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anonymous

  • one year ago

square ABCD has sides of length 12 units each. Points W, X, Y, and Z lie on sides AB, BC, CD, and DA, respectively, so that AW = 1/2 AB, BX = 1/2 BC, CY = 1/3 CD, and AZ = 1/4 DA. What is the area of the quadrilateral WXYZ?

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  1. anonymous
    • one year ago
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    |dw:1440267886154:dw| I need your help

  2. mathmate
    • one year ago
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    |dw:1440268312680:dw| Area of square, As= (12x)^2 = 144x^2 Calculate the total area of the 4 triangles at the four corners (A=bh/2) At=6x*6x+6x*4x+8x*9x+3x*6x=.... Subtract At from As to get the area of the quadrilateral. Aq=As-At

  3. mathmate
    • one year ago
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    Correction: At=(6x*6x+6x*4x+8x*9x+3x*6x)/2=....

  4. anonymous
    • one year ago
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    ????? I'm just in Junior high...

  5. anonymous
    • one year ago
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    @mathmate

  6. anonymous
    • one year ago
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    Did he answer the question??

  7. anonymous
    • one year ago
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    IDK

  8. mathmate
    • one year ago
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    If you are confused with the 12x and 144x^2 (which is a general case), set x=1, so the side of the square is 12 units (as given in the question.

  9. mathmate
    • one year ago
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    |dw:1440269114269:dw|

  10. mathmate
    • one year ago
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    Then calculate area of the four corner triangles: |dw:1440269166364:dw| The area of the square is 12*12=144 u^2 Area of the quadrilateral is 144-(sum of the areas of the four triangles). Can you take it from here?

  11. anonymous
    • one year ago
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    So if each sides are 12. The area of the rectangle is 144. then add all of the triangles together subtract that from 144.

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