Students
Teachers & SchoolsStudents
Teachers & SchoolsSAT Math 7.4 Geometry and Measurement
Area and Volume | Area |
Circles | Area Equations of Circles |
Geometry | Area and Volume Circles |
Problem Solving and Data Analysis | Ratios, rates, and proportions |
Product Type | SAT Math |
SAT Math | Geometry and Measurement |
Okay, so we’ve got three circles here…and we want to know how many of the itty-bitty
circles we could fit inside the big circle.
Well, all right…to be more technically accurate, we want a ratio of the area.
And all we can use to get there is a couple of midpoints.
These guys had better be good. Okay… here’s another problem where we’re
going to have to go ahead and just call something “x.”
In this case, we know that ED is a radius of the smallest circle… so we’ll call
that guy “x.”
That would make FD, the smallest circle's diameter, 2x.
And, because FD is also a radius of the medium circle…GD is going to be 4x.
Now that we have our radii, it’s time to whip out the formula for the area of a circle,
and go to work… For our smallest circle, we’ll take the formula
Area equals pi times r squared and plug in our… r…
…to get pi x squared.
For the big circle, we get pi times the quantity 4x… squared… or pi sixteen x squared.
We’re looking for the ratio, which is always the same as a fraction…
…and in this case is pi times x squared over pi times sixteen x squared.
The pi and the x squared cancel out…leaving us with just 1 over 16…
…which is equivalent to option A.