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