1 / 11

Week 22, Problem 2 By: Fabian S.

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….

Download Presentation

Week 22, Problem 2 By: Fabian S.

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


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

  2. 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)

  3. 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

  4. 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

  5. My work… To make my life easier I decided to use a circle so I can imagine how it would look r r+1

  6. More work… r r+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)

  7. 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

  8. 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

  9. 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…

  10. 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

  11. My final answer……… 5

More Related