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ZyairaBoyd

Can anyone solve this? (still not solved)

  • 2 years ago
  • 2 years ago

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  1. ZyairaBoyd
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    • 2 years ago
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  2. pre-algebra
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    Yes, but it is too cumbersome.

    • 2 years ago
  3. ZyairaBoyd
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    i know -.-

    • 2 years ago
  4. Shayaan_Mustafa
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    hi ZyairaBoyd :) Is it regular pentagon?

    • 2 years ago
  5. ZyairaBoyd
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    Pretty sure.....

    • 2 years ago
  6. Shayaan_Mustafa
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    it seems too. So just ignore EB and EC. As you know regular pentagon always have same sides and here one side is given that is 10cm. If it is regular pentagon then it must have all sides with 10cm. Right? And area of regular pentagon is\[\large A=1.72x^2\] substitute x=10cm\[\large A=1.72(10cm)^2\]\[\large A=1.72(100cm^2)\]\[\large A=172cm^2\]If and only if it is a regular pentagon. Otherwise it is wrong..

    • 2 years ago
  7. pre-algebra
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    It's not a regular pentagon.

    • 2 years ago
  8. ZyairaBoyd
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    Ok.

    • 2 years ago
  9. Shayaan_Mustafa
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    OK. Let us see if it is not a regular pentagon. Be with me.

    • 2 years ago
  10. ZyairaBoyd
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    Man got to go. See ya.

    • 2 years ago
  11. Shayaan_Mustafa
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    @ZyairaBoyd Is that enough for you?

    • 2 years ago
  12. ZyairaBoyd
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    Huh? No....

    • 2 years ago
  13. ZyairaBoyd
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    @Shayaan_Mustafa

    • 2 years ago
  14. KingGeorge
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    They want you to solve this with triangles. First, notice that solving for triangle ECB is the easiest. It's just \[{1 \over2}\cdot 10\cdot 13=65\]

    • 2 years ago
  15. ZyairaBoyd
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    Yeah so i have to like do it one by one?

    • 2 years ago
  16. katielong
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    i'll be 2 mins.. i can do it though and il give you a good method!!

    • 2 years ago
  17. ZyairaBoyd
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    I dont mind.

    • 2 years ago
  18. KingGeorge
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    Then, since EB=EC=4, the areas of triangles EBA and triangle ECD are the same. and are given by \[{1 \over 2}\cdot 15\cdot 4=30\]Since there are two of these triangles, we have an area of 60.

    • 2 years ago
  19. KingGeorge
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    Now we add \(60+65=125\) To get the overall area.

    • 2 years ago
  20. ZyairaBoyd
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    oh ok.

    • 2 years ago
  21. KingGeorge
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    We have three triangles with area 65, 30, and 30. If we add them, we get 125.

    • 2 years ago
  22. ZyairaBoyd
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    Ty for breaking it down...you saved me alot of time.

    • 2 years ago
  23. ZyairaBoyd
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    theres another one just like that with 3 triangles

    • 2 years ago
  24. ZyairaBoyd
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    Did i do this right so far?

    • 2 years ago
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  25. KingGeorge
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    That looks correct to me.

    • 2 years ago
  26. ZyairaBoyd
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    ok so now all i have to do is add them all together correct?

    • 2 years ago
  27. KingGeorge
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    Correct.

    • 2 years ago
  28. ZyairaBoyd
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    Ok.

    • 2 years ago
  29. katielong
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    |dw:1334855745714:dw| here you see i worked out all of the sides by doing some halving. so... you can see the top triangles are 4cmx7.5cm to find the area times them which equals 30 if you notice, it now needs halving but 30 is the equivilant of the area of two of those triangles so you dont need to half it. its the same with the 13cmx5cm triangle 13x5=65 then you need to finally find the area of the last one... which is the same as the first... so to find the area of the shape you need to do... 30+65+30=........answer.....=125!!!!!!!!! hope it helped?

    • 2 years ago
  30. ZyairaBoyd
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    i got 22.

    • 2 years ago
  31. ZyairaBoyd
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    Ty @katielong

    • 2 years ago
  32. katielong
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    no probss =) do you understand it??

    • 2 years ago
  33. ZyairaBoyd
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    yes.

    • 2 years ago
  34. ZyairaBoyd
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    |dw:1334856366881:dw|

    • 2 years ago
  35. ZyairaBoyd
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    @KingGeorge & @katielong

    • 2 years ago
  36. KingGeorge
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    You're very welcome.

    • 2 years ago
  37. katielong
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    youre welcommeeeee =)

    • 2 years ago
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