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Aizhalee

  • one year ago

(Fan and Medal will be given)! Help with Area Find the area of the image below:

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  1. aizhalee
    • one year ago
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  2. Michele_Laino
    • one year ago
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    hint: the area of your geometrical shape, is given by the sum of the areas of these triangles:

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  3. aizhalee
    • one year ago
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    so what exactly do I do ? Im not too well with this benchmark D: .

  4. Michele_Laino
    • one year ago
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    we have: area of triangle ABE: \[A = \frac{{BE \times AK}}{2} = \frac{{8 \times 5}}{2} = ...\]

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  5. aizhalee
    • one year ago
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    so 8 * 5 = 40 an d 40 divided by 2 is 20 .

  6. Michele_Laino
    • one year ago
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    ok!

  7. Michele_Laino
    • one year ago
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    area of triangle CDE: \[A = \frac{{CE \times DK}}{2} = \frac{{5 \times 4}}{2} = ...\]

  8. aizhalee
    • one year ago
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    5 * 4 equals 20 and 20 divded by 2 is 10

  9. Michele_Laino
    • one year ago
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    ok!

  10. Michele_Laino
    • one year ago
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    finally: area of triangle CBE: \[A = \frac{{BE \times CE}}{2} = \frac{{8 \times 5}}{2} = ...\]

  11. aizhalee
    • one year ago
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    8*5 equals 40 again divided by two is 20 :D

  12. Michele_Laino
    • one year ago
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    ok!

  13. Michele_Laino
    • one year ago
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    so total area is: 20+10+20=...

  14. aizhalee
    • one year ago
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    50

  15. Michele_Laino
    • one year ago
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    yes! nevertheless that answer is not an option of yours

  16. aizhalee
    • one year ago
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    wow , :/

  17. Michele_Laino
    • one year ago
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    my reasoning is right, I don't understand maybe is there a scale factor, for your drawing?

  18. Michele_Laino
    • one year ago
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    namely, how many units are the area of the littlest square, of your drawing?

  19. aizhalee
    • one year ago
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    can u help me find the perimeter for it please michele , D: its not area , its perimeter. sorry ive made a mistake

  20. Michele_Laino
    • one year ago
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    ok!

  21. Michele_Laino
    • one year ago
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    then we have to apply the theorem of Pitagora repeatedly

  22. Michele_Laino
    • one year ago
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    so we can write this: \[BC = \sqrt {C{E^2} + B{E^2}} = \sqrt {{5^2} + {8^2}} = \sqrt {25 + 64} = ...?\]

  23. aizhalee
    • one year ago
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    ok is that is 89

  24. Michele_Laino
    • one year ago
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    yes! and what is:sqrt(89)=...?

  25. aizhalee
    • one year ago
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    9.433

  26. Michele_Laino
    • one year ago
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    that's right!

  27. Michele_Laino
    • one year ago
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    now we have: \[AB = \sqrt {A{H^2} + B{H^2}} = \sqrt {{5^2} + {6^2}} = \sqrt {25 + 36} = ...?\]

  28. Michele_Laino
    • one year ago
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  29. aizhalee
    • one year ago
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    its 61

  30. Michele_Laino
    • one year ago
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    and what is: sqrt(61)=...?

  31. aizhalee
    • one year ago
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    the square root of 61is 7.81

  32. Michele_Laino
    • one year ago
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    perfect!

  33. aizhalee
    • one year ago
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    :D

  34. Michele_Laino
    • one year ago
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    now we have: \[AE = \sqrt {A{H^2} + E{H^2}} = \sqrt {{5^2} + {2^2}} = \sqrt {25 + 4} = ...?\]

  35. aizhalee
    • one year ago
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    25 plus 4 is 29 and the square root of 29 is 5.38

  36. Michele_Laino
    • one year ago
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    yes! correct!

  37. Michele_Laino
    • one year ago
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    next we have: \[ED = \sqrt {K{D^2} + E{K^2}} = \sqrt {{4^2} + {2^2}} = \sqrt {16 + 4} = ...?\]

  38. aizhalee
    • one year ago
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    16 + 4 = 20 and square root of 20 is 4.47

  39. Michele_Laino
    • one year ago
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    ok!

  40. Michele_Laino
    • one year ago
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    finally, we have: \[CD = \sqrt {K{D^2} + C{K^2}} = \sqrt {{4^2} + {3^2}} = \sqrt {16 + 9} = ...?\]

  41. aizhalee
    • one year ago
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    5 :D

  42. Michele_Laino
    • one year ago
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    perfect!

  43. Michele_Laino
    • one year ago
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    so the requested perimeter, is: BC+AB+AE+ED+CD= =9.43+7.81+5.39+4.47+5=...?

  44. aizhalee
    • one year ago
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    32.1 :D yipee

  45. Michele_Laino
    • one year ago
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    that's right! :)

  46. aizhalee
    • one year ago
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    Thank you sooo much :D i will be adding these into my notes thanks a lot I appreciate it ! God bless.

  47. Michele_Laino
    • one year ago
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    Thanks!! :) :)

  48. aizhalee
    • one year ago
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    Yw :)

  49. Michele_Laino
    • one year ago
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    :)

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