pairs of lines and angles

Parallel lines and transversals

Parallel line proofs

Perpendicular line proofs

NL.SO.1202M1

Metric systems

Add and Subtract Mixed Units

Metric Length Conversion

Add or Subtract Different Metric Units Convert to CM

Add or Subtract Different Metric Units Convert to M

Add or Subtract Different Metric Units Convert to MM

Conversions involving squares and cubic

Imperial to Metric Conversions — Squares and Cubes

Imperial Conversion — Squares and Cubes

Metric to Imperial Conversion — Squares and Cubes

Metric Area Conversion FT to CM

Metric Area Conversion Foot to Inch

Metric Area Conversion M/CM to Inch

NL.SO.1202M2

Conversions between metric and imperial systems

Metric to Imperial Conversions

Imperial to Metric Conversions

Conversions Between Metric and Imperial Systems Speed KM to MI

Conversions Between Metric and Imperial Systems Speed  MI to M

Metric-Imperial Length Conversion Inch To Cm

Metric-Imperial Length Conversion MM/CM/M/KM To FT

Metric-Imperial Length Conversion MM/CM/M/KM To MI

Imperial systems

Imperial Length Conversion

NL.SO.1202M3

Upper and lower bound

Upper and Lower Bound

Converting repeating decimals to fractions

NL.SO.1202M4

Surface area and volume of prisms

Surface area and volume of pyramids

Surface area and volume of cylinders

Surface area and volume of cones

Surface area and volume of spheres

NL.SO.1202N1

Taxes, discounts, tips and more

Discounts — Word Problem

Taxes — Word Problem

Percentage - Applying a Discount Price– Word Problem

Percentage - Applying a Discount Percent – Word Problem

Percentage - Applying a Tax Total – Word Problem

Percentage - Applying a tax Percent – Word Problem

Direct variation

Identifying direct variations

Inverse variation

Joint and combined variation

Money related questions in linear equations

Applications of linear relations

Ratios

Ratios - Find The Missing Value LHS

Ratios - Find The Missing Value RHS

Equivalent Ratios

Rates

Find the Biggest Rate 

Find the Unit Price 

Unit Rate Fill in Box

Find the Quickest Rate 

Unit Rate

Proportions

Proportions — Missing Denominator

Proportions — Missing Numerator

Proportions - Find the Missing Denominator LHS

Proportions - Find the Missing Denominator RHS

Proportions - Find the Missing Numerator LHS

Proportions - Find the Missing Numerator RHS

Percent of a number

Percent of a Number

percent of a number – Word Problem

Percent of a Number  Fraction Percent

Percent of a Number Integer Percent 

Percent of a number number With Decimal Percent 

Percent of a Number  Three Digits Number Percent

Adding and multiplying percents

Add and Multiply Percents — Word Problem

Add and Multiply Percents – Word Problem

Applied Mathematics 1202

Measurement

Number

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.

Finding a circle's diameter using chord lengths and the Pythagorean theorem

Finding the center of a round table using tangent lines

Finding angle CED using the inscribed angle in a semicircle theorem

Finding chord length CE using the Pythagorean theorem

Finding chord segment BE using the Pythagorean theorem

Intro to chord and tangent properties in circle geometry

Key properties of chords, inscribed angles, and central angles

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. 