Find area of quadrilateral in triangle. [closed]
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What is the area of $HIJK$ quadrilateral, if the area of $ABC$ triangle is $70$, $BE=ED=DA$, and $BF=FG= GC$?
geometry euclidean-geometry triangles area quadrilateral
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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.
add a comment |
$begingroup$
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
$endgroup$
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.
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I got $frac{50}{3}.$
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– Michael Rozenberg
Jan 8 at 19:59
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There should be a solution using Ceva's thm or sth.
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– user614671
Jan 8 at 20:02
1
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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 $frac92$.
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– achille hui
Jan 9 at 14:28
add a comment |
$begingroup$
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
$endgroup$
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
geometry euclidean-geometry triangles area quadrilateral
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.
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I got $frac{50}{3}.$
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– 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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
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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.
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Thanks! it seems I needed to be more specific to wolfram about what I was trying to do.
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– ShapeOfMatter
Jan 9 at 15:04
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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.
edited Jan 9 at 15:03
answered Jan 8 at 19:21
ShapeOfMatterShapeOfMatter
1794
1794
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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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
answered Jan 8 at 18:09
Peter ForemanPeter Foreman
5,3641216
5,3641216
add a comment |
add a comment |
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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