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SAT Math 5.4 Geometry and Measurement
235 Views

SAT Math 5.4 Geometry and Measurement

1
Attributes of Circles
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This video can run circles around other videos about circles. Sir Cumference, of Round Table fame, explains all the properties of… well, circles....

2
Perimeter and Circumference
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It's super useful to be able to find the distance around stuff. Like when you’re being pursued by authorities while running around the base of an...

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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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