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Week 22, Problem 2 By: Fabian S.

Week 22, Problem 2 By: Fabian S

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Week 22, Problem 2 By: Fabian S. The Problem…. When the diameter of a pizza increases by 2 inches, the area increases by 44%. What was the original the area, in square inches, of the original pizza? Express your answers in terms of pii (3.14). Important things to remember…. - PowerPoint PPT Presentation

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Page 1: Week 22, Problem 2       By: Fabian S

Week 22, Problem 2

By: Fabian S.

Page 2: Week 22, Problem 2       By: Fabian S

The Problem…

When the diameter of a pizza increases by 2 inches, the area increases by 44%. What was the original the area, in square inches, of the original pizza? Express your answers in terms of pii (3.14)

Page 3: Week 22, Problem 2       By: Fabian S

Important things to remember…

So the area increases by 44%, And the question asks what was the area, in square inches, of the original Pizza. So the best way to go is to change 44% into a decimal, so then it would be, .44

Remember that the diameter increases 2 inches

Page 4: Week 22, Problem 2       By: Fabian S

My Explanation

So what I know is that the formula to find the area of the original pizza is, A= Pii (3.14) * r squared. I also squared it because the diameter doubles and the area increases

Page 5: Week 22, Problem 2       By: Fabian S

My work…To make my life easier I decided to use a circle so I can imagine how it would look

r r+1

Page 6: Week 22, Problem 2       By: Fabian S

More work…

rr+1

Now I need to find the areas of these 2 circles so for the first circle my area would equal A= pii *r squared and the second circle would be A= Pii * (r+1) squared, A= 1.44(pii * r squared)

Page 7: Week 22, Problem 2       By: Fabian S

Even more work…

Now I have my two equations that are 1.44( pii* r squared) and 1.44= pii (r squared+ 2r+1). In order to find my answer these two equations need to be equal to each other . So first off I could distribute pii to my equation that is pii (r squared+2r+1) so then it would be…

Pii r squared+2pii r+ Pii

And then it would have to be equal to…

1.44pii r squared

1.44pii r squared= pii r squared+2pii r+ Pii

Page 8: Week 22, Problem 2       By: Fabian S

More, and more work…

Now I could just simplify more by eliminating my like terms so I could start by eliminating pii r squared…

1.44pii r squared= Pii r squared+ 2pii squared+ pii

So then I will only have…

.44pii r 2= 2 pii r+ pii

Now that I ended up with this I could just divide everything by pii

Page 9: Week 22, Problem 2       By: Fabian S

The final steps……

So I divide everything by pii…

.44pii r squared= 2pii r+ pii pii pii

So then I will end up getting..

.44r squared= 2r+ pii

Now the final thing to do is check if its equal so my best thing to do is guess and check…

Page 10: Week 22, Problem 2       By: Fabian S

One last thing…

So when I did my guess and check I tried substituting 5 for r and I was lucky because when I plugged it it both equations equaled 11

.44* 5 squared= 11 2*5+1= 11

So my final answer is 5

Page 11: Week 22, Problem 2       By: Fabian S

My final answer………

5