XP Math Games Arcade

06-16-2008

I am having some trouble with this problem. I know the answer is 7.

This is the problem...

The radius of a large pizza is 1in. less then twice the radius of a small pizza.The difference between the areas of the pizzas is 33pi in^2.Find the radius of the large pizza.

So this must be a quadratic equation.

This is the part I don't get. Should it be like this...

-33pi in^2 + 2r -1=0

Sorry couldn't find the symbol for pie. But would I multiply the -33 first? Or is my formula completely wrong.

Or maybe putting the 33 on 2 sets of formulas. One that represents the small and one the large? Then subtracting the difference? Just need a shove in the right direction.

**Sillysidley** · 06-16-2008

Well, let the radius of the smaller circle be r, so the radius of the larger circle is 2r-1. We then have $(2r-1)^2-r^2=33$ , since we just divided out the pi. Now, I just eyed it out and found r=4, so our answer is 7. If you can't see that, you can expand to get $4r^2-4r+1-r^2=33\rightarrow\3r^2-4r+1=33\rightarrow\3r^2-4r-32=0\rightarrow\(3r+8)(r-4)=0$ , so we know r must be 4.
Or, after getting $(2r-1)^2-r^2=33$ , you can apply difference of squares to get $(3r-1)(r-1)=33$ and you can either make it a quadratic again, or in this case it's easy to examine factors and find r=4.

06-17-2008

Ok. Thanks.You gave me more help then I wanted, but I was doing it backwards. So, if I originally left the 33pi^2 on the right and then divided out the square then I would of had the (2r-1)^2? This is where that square came from? Something like using a literal equation?We just started quadratic equations and this was the last question in my homework. Thanks a lot for your help,

it is appreciated!

06-17-2008

One more thing is the r^2 also the result of dividing out the PI?

**Sillysidley** · 06-17-2008

Well, since we let the radius of the smaller circle be r so the bigger one is 2r-1, we can convert what it says to an equation: Since the difference of the areas is 33 pi, we have $(2r-1)^2\pi-r^2\pi=33\pi$ , so we can divide out pi.

06-17-2008

Thanks...thats what I was looking for.

**Scion** · 09-21-2008

how do you know the answer?

06-16-2008	#1
Equality Guest Posts: n/a	Word Problem I am having some trouble with this problem. I know the answer is 7. This is the problem... The radius of a large pizza is 1in. less then twice the radius of a small pizza.The difference between the areas of the pizzas is 33pi in^2.Find the radius of the large pizza. So this must be a quadratic equation. This is the part I don't get. Should it be like this... -33pi in^2 + 2r -1=0 Sorry couldn't find the symbol for pie. But would I multiply the -33 first? Or is my formula completely wrong. Or maybe putting the 33 on 2 sets of formulas. One that represents the small and one the large? Then subtracting the difference? Just need a shove in the right direction. Last edited by Equality; 06-16-2008 at 09:12 PM.. Reason: Another Thought

06-16-2008	#2
Sillysidley Join Date: Oct 2006 Posts: 822	Well, let the radius of the smaller circle be r, so the radius of the larger circle is 2r-1. We then have $(2r-1)^2-r^2=33$ , since we just divided out the pi. Now, I just eyed it out and found r=4, so our answer is 7. If you can't see that, you can expand to get $4r^2-4r+1-r^2=33\rightarrow\3r^2-4r+1=33\rightarrow\3r^2-4r-32=0\rightarrow\(3r+8)(r-4)=0$ , so we know r must be 4. Or, after getting $(2r-1)^2-r^2=33$ , you can apply difference of squares to get $(3r-1)(r-1)=33$ and you can either make it a quadratic again, or in this case it's easy to examine factors and find r=4. __________________ .

06-17-2008	#5
Sillysidley Join Date: Oct 2006 Posts: 822	Well, since we let the radius of the smaller circle be r so the bigger one is 2r-1, we can convert what it says to an equation: Since the difference of the areas is 33 pi, we have $(2r-1)^2\pi-r^2\pi=33\pi$ , so we can divide out pi. __________________ .

09-21-2008	#7
Scion Join Date: Nov 2006 Posts: 437	how do you know the answer? __________________

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