How to compute center of gravity of trapezoid












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Hot to compute the mediana / center of gravity of trapezoid in analytical geometry?



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  • 30;1? Is the 1 correct?
    – Elsa
    Nov 28 '18 at 2:52










  • Quoting wiki's entry of centroid of a polygon, $$ begin{align} C_x &= frac{1}{6A} sum_{i=0}^{n-1}(x_i + x_{i+1})(x_i y_{i+1} - x_{i+1}y_i)\ C_y &= frac{1}{6A} sum_{i=0}^{n-1}(y_i + y_{i+1})(x_i y_{i+1} - x_{i+1}y_i) end{align} $$ where $A = frac12sumlimits_{i=0}^{n-1}(x_i y_{i+1} - x_{i+1}y_i)$ is polygon's signed area.
    – achille hui
    Nov 28 '18 at 4:08
















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Hot to compute the mediana / center of gravity of trapezoid in analytical geometry?



IMAGE



Greetings from Poland!










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  • 30;1? Is the 1 correct?
    – Elsa
    Nov 28 '18 at 2:52










  • Quoting wiki's entry of centroid of a polygon, $$ begin{align} C_x &= frac{1}{6A} sum_{i=0}^{n-1}(x_i + x_{i+1})(x_i y_{i+1} - x_{i+1}y_i)\ C_y &= frac{1}{6A} sum_{i=0}^{n-1}(y_i + y_{i+1})(x_i y_{i+1} - x_{i+1}y_i) end{align} $$ where $A = frac12sumlimits_{i=0}^{n-1}(x_i y_{i+1} - x_{i+1}y_i)$ is polygon's signed area.
    – achille hui
    Nov 28 '18 at 4:08














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Hot to compute the mediana / center of gravity of trapezoid in analytical geometry?



IMAGE



Greetings from Poland!










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Hot to compute the mediana / center of gravity of trapezoid in analytical geometry?



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Greetings from Poland!







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edited Nov 28 '18 at 2:17









Rócherz

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asked Nov 28 '18 at 2:15









DaveG

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  • 30;1? Is the 1 correct?
    – Elsa
    Nov 28 '18 at 2:52










  • Quoting wiki's entry of centroid of a polygon, $$ begin{align} C_x &= frac{1}{6A} sum_{i=0}^{n-1}(x_i + x_{i+1})(x_i y_{i+1} - x_{i+1}y_i)\ C_y &= frac{1}{6A} sum_{i=0}^{n-1}(y_i + y_{i+1})(x_i y_{i+1} - x_{i+1}y_i) end{align} $$ where $A = frac12sumlimits_{i=0}^{n-1}(x_i y_{i+1} - x_{i+1}y_i)$ is polygon's signed area.
    – achille hui
    Nov 28 '18 at 4:08


















  • 30;1? Is the 1 correct?
    – Elsa
    Nov 28 '18 at 2:52










  • Quoting wiki's entry of centroid of a polygon, $$ begin{align} C_x &= frac{1}{6A} sum_{i=0}^{n-1}(x_i + x_{i+1})(x_i y_{i+1} - x_{i+1}y_i)\ C_y &= frac{1}{6A} sum_{i=0}^{n-1}(y_i + y_{i+1})(x_i y_{i+1} - x_{i+1}y_i) end{align} $$ where $A = frac12sumlimits_{i=0}^{n-1}(x_i y_{i+1} - x_{i+1}y_i)$ is polygon's signed area.
    – achille hui
    Nov 28 '18 at 4:08
















30;1? Is the 1 correct?
– Elsa
Nov 28 '18 at 2:52




30;1? Is the 1 correct?
– Elsa
Nov 28 '18 at 2:52












Quoting wiki's entry of centroid of a polygon, $$ begin{align} C_x &= frac{1}{6A} sum_{i=0}^{n-1}(x_i + x_{i+1})(x_i y_{i+1} - x_{i+1}y_i)\ C_y &= frac{1}{6A} sum_{i=0}^{n-1}(y_i + y_{i+1})(x_i y_{i+1} - x_{i+1}y_i) end{align} $$ where $A = frac12sumlimits_{i=0}^{n-1}(x_i y_{i+1} - x_{i+1}y_i)$ is polygon's signed area.
– achille hui
Nov 28 '18 at 4:08




Quoting wiki's entry of centroid of a polygon, $$ begin{align} C_x &= frac{1}{6A} sum_{i=0}^{n-1}(x_i + x_{i+1})(x_i y_{i+1} - x_{i+1}y_i)\ C_y &= frac{1}{6A} sum_{i=0}^{n-1}(y_i + y_{i+1})(x_i y_{i+1} - x_{i+1}y_i) end{align} $$ where $A = frac12sumlimits_{i=0}^{n-1}(x_i y_{i+1} - x_{i+1}y_i)$ is polygon's signed area.
– achille hui
Nov 28 '18 at 4:08










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Hint: break the shape into two parts and take moments about a point. I am happy to provide a full solution if you still need help.






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    1 Answer
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    1 Answer
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    Hint: break the shape into two parts and take moments about a point. I am happy to provide a full solution if you still need help.






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      Hint: break the shape into two parts and take moments about a point. I am happy to provide a full solution if you still need help.






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        Hint: break the shape into two parts and take moments about a point. I am happy to provide a full solution if you still need help.






        share|cite|improve this answer












        Hint: break the shape into two parts and take moments about a point. I am happy to provide a full solution if you still need help.







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        answered Nov 28 '18 at 2:44









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