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SAT Math 7.4 Geometry and Measurement


Transcript

00:02

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

00:06

Apparently, it has some kind of restraining order against the circumference.

00:17

F is the midpoint of DG and E is the midpoint of FD.

00:21

What is the ratio of the area of the circle with center E to the area of the circle with center G?

00:29

Here are the potential answers...

00:33

Okay, so we’ve got three circles here…and we want to know how many of the itty-bitty

00:37

circles we could fit inside the big circle.

00:40

Well, all right…to be more technically accurate, we want a ratio of the area.

00:45

And all we can use to get there is a couple of midpoints.

00:48

These guys had better be good. Okay… here’s another problem where we’re

00:48

going to have to go ahead and just call something “x.”

00:48

In this case, we know that ED is a radius of the smallest circle… so we’ll call

00:53

that guy “x.”

00:54

That would make FD, the smallest circle's diameter, 2x.

00:59

And, because FD is also a radius of the medium circle…GD is going to be 4x.

01:05

Now that we have our radii, it’s time to whip out the formula for the area of a circle,

01:09

and go to work… For our smallest circle, we’ll take the formula

01:13

Area equals pi times r squared and plug in our… r…

01:20

…to get pi x squared.

01:22

For the big circle, we get pi times the quantity 4x… squared… or pi sixteen x squared.

01:28

We’re looking for the ratio, which is always the same as a fraction…

01:32

…and in this case is pi times x squared over pi times sixteen x squared.

01:38

The pi and the x squared cancel out…leaving us with just 1 over 16…

01:43

…which is equivalent to option A.

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