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SAT Math Videos 171 videos

SAT Math 1.1 Algebra and Functions
315 Views

SAT Math 1.1 Algebra and Functions. Find an algebraic equation to correspond with the data.

SAT Math 1.1 Geometry and Measurement
719 Views

SAT Math 1.1 Geometry and Measurement. What is the circumference of the circle?

SAT Math 1.1 Numbers and Operations
280 Views

SAT Math 1.1 Numbers and Operations. How many combinations of beverage and cereal can be made?

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SAT Math 5.4 Geometry and Measurement 235 Views


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SAT Math 5.4 Geometry and Measurement

Language:
English Language

Transcript

00:03

Here’s your shmoop du jour, brought to you by chords.

00:06

Let’s make sweet, sweet music together.

00:09

The diameter of this circle is 14.

00:12

Lines JN and MK are perpendicular to each other.  What is the length of the chord JK? 

00:20

And here are the potential answers...

00:25

We’re not sure where this circle is headed, but given the bow tie, it must be someplace fancy.

00:31

Okay, so this problem makes a big deal out of the fact that JN and MK are perpendicular…

00:36

What do the lines’ perpendicularity tell us?

00:38

It tells us that, because L is the midpoint of the circle…

00:41

…angle JLK is, by rule, a right angle.

00:45

The diameter of the circle is 14, which makes each radius 7…

00:48

…so we now also know the length of both JL and KL.

00:52

And whaddya know…by looking at our clock, it appears to be Pythagorean theorem time again…

00:58

Our “a” and “b” are each 7…

01:01

So when we take a squared plus b squared equals c squared…

01:04

…we get 7 squared plus 7 squared equals c squared…

01:09

49 plus 49 is 98…and the square root of 98 can simplify to 49 times 2…

01:15

…or 7 square root of 2.

01:17

Answer D.

01:18

There’s actually another way we could have gone about things…

01:21

We know that JKL is an isosceles triangle, given that it’s a right triangle and the

01:26

length of both sides is equal.

01:28

So we could have just used what we know about the ratio of sides in an isosceles triangle

01:33

to tell us that the hypotenuse must be x square root of 2…

01:37

…or, in this case, 7 square root of 2. Either way. It’s answer D.

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