Find area of quadrilateral in triangle. [closed]












-1












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What is the area of $HIJK$ quadrilateral, if the area of $ABC$ triangle is $70$, $BE=ED=DA$, and $BF=FG= GC$?












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closed as off-topic by Crostul, verret, Alexander Gruber Jan 8 at 22:04


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Crostul, verret, Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    I got $frac{50}{3}.$
    $endgroup$
    – Michael Rozenberg
    Jan 8 at 19:59










  • $begingroup$
    There should be a solution using Ceva's thm or sth.
    $endgroup$
    – user614671
    Jan 8 at 20:02






  • 1




    $begingroup$
    The area is $9$. Tthe points $B,I,K$ are collinear and the line containing them intersect $AC$ at its midpoint $M$. $triangle HIK$ is bounded by 3 cevians $AF$, $BM$ and $CD$ with $$(x,y,z) stackrel{def}{=}left(frac{CF}{BF}, frac{AM}{MC}, frac{BD}{AD}right) = (2,1,2)$$ By Routh's theorem, area of $triangle HIK$ is $$frac{(xyz-1)^2}{(xy+y+1)(yz+z+1)(zx+x+1)}cdot 70 = frac{9}{140}cdot 70 = frac92$$ By a similar argument, area of $triangle IJK$ is also $frac92$.
    $endgroup$
    – achille hui
    Jan 9 at 14:28
















-1












$begingroup$


What is the area of $HIJK$ quadrilateral, if the area of $ABC$ triangle is $70$, $BE=ED=DA$, and $BF=FG= GC$?












share|cite|improve this question











$endgroup$



closed as off-topic by Crostul, verret, Alexander Gruber Jan 8 at 22:04


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Crostul, verret, Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    I got $frac{50}{3}.$
    $endgroup$
    – Michael Rozenberg
    Jan 8 at 19:59










  • $begingroup$
    There should be a solution using Ceva's thm or sth.
    $endgroup$
    – user614671
    Jan 8 at 20:02






  • 1




    $begingroup$
    The area is $9$. Tthe points $B,I,K$ are collinear and the line containing them intersect $AC$ at its midpoint $M$. $triangle HIK$ is bounded by 3 cevians $AF$, $BM$ and $CD$ with $$(x,y,z) stackrel{def}{=}left(frac{CF}{BF}, frac{AM}{MC}, frac{BD}{AD}right) = (2,1,2)$$ By Routh's theorem, area of $triangle HIK$ is $$frac{(xyz-1)^2}{(xy+y+1)(yz+z+1)(zx+x+1)}cdot 70 = frac{9}{140}cdot 70 = frac92$$ By a similar argument, area of $triangle IJK$ is also $frac92$.
    $endgroup$
    – achille hui
    Jan 9 at 14:28














-1












-1








-1





$begingroup$


What is the area of $HIJK$ quadrilateral, if the area of $ABC$ triangle is $70$, $BE=ED=DA$, and $BF=FG= GC$?












share|cite|improve this question











$endgroup$




What is the area of $HIJK$ quadrilateral, if the area of $ABC$ triangle is $70$, $BE=ED=DA$, and $BF=FG= GC$?









geometry euclidean-geometry triangles area quadrilateral






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share|cite|improve this question













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edited Jan 8 at 19:33







user593746

















asked Jan 8 at 17:27









Nameless AlienNameless Alien

114




114




closed as off-topic by Crostul, verret, Alexander Gruber Jan 8 at 22:04


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Crostul, verret, Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Crostul, verret, Alexander Gruber Jan 8 at 22:04


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Crostul, verret, Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    I got $frac{50}{3}.$
    $endgroup$
    – Michael Rozenberg
    Jan 8 at 19:59










  • $begingroup$
    There should be a solution using Ceva's thm or sth.
    $endgroup$
    – user614671
    Jan 8 at 20:02






  • 1




    $begingroup$
    The area is $9$. Tthe points $B,I,K$ are collinear and the line containing them intersect $AC$ at its midpoint $M$. $triangle HIK$ is bounded by 3 cevians $AF$, $BM$ and $CD$ with $$(x,y,z) stackrel{def}{=}left(frac{CF}{BF}, frac{AM}{MC}, frac{BD}{AD}right) = (2,1,2)$$ By Routh's theorem, area of $triangle HIK$ is $$frac{(xyz-1)^2}{(xy+y+1)(yz+z+1)(zx+x+1)}cdot 70 = frac{9}{140}cdot 70 = frac92$$ By a similar argument, area of $triangle IJK$ is also $frac92$.
    $endgroup$
    – achille hui
    Jan 9 at 14:28


















  • $begingroup$
    I got $frac{50}{3}.$
    $endgroup$
    – Michael Rozenberg
    Jan 8 at 19:59










  • $begingroup$
    There should be a solution using Ceva's thm or sth.
    $endgroup$
    – user614671
    Jan 8 at 20:02






  • 1




    $begingroup$
    The area is $9$. Tthe points $B,I,K$ are collinear and the line containing them intersect $AC$ at its midpoint $M$. $triangle HIK$ is bounded by 3 cevians $AF$, $BM$ and $CD$ with $$(x,y,z) stackrel{def}{=}left(frac{CF}{BF}, frac{AM}{MC}, frac{BD}{AD}right) = (2,1,2)$$ By Routh's theorem, area of $triangle HIK$ is $$frac{(xyz-1)^2}{(xy+y+1)(yz+z+1)(zx+x+1)}cdot 70 = frac{9}{140}cdot 70 = frac92$$ By a similar argument, area of $triangle IJK$ is also $frac92$.
    $endgroup$
    – achille hui
    Jan 9 at 14:28
















$begingroup$
I got $frac{50}{3}.$
$endgroup$
– Michael Rozenberg
Jan 8 at 19:59




$begingroup$
I got $frac{50}{3}.$
$endgroup$
– Michael Rozenberg
Jan 8 at 19:59












$begingroup$
There should be a solution using Ceva's thm or sth.
$endgroup$
– user614671
Jan 8 at 20:02




$begingroup$
There should be a solution using Ceva's thm or sth.
$endgroup$
– user614671
Jan 8 at 20:02




1




1




$begingroup$
The area is $9$. Tthe points $B,I,K$ are collinear and the line containing them intersect $AC$ at its midpoint $M$. $triangle HIK$ is bounded by 3 cevians $AF$, $BM$ and $CD$ with $$(x,y,z) stackrel{def}{=}left(frac{CF}{BF}, frac{AM}{MC}, frac{BD}{AD}right) = (2,1,2)$$ By Routh's theorem, area of $triangle HIK$ is $$frac{(xyz-1)^2}{(xy+y+1)(yz+z+1)(zx+x+1)}cdot 70 = frac{9}{140}cdot 70 = frac92$$ By a similar argument, area of $triangle IJK$ is also $frac92$.
$endgroup$
– achille hui
Jan 9 at 14:28




$begingroup$
The area is $9$. Tthe points $B,I,K$ are collinear and the line containing them intersect $AC$ at its midpoint $M$. $triangle HIK$ is bounded by 3 cevians $AF$, $BM$ and $CD$ with $$(x,y,z) stackrel{def}{=}left(frac{CF}{BF}, frac{AM}{MC}, frac{BD}{AD}right) = (2,1,2)$$ By Routh's theorem, area of $triangle HIK$ is $$frac{(xyz-1)^2}{(xy+y+1)(yz+z+1)(zx+x+1)}cdot 70 = frac{9}{140}cdot 70 = frac92$$ By a similar argument, area of $triangle IJK$ is also $frac92$.
$endgroup$
– achille hui
Jan 9 at 14:28










2 Answers
2






active

oldest

votes


















1












$begingroup$

Let's put the area of 70 on hold for a moment. Furthermore, given the listed constraints and the implicit assumption that the problem is solvable using only those constraints, we choose any proportions for the outer triangle we like.



Let's examine the case of a right isosceles triangle with hypotenuse $BC$ and a base length of 3. It has an area of $frac{9}{2}$.



Letting the origin $(0,0)$ rest at $A$





  • $B$ is at $(0,3)$.


  • $C$ is at $(3,0)$.


  • $D$ is at $(0,1)$.


  • $E$ is at $(0,2)$.


  • $F$ is at $(1,2)$.


  • $G$ is at $(2,1)$.


This gives us the following lines





  • $AF$ is $y=2x$


  • $AG$ is $y=frac{1}{2}x$


  • $DC$ is $y=-frac{1}{3}x+1$


  • $EC$ is $y=-frac{2}{3}x+2$


Considering the four batches of two equations with two unknowns yields the coordinates of the remaining four points.





  • $H$, from $y=2x,;y=-frac{1}{3}x+1$ is $(frac{3}{7},frac{6}{7})$.


  • $I$, from $y=2x,;y=-frac{2}{3}x+2$ is $(frac{3}{4},frac{3}{2})$.


  • $J$, from $y=frac{1}{2}x,;y=-frac{2}{3}x+2$ is $(frac{12}{7},frac{6}{7})$.


  • $K$, from $y=frac{1}{2}x,;y=-frac{1}{3}x+1$ is $(frac{6}{5},frac{3}{5})$.


We can express the general form for the area of a quadrilateral based on it's vertices using vector arithmetic
$$A=frac{1}{2}vert(vec{J}-vec{H})times(vec{K}-vec{I})vert$$
We can plug this into Wolfram Alpha as



abs(Cross[([12/7,6/7]-[3/7,6/7]),([6/5,3/5]-[3/4,3/2])])*.5


Scaling that for our area 70 triangle, gives a final area of 9.



This differs from Peter Foreman's answer, so one of us must have made a typo or a false assumption. Frankly mine just looks wrong, judging from your diagram. In any case he deserves credit for advocating such a brute-force approach.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks! it seems I needed to be more specific to wolfram about what I was trying to do.
    $endgroup$
    – ShapeOfMatter
    Jan 9 at 15:04



















0












$begingroup$

The area is $frac{741}{68}$ I'm quite sure. You can find the coordinates of $H,I,J,K$ by using the equations of each straight line and taking $A$ as the origin. I then used Wolfram: Alpha to find the area given the coordinates of each vertex.






share|cite|improve this answer









$endgroup$




















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Let's put the area of 70 on hold for a moment. Furthermore, given the listed constraints and the implicit assumption that the problem is solvable using only those constraints, we choose any proportions for the outer triangle we like.



    Let's examine the case of a right isosceles triangle with hypotenuse $BC$ and a base length of 3. It has an area of $frac{9}{2}$.



    Letting the origin $(0,0)$ rest at $A$





    • $B$ is at $(0,3)$.


    • $C$ is at $(3,0)$.


    • $D$ is at $(0,1)$.


    • $E$ is at $(0,2)$.


    • $F$ is at $(1,2)$.


    • $G$ is at $(2,1)$.


    This gives us the following lines





    • $AF$ is $y=2x$


    • $AG$ is $y=frac{1}{2}x$


    • $DC$ is $y=-frac{1}{3}x+1$


    • $EC$ is $y=-frac{2}{3}x+2$


    Considering the four batches of two equations with two unknowns yields the coordinates of the remaining four points.





    • $H$, from $y=2x,;y=-frac{1}{3}x+1$ is $(frac{3}{7},frac{6}{7})$.


    • $I$, from $y=2x,;y=-frac{2}{3}x+2$ is $(frac{3}{4},frac{3}{2})$.


    • $J$, from $y=frac{1}{2}x,;y=-frac{2}{3}x+2$ is $(frac{12}{7},frac{6}{7})$.


    • $K$, from $y=frac{1}{2}x,;y=-frac{1}{3}x+1$ is $(frac{6}{5},frac{3}{5})$.


    We can express the general form for the area of a quadrilateral based on it's vertices using vector arithmetic
    $$A=frac{1}{2}vert(vec{J}-vec{H})times(vec{K}-vec{I})vert$$
    We can plug this into Wolfram Alpha as



    abs(Cross[([12/7,6/7]-[3/7,6/7]),([6/5,3/5]-[3/4,3/2])])*.5


    Scaling that for our area 70 triangle, gives a final area of 9.



    This differs from Peter Foreman's answer, so one of us must have made a typo or a false assumption. Frankly mine just looks wrong, judging from your diagram. In any case he deserves credit for advocating such a brute-force approach.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks! it seems I needed to be more specific to wolfram about what I was trying to do.
      $endgroup$
      – ShapeOfMatter
      Jan 9 at 15:04
















    1












    $begingroup$

    Let's put the area of 70 on hold for a moment. Furthermore, given the listed constraints and the implicit assumption that the problem is solvable using only those constraints, we choose any proportions for the outer triangle we like.



    Let's examine the case of a right isosceles triangle with hypotenuse $BC$ and a base length of 3. It has an area of $frac{9}{2}$.



    Letting the origin $(0,0)$ rest at $A$





    • $B$ is at $(0,3)$.


    • $C$ is at $(3,0)$.


    • $D$ is at $(0,1)$.


    • $E$ is at $(0,2)$.


    • $F$ is at $(1,2)$.


    • $G$ is at $(2,1)$.


    This gives us the following lines





    • $AF$ is $y=2x$


    • $AG$ is $y=frac{1}{2}x$


    • $DC$ is $y=-frac{1}{3}x+1$


    • $EC$ is $y=-frac{2}{3}x+2$


    Considering the four batches of two equations with two unknowns yields the coordinates of the remaining four points.





    • $H$, from $y=2x,;y=-frac{1}{3}x+1$ is $(frac{3}{7},frac{6}{7})$.


    • $I$, from $y=2x,;y=-frac{2}{3}x+2$ is $(frac{3}{4},frac{3}{2})$.


    • $J$, from $y=frac{1}{2}x,;y=-frac{2}{3}x+2$ is $(frac{12}{7},frac{6}{7})$.


    • $K$, from $y=frac{1}{2}x,;y=-frac{1}{3}x+1$ is $(frac{6}{5},frac{3}{5})$.


    We can express the general form for the area of a quadrilateral based on it's vertices using vector arithmetic
    $$A=frac{1}{2}vert(vec{J}-vec{H})times(vec{K}-vec{I})vert$$
    We can plug this into Wolfram Alpha as



    abs(Cross[([12/7,6/7]-[3/7,6/7]),([6/5,3/5]-[3/4,3/2])])*.5


    Scaling that for our area 70 triangle, gives a final area of 9.



    This differs from Peter Foreman's answer, so one of us must have made a typo or a false assumption. Frankly mine just looks wrong, judging from your diagram. In any case he deserves credit for advocating such a brute-force approach.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks! it seems I needed to be more specific to wolfram about what I was trying to do.
      $endgroup$
      – ShapeOfMatter
      Jan 9 at 15:04














    1












    1








    1





    $begingroup$

    Let's put the area of 70 on hold for a moment. Furthermore, given the listed constraints and the implicit assumption that the problem is solvable using only those constraints, we choose any proportions for the outer triangle we like.



    Let's examine the case of a right isosceles triangle with hypotenuse $BC$ and a base length of 3. It has an area of $frac{9}{2}$.



    Letting the origin $(0,0)$ rest at $A$





    • $B$ is at $(0,3)$.


    • $C$ is at $(3,0)$.


    • $D$ is at $(0,1)$.


    • $E$ is at $(0,2)$.


    • $F$ is at $(1,2)$.


    • $G$ is at $(2,1)$.


    This gives us the following lines





    • $AF$ is $y=2x$


    • $AG$ is $y=frac{1}{2}x$


    • $DC$ is $y=-frac{1}{3}x+1$


    • $EC$ is $y=-frac{2}{3}x+2$


    Considering the four batches of two equations with two unknowns yields the coordinates of the remaining four points.





    • $H$, from $y=2x,;y=-frac{1}{3}x+1$ is $(frac{3}{7},frac{6}{7})$.


    • $I$, from $y=2x,;y=-frac{2}{3}x+2$ is $(frac{3}{4},frac{3}{2})$.


    • $J$, from $y=frac{1}{2}x,;y=-frac{2}{3}x+2$ is $(frac{12}{7},frac{6}{7})$.


    • $K$, from $y=frac{1}{2}x,;y=-frac{1}{3}x+1$ is $(frac{6}{5},frac{3}{5})$.


    We can express the general form for the area of a quadrilateral based on it's vertices using vector arithmetic
    $$A=frac{1}{2}vert(vec{J}-vec{H})times(vec{K}-vec{I})vert$$
    We can plug this into Wolfram Alpha as



    abs(Cross[([12/7,6/7]-[3/7,6/7]),([6/5,3/5]-[3/4,3/2])])*.5


    Scaling that for our area 70 triangle, gives a final area of 9.



    This differs from Peter Foreman's answer, so one of us must have made a typo or a false assumption. Frankly mine just looks wrong, judging from your diagram. In any case he deserves credit for advocating such a brute-force approach.






    share|cite|improve this answer











    $endgroup$



    Let's put the area of 70 on hold for a moment. Furthermore, given the listed constraints and the implicit assumption that the problem is solvable using only those constraints, we choose any proportions for the outer triangle we like.



    Let's examine the case of a right isosceles triangle with hypotenuse $BC$ and a base length of 3. It has an area of $frac{9}{2}$.



    Letting the origin $(0,0)$ rest at $A$





    • $B$ is at $(0,3)$.


    • $C$ is at $(3,0)$.


    • $D$ is at $(0,1)$.


    • $E$ is at $(0,2)$.


    • $F$ is at $(1,2)$.


    • $G$ is at $(2,1)$.


    This gives us the following lines





    • $AF$ is $y=2x$


    • $AG$ is $y=frac{1}{2}x$


    • $DC$ is $y=-frac{1}{3}x+1$


    • $EC$ is $y=-frac{2}{3}x+2$


    Considering the four batches of two equations with two unknowns yields the coordinates of the remaining four points.





    • $H$, from $y=2x,;y=-frac{1}{3}x+1$ is $(frac{3}{7},frac{6}{7})$.


    • $I$, from $y=2x,;y=-frac{2}{3}x+2$ is $(frac{3}{4},frac{3}{2})$.


    • $J$, from $y=frac{1}{2}x,;y=-frac{2}{3}x+2$ is $(frac{12}{7},frac{6}{7})$.


    • $K$, from $y=frac{1}{2}x,;y=-frac{1}{3}x+1$ is $(frac{6}{5},frac{3}{5})$.


    We can express the general form for the area of a quadrilateral based on it's vertices using vector arithmetic
    $$A=frac{1}{2}vert(vec{J}-vec{H})times(vec{K}-vec{I})vert$$
    We can plug this into Wolfram Alpha as



    abs(Cross[([12/7,6/7]-[3/7,6/7]),([6/5,3/5]-[3/4,3/2])])*.5


    Scaling that for our area 70 triangle, gives a final area of 9.



    This differs from Peter Foreman's answer, so one of us must have made a typo or a false assumption. Frankly mine just looks wrong, judging from your diagram. In any case he deserves credit for advocating such a brute-force approach.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 9 at 15:03

























    answered Jan 8 at 19:21









    ShapeOfMatterShapeOfMatter

    1794




    1794












    • $begingroup$
      Thanks! it seems I needed to be more specific to wolfram about what I was trying to do.
      $endgroup$
      – ShapeOfMatter
      Jan 9 at 15:04


















    • $begingroup$
      Thanks! it seems I needed to be more specific to wolfram about what I was trying to do.
      $endgroup$
      – ShapeOfMatter
      Jan 9 at 15:04
















    $begingroup$
    Thanks! it seems I needed to be more specific to wolfram about what I was trying to do.
    $endgroup$
    – ShapeOfMatter
    Jan 9 at 15:04




    $begingroup$
    Thanks! it seems I needed to be more specific to wolfram about what I was trying to do.
    $endgroup$
    – ShapeOfMatter
    Jan 9 at 15:04











    0












    $begingroup$

    The area is $frac{741}{68}$ I'm quite sure. You can find the coordinates of $H,I,J,K$ by using the equations of each straight line and taking $A$ as the origin. I then used Wolfram: Alpha to find the area given the coordinates of each vertex.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The area is $frac{741}{68}$ I'm quite sure. You can find the coordinates of $H,I,J,K$ by using the equations of each straight line and taking $A$ as the origin. I then used Wolfram: Alpha to find the area given the coordinates of each vertex.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The area is $frac{741}{68}$ I'm quite sure. You can find the coordinates of $H,I,J,K$ by using the equations of each straight line and taking $A$ as the origin. I then used Wolfram: Alpha to find the area given the coordinates of each vertex.






        share|cite|improve this answer









        $endgroup$



        The area is $frac{741}{68}$ I'm quite sure. You can find the coordinates of $H,I,J,K$ by using the equations of each straight line and taking $A$ as the origin. I then used Wolfram: Alpha to find the area given the coordinates of each vertex.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 8 at 18:09









        Peter ForemanPeter Foreman

        5,3641216




        5,3641216















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