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anonymous

  • one year ago

MEDAL AND FAN!!!!! PLEASE HELP!!!!! Referring to the figure, find the area of the quadrilateral shown.

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  1. anonymous
    • one year ago
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    |dw:1436931454185:dw|

  2. anonymous
    • one year ago
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    Would the answer be 12 in. ^2

  3. jtvatsim
    • one year ago
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    Yes! You are correct! Actually, you can see that one corner fits perfectly into the other to make a rectangle. |dw:1436932124281:dw| The area is then given by 6in x 2in = 12 in ^2.

  4. anonymous
    • one year ago
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    |dw:1436931904321:dw|

  5. anonymous
    • one year ago
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    Do you know how I can solve this?

  6. anonymous
    • one year ago
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    @jtvatsim

  7. jtvatsim
    • one year ago
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    Well, if you stare at the picture for a bit, you'll realize that you actually have two triangles. |dw:1436932291925:dw|

  8. jtvatsim
    • one year ago
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    Do you know how to find the area of a triangle?

  9. anonymous
    • one year ago
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    \[base \times height \div 2\]

  10. anonymous
    • one year ago
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    @jtvatsim Would it be 7 cm x 4cm?

  11. jtvatsim
    • one year ago
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    You are close. There are actually a lot of triangles that you could find the area of, after thinking about these these two are easier to work with... |dw:1436932671767:dw| The big triangle on the right has base 4cm and height 7cm. The big triangle on the left has base 4 cm and height 3 cm. Add these two areas together to get: 1/2(4cm)(7cm) + 1/2(4cm)(3cm) = 14cm^2 + 6cm^2 = 20cm^2.

  12. jtvatsim
    • one year ago
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    Does that make sense? Any questions? :)

  13. anonymous
    • one year ago
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    4x3=12 12/2=6

  14. anonymous
    • one year ago
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    7x4=28 28/2=14 So..... the answer is 20 cm^2

  15. jtvatsim
    • one year ago
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    That is correct!

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