Question

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.

Answer 1

the area if the larger - area of the smaller

Answer 2

33pi = something derived from the word problem lol

Answer 3

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

Answer 4

rL = 1 less than the diam of the smaller

Answer 5

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

Answer 6

\[Rs = Rs\]
\[Rl = 2(Rs)-1\]

\[33pi = pi(1Rs-1)^2 - (Rs)^2\]

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Radius of larger = 7

Answer 11

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

Answer 12

Largo, sounds like another floridian

Answer 13

we'll see what PJW says

Answer 14

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

Answer 15

which is 7 and makes sense to me :)

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

i was just supervising ;)

Answer 20

amistre got it algebraically faster than me :) gj amistre