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ZyairaBoyd

  • 2 years ago

Can anyone solve this? (still not solved)

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

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

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

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

  6. Shayaan_Mustafa
    • 2 years ago
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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..

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

  8. ZyairaBoyd
    • 2 years ago
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    Ok.

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

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

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

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

  13. ZyairaBoyd
    • 2 years ago
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    @Shayaan_Mustafa

  14. KingGeorge
    • 2 years ago
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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 \[{1 \over2}\cdot 10\cdot 13=65\]

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

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

  17. ZyairaBoyd
    • 2 years ago
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    I dont mind.

  18. KingGeorge
    • 2 years ago
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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.

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

  20. ZyairaBoyd
    • 2 years ago
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    oh ok.

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

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

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

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

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

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

  27. KingGeorge
    • 2 years ago
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    Correct.

  28. ZyairaBoyd
    • 2 years ago
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    Ok.

  29. katielong
    • 2 years ago
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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?

  30. ZyairaBoyd
    • 2 years ago
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    i got 22.

  31. ZyairaBoyd
    • 2 years ago
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    Ty @katielong

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

  33. ZyairaBoyd
    • 2 years ago
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    yes.

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

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

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

  37. katielong
    • 2 years ago
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    youre welcommeeeee =)

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