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Geometry and Measurement Videos 48 videos

SAT Math 6.1 Geometry and Measurement
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SAT Math 6.1 Geometry and Measurement

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SAT Math 3.3 Geometry and Measurement
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SAT Math 3.3 Geometry and Measurement

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SAT Math 3.1 Geometry and Measurement 194 Views


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Description:

SAT Math 3.1 Geometry and Measurement

Language:
English Language

Transcript

00:02

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

00:07

Leave your dirty minds at the door, please.

00:11

In this figure, O1 and O2 are circles with equal radii of 4 centimeters.

00:17

Find the length of the tangent O2P.

00:20

And here are the potential answers:

00:23

O2P or not to pee, that is the question.

00:28

So no matter how much this problem tries to fool us into thinking it’s about circles…

00:32

…we know it’s really a triangle problem.

00:35

Okay, so it’s missing a leg, so let’s go ahead and draw in that missing leg right now…

00:39

but be careful, we’re not drawing it straight up to the end of the line… only to the letter “P”…

00:44

Now, by property of a tangent, the radius O1 P has to be perpendicular to O2 P…

00:51

…making O1 P O2 a right triangle.

00:56

And whenever we have a right triangle… we can always call in our good friend Pythagoras

00:59

to help us out of our jam.

01:02

So (O1P)2 + (O2P)2 = (O1O2)2

01:06

We’re looking for O2 P, so we don’t know that one yet… but we can plug in the other

01:11

values because we’re told that the radius of both circles is 4.

01:16

So… O2 P squared plus 4 squared = 8 squared…or O2 P squared plus 16 = 64.

01:23

Subtract 16 from both sides, and O2 P squared = 48…

01:27

…which, after we take the square root of both sides,

01:29

reduces to the square root of 3 times 16, and we can take the root of 16 and pull it out of there….

01:37

…leaves us with 4 square root of 3 – answer C.

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