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anonymous

  • 5 years ago

Is it possible to change a figure with an area of 18 and a perimeter of 18 to a perimeter of 20, but with the same area?

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  1. amistre64
    • 5 years ago
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    P=18 and A = 18 P = 20 and A = 18 that what your asking for?

  2. amistre64
    • 5 years ago
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    its possible, even easy...

  3. anonymous
    • 5 years ago
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    I need to change the perimeter, but leave the area the same

  4. amistre64
    • 5 years ago
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    what shape are you needing to change? rectangle, triangle, pentaangle?

  5. anonymous
    • 5 years ago
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    Rectangle

  6. amistre64
    • 5 years ago
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    i was hoping it was a rectangle :) ok.... both areas need to be 18. 18 = lw right?

  7. anonymous
    • 5 years ago
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    yeah

  8. amistre64
    • 5 years ago
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    we can get a "value" for either l or w from this to use in the perimeter equations. Lets say: 18/w = l good ?

  9. anonymous
    • 5 years ago
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    yeah, thanks

  10. amistre64
    • 5 years ago
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    p = 2l + 2w 4 sides all together, 2 of each right?

  11. anonymous
    • 5 years ago
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    yeah

  12. amistre64
    • 5 years ago
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    so: just use our "value" in this equation like this: 18 = 2(18/w) + 2w

  13. anonymous
    • 5 years ago
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    ok

  14. amistre64
    • 5 years ago
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    18 = 36/w + 2w.... lets get a common denominator to make life easy 18 = 36 + 2w^2 ---------- still with me? w

  15. anonymous
    • 5 years ago
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    yeah, I'm with you

  16. amistre64
    • 5 years ago
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    tell me when I do something stupid :) we can multiply that bottom "w" across to the 18 right? 18w = 36 + 2w^2

  17. anonymous
    • 5 years ago
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    yeah. I believe so

  18. amistre64
    • 5 years ago
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    we can... now put all like terms to one side 0 = 2w^2 -18w +36 we are ready to find out what w can be right?

  19. amistre64
    • 5 years ago
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    all terms to one side is what I meant to type :)

  20. anonymous
    • 5 years ago
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    ok Yeah, we're ready =)

  21. amistre64
    • 5 years ago
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    factor out the 2, its in our way 0 = 2(w^2 -9w +18) now we know that anything times zero = zero, our only choice is to make that left part equal to zero. w^2 -9w +18 = 0 (w-6) (w-3) = 0 when w = 6 or 3 are good. actually w can stand for length or width here so we found both numbers :) lets try it our: A = 6(3) = 18 P = 2(6) + 2(3) P = 12 + 6 = 18 were good for that one right?

  22. amistre64
    • 5 years ago
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    and by left I mean right..... its like haveing a stroke....

  23. anonymous
    • 5 years ago
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    ok

  24. amistre64
    • 5 years ago
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    lets go for 20 :) same steps but I aint gonna talk so much...

  25. anonymous
    • 5 years ago
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    ok

  26. amistre64
    • 5 years ago
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    20 = 2(18/w) + 2w 20 = 36/w + 2w 20 = 36 + 2w^2 ---------- w 20w = 36 +2w^2 0 = 2w^2 -20w +36 0 = 2(w^2 -10w +18) w^2 -10w +18 = 0 now this one we use the quadratic formula on because it wont behave for us :)

  27. amistre64
    • 5 years ago
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    10 +- sqrt(100 -(4)(18)) --------------------- 2 10 +- sqrt(28) ------------ 2 w = 5 + sqrt(7) and l = 5-sqrt(7) lets check them out

  28. amistre64
    • 5 years ago
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    A = (5-sqrt(7)) (5+sqrt(7)) A = 25 - 7 A = 18

  29. anonymous
    • 5 years ago
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    can i get you to help me with my problems

  30. amistre64
    • 5 years ago
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    sure, just on sec please

  31. amistre64
    • 5 years ago
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    P = 2(5-sqrt(7)) + 2(5+sqrt(7)) P = 10 -2sqrt(7) + 10 + 2sqrt(7) P = 20 +2sqrt(7) -2sqrt(7) P = 20

  32. amistre64
    • 5 years ago
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    howd we do?

  33. anonymous
    • 5 years ago
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    good, thanks!

  34. amistre64
    • 5 years ago
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    lol... enjoy :)

  35. anonymous
    • 5 years ago
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    :)

  36. anonymous
    • 5 years ago
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    i have to go now. Bye, and thanks again!

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