OpenStudy (anonymous):

the radius of a large pizza is 1 inch less than twice the radius of a small pizza. The difference between the areas of the two pizzas is 33 pi in^2. find the radius of the large pizza.

6 years ago
OpenStudy (amistre64):

the area if the larger - area of the smaller

6 years ago
OpenStudy (amistre64):

33pi = something derived from the word problem lol

6 years ago
OpenStudy (anonymous):

Area of larger pizza = 49 pi Area of smaller pizza = 16 pi Radius of large pizza = 7

6 years ago
OpenStudy (amistre64):

rL = 1 less than the diam of the smaller

6 years ago
OpenStudy (anonymous):

@PJW - where did you get those numbers? Is that all from 33 pi in^2?

6 years ago
OpenStudy (amistre64):

$Rs = Rs$ $Rl = 2(Rs)-1$ $33pi = pi(1Rs-1)^2 - (Rs)^2$

6 years ago
OpenStudy (amistre64):

$33 = 4(Rs)^2 +1 -4(Rs) -(Rs)^2$ $33 = 3(Rs)^2 -4(Rs) +1$ $0 = 3(Rs)^2 -4(Rs) -32$ maybe? lol

6 years ago
OpenStudy (anonymous):

amistre64 - u can't ask me! I have no idea! I'm still at 2r=diameter and trying to build on that LOL

6 years ago
OpenStudy (amistre64):

$[(Rs)-12/3] [(Rs)+8/3]$ $[(Rs-4)][(3Rs+8)]$ $Rs = 4$ if I did it right

6 years ago
OpenStudy (amistre64):

Area of smaller then = 16pi Area of larger = 16pi + 33pi = 49pi Radius of larger = 7

6 years ago
OpenStudy (amistre64):

it matches PJs so I guess I did it right ;) or got lucky

6 years ago
OpenStudy (amistre64):

Largo, sounds like another floridian

6 years ago
OpenStudy (anonymous):

we'll see what PJW says

6 years ago
OpenStudy (amistre64):

4+4 = 8; which matches the description of the larger radius = 1 less the the diam of the smaller

6 years ago
OpenStudy (anonymous):

which is 7 and makes sense to me :)

6 years ago
OpenStudy (amistre64):

I coulda picked better variable names methinks ... but at least I kept up with my typos this time ;)

6 years ago
OpenStudy (anonymous):

if you want it algebraically then suppose the area of the larger pizza is $\pi*x^2$ and the area of the smaller pizza is $\pi*y^2$ so $\pi*x^2$-$\pi*y^2$=33$\pi$ $x+1=2y$ $x=2y-1$ substitute $(4y^2-4y+1)*\pi + \pi*y^2=33*\pi$ $(3y^2-4y-32)*\pi=0$ $(3y+8)*(y-4)*\pi=0$ $y=4$ substitute $x=2y-1$ $x=7$ sorry this took forever to make

6 years ago
OpenStudy (anonymous):

Thank you PJW! I really appreciate your help! And of course the help of amistre

6 years ago
OpenStudy (amistre64):

i was just supervising ;)

6 years ago
OpenStudy (anonymous):

amistre got it algebraically faster than me :) gj amistre

6 years ago