SAT prep question.

S

sayruhhsupp

New member
Joined
Aug 4, 2009
Messages
3
Hi, I'm doing an SAT prep packet and one of the problems I'm stuck on I can't figure out how to even start it. Here's how it goes...


44384189d8b6ba88c715c266b67885bd216e3b1.jpg


In the figure above, the radius of the inner circle of the tire is twice the width of the tire itself. To the nearest integer, what percent of the entire area is the area of the tire?


I have no work since I can't figure out how to begin this problem. Could anyone help me?
 
sayruhhsupp said:
Hi, I'm doing an SAT prep packet and one of the problems I'm stuck on I can't figure out how to even start it. Here's how it goes...


44384189d8b6ba88c715c266b67885bd216e3b1.jpg


In the figure above, the radius of the inner circle of the tire is twice the width of the tire itself. To the nearest integer, what percent of the entire area is the area of the tire?


I have no work since I can't figure out how to begin this problem. Could anyone help me?
Click to expand...

We will assume that although it was called tire - it is actually annular disk of negligible thickness

Let the width of the disk = w

What is the inner radius (r[sub:3p5ankt3]i[/sub:3p5ankt3]) of the disk? radius of the inner circle of the tire is twice the width

What is the outer radius (r[sub:3p5ankt3]o[/sub:3p5ankt3]) of the disk (as a function of w) ? r[sub:3p5ankt3]o[/sub:3p5ankt3] = r[sub:3p5ankt3]i[/sub:3p5ankt3] + w

What is area the outer circle (A[sub:3p5ankt3]o[/sub:3p5ankt3]) of the disk (as a function of w) ? = (pi)(r[sub:3p5ankt3]o[/sub:3p5ankt3][sup:3p5ankt3]2[/sup:3p5ankt3])

What is area the inner circle (A[sub:3p5ankt3]i[/sub:3p5ankt3]) of the disk (as a function of w) ?= (pi)(r[sub:3p5ankt3]i[/sub:3p5ankt3][sup:3p5ankt3]2[/sup:3p5ankt3])

Now find the ratio.....
 
r[sub:24qna3ba]i[/sub:24qna3ba]/r[sub:24qna3ba]i[/sub:24qna3ba]+w = (pi)(r[sub:24qna3ba]i[/sub:24qna3ba][sup:24qna3ba]2[/sup:24qna3ba])/(pi)(r[sub:24qna3ba]o[/sub:24qna3ba][sup:24qna3ba]2[/sup:24qna3ba])

(pi)(r[sub:24qna3ba]i[/sub:24qna3ba][sup:24qna3ba]2[/sup:24qna3ba]) = w * (pi)(r[sub:24qna3ba]o[/sub:24qna3ba][sup:24qna3ba]2[/sup:24qna3ba])

am i on the right track so far?
 
Why are you making it so complicated and hard to follow?

w = width of tire ; then radius inner circle = 2w and radius large circle = 3w

area tire = area large circle - area inner circle = pi(9w^2) - pi(4w^2)
note that that simplifies to 5(pi)w^2

You ok now?
 
You must log in or register to reply here.
Top