Relating the areas of regions determined by joining midpoints of two sides of a convex quadrilateral to...
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I have tried solve the following problem, but I can not see how to approach it:
In the figure, $ABCD$ is an arbitrary convex quadrilateral, $M$ and $N$ are, respectively, the middle points of $overline{AB}$ and $overline{CD}$, and $S,S_1,S_2$ are the area of the shaded regions. Prove that $$S=S_1+S_2$$
Any hints are welcome!
euclidean-geometry problem-solving area
asked Jun 12 at 18:33
DiegoMath
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I have tried solve the following problem, but I can not see how to approach it:
In the figure, $ABCD$ is an arbitrary convex quadrilateral, $M$ and $N$ are, respectively, the middle points of $overline{AB}$ and $overline{CD}$, and $S,S_1,S_2$ are the area of the shaded regions. Prove that $$S=S_1+S_2$$
Any hints are welcome!
euclidean-geometry problem-solving area
asked Jun 12 at 18:33
DiegoMath
2,0491021
Do you mean "an arbitrary convex quadrilateral"?
– ajotatxe
Jun 12 at 18:51
Yes, It should be a convex quadrilateral. I will change the statement.
– DiegoMath
Jun 12 at 18:57
add a comment |
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0
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up vote
0
down vote
favorite
I have tried solve the following problem, but I can not see how to approach it:
In the figure, $ABCD$ is an arbitrary convex quadrilateral, $M$ and $N$ are, respectively, the middle points of $overline{AB}$ and $overline{CD}$, and $S,S_1,S_2$ are the area of the shaded regions. Prove that $$S=S_1+S_2$$
Any hints are welcome!
euclidean-geometry problem-solving area
asked Jun 12 at 18:33
DiegoMath
2,0491021
I have tried solve the following problem, but I can not see how to approach it:
In the figure, $ABCD$ is an arbitrary convex quadrilateral, $M$ and $N$ are, respectively, the middle points of $overline{AB}$ and $overline{CD}$, and $S,S_1,S_2$ are the area of the shaded regions. Prove that $$S=S_1+S_2$$
Any hints are welcome!
euclidean-geometry problem-solving area
euclidean-geometry problem-solving area
asked Jun 12 at 18:33
DiegoMath
2,0491021
asked Jun 12 at 18:33
DiegoMath
2,0491021
edited Jun 12 at 18:58
asked Jun 12 at 18:33
DiegoMath
2,0491021
asked Jun 12 at 18:33
DiegoMath
2,0491021
asked Jun 12 at 18:33
DiegoMath
2,0491021
2,0491021
Do you mean "an arbitrary convex quadrilateral"?
– ajotatxe
Jun 12 at 18:51
Yes, It should be a convex quadrilateral. I will change the statement.
– DiegoMath
Jun 12 at 18:57
add a comment |
Do you mean "an arbitrary convex quadrilateral"?
– ajotatxe
Jun 12 at 18:51
Yes, It should be a convex quadrilateral. I will change the statement.
– DiegoMath
Jun 12 at 18:57
Do you mean "an arbitrary convex quadrilateral"?
– ajotatxe
Jun 12 at 18:51
Do you mean "an arbitrary convex quadrilateral"?
– ajotatxe
Jun 12 at 18:51
Yes, It should be a convex quadrilateral. I will change the statement.
– DiegoMath
Jun 12 at 18:57
Yes, It should be a convex quadrilateral. I will change the statement.
– DiegoMath
Jun 12 at 18:57
add a comment |
2 Answers
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I could not draw the diagram but here is the solution. Draw a dotted line from A parallel to DC. Draw dotted perpendiculars from A to DC, M to DC and also from B to DC, they intersect DC at K, G and H respectively (say). Now MG and BH also intersect the line drawn from A parallel to DC say at point E and F respectively. Now as AM = MB so AE = EF and ME = (1/2) BF, let ME = x, BF = 2x. Now say AN and MD intersect at S1 and MC and BN intersect at S2. Now
$Delta ADN = (1/2){overline{DN}}.{overline{AK}}$,
$Delta MDN = (1/2){overline{DN}}.{overline{MG}} = (1/2){overline{DN}}.({overline{AK}} + x) $,
$Delta MNC = (1/2){overline{NC}}.({overline{AK}} + x) = (1/2){overline{DN}}.({overline{AK}} + x)$,
$Delta BNC = (1/2){overline{DN}}.({overline{AK}} + 2x)$,
Now $Delta ADN = Delta ADS_1+ Delta S_1DN$,
$Delta MDN = Delta S_1DN + Delta MS_1N$,
$Delta MNC = Delta MS_2N + Delta S_2NC$,
$Delta BNC = Delta S_2NC + Delta BS_2C$
So, $Delta MDN - Delta ADN = Delta MS_1N - Delta ADS_1 = (1/2){overline{DN}}x$
$Delta BNC - Delta MNC = Delta BS_2C - Delta MS_2N = (1/2){overline{DN}}x$
So $Delta MS_1N + Delta MS_2N = Delta BS_2C + Delta ADS_1$
Note: I assumed BH > AK , but anyway it does not matter.
answered Jun 12 at 20:19
AMITAVA
713
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Using Shoe Lace formulas for quadrilateral and triangle, you can verify algebrically the sum of areas of triangle is area of the quadrilateral.
A = [x1,y1]; B = [x2,y2]; C = [x3,y3]; D = [x4,y4]
M = (1/2*x1 + 1/2*x2, 1/2*y1 + 1/2*y2)
N = (1/2*x3 + 1/2*x4, 1/2*y3 + 1/2*y4)
Q = [(x1*x3*y1 + x2*x3*y1 - x1*x4*y1 + x2*x4*y1 - x3*x4*y1 - x4^2*y1 - 2*x1*x4*y2 + x3*x4*y2 + x4^2*y2 - x1^2*y3 - x1*x2*y3 + 2*x1*x4*y3 + x1^2*y4 + x1*x2*y4 - x1*x3*y4 - x2*x3*y4 + x1*x4*y4 - x2*x4*y4)/(2*x2*y1 + x3*y1 - 3*x4*y1 - 2*x1*y2 + x3*y2 + x4*y2 - x1*y3 - x2*y3 + 2*x4*y3 + 3*x1*y4 - x2*y4 - 2*x3*y4), (x3*y1^2 - x4*y1^2 + x3*y1*y2 - x4*y1*y2 - x1*y1*y3 + x4*y1*y3 - x1*y2*y3 + x4*y2*y3 + x1*y1*y4 + 2*x2*y1*y4 - 2*x3*y1*y4 - x4*y1*y4 - x1*y2*y4 + x4*y2*y4 + x1*y3*y4 - x2*y3*y4 + x1*y4^2 - x2*y4^2)/(2*x2*y1 + x3*y1 - 3*x4*y1 - 2*x1*y2 + x3*y2 + x4*y2 - x1*y3 - x2*y3 + 2*x4*y3 + 3*x1*y4 - x2*y4 - 2*x3*y4)]
R = [(2*x2*x3*y1 - x3^2*y1 - x3*x4*y1 - x1*x3*y2 + x2*x3*y2 + x3^2*y2 - x1*x4*y2 - x2*x4*y2 + x3*x4*y2 - x1*x2*y3 - x2^2*y3 + x1*x3*y3 - x2*x3*y3 + x1*x4*y3 + x2*x4*y3 + x1*x2*y4 + x2^2*y4 - 2*x2*x3*y4)/(2*x2*y1 - x3*y1 - x4*y1 - 2*x1*y2 + 3*x3*y2 - x4*y2 + x1*y3 - 3*x2*y3 + 2*x4*y3 + x1*y4 + x2*y4 - 2*x3*y4), (x3*y1*y2 - x4*y1*y2 + x3*y2^2 - x4*y2^2 + x2*y1*y3 - x3*y1*y3 - 2*x1*y2*y3 - x2*y2*y3 + x3*y2*y3 + 2*x4*y2*y3 + x1*y3^2 - x2*y3^2 + x2*y1*y4 - x3*y1*y4 + x2*y2*y4 - x3*y2*y4 + x1*y3*y4 - x2*y3*y4)/(2*x2*y1 - x3*y1 - x4*y1 - 2*x1*y2 + 3*x3*y2 - x4*y2 + x1*y3 - 3*x2*y3 + 2*x4*y3 + x1*y4 + x2*y4 - 2*x3*y4)]
Area_MRNQ = QuadrilateralShoeLaceArea(M,R,N,Q)
Area_MRNQ = -1/2*(2*x2*y1 - x3*y1 - x4*y1 - 2*x1*y2 + x3*y2 + x4*y2 + x1*y3 - x2*y3 + x1*y4 - x2*y4)*(x2*y1 - x4*y1 - x1*y2 + x3*y2 - x2*y3 + x4*y3 + x1*y4 - x3*y4)*(x3*y1 - x4*y1 + x3*y2 - x4*y2 - x1*y3 - x2*y3 + 2*x4*y3 + x1*y4 + x2*y4 - 2*x3*y4)/((2*x2*y1 + x3*y1 - 3*x4*y1 - 2*x1*y2 + x3*y2 + x4*y2 - x1*y3 - x2*y3 + 2*x4*y3 + 3*x1*y4 - x2*y4 - 2*x3*y4)*(2*x2*y1 - x3*y1 - x4*y1 - 2*x1*y2 + 3*x3*y2 - x4*y2 + x1*y3 - 3*x2*y3 + 2*x4*y3 + x1*y4 + x2*y4 - 2*x3*y4))
Area_AQD = TriangleShoeLaceArea(A,Q,D)
Area_AQD = -1/2*(x2*y1 - x4*y1 - x1*y2 + x4*y2 + x1*y4 - x2*y4)*(x3*y1 - x4*y1 - x1*y3 + x4*y3 + x1*y4 - x3*y4)/(2*x2*y1 + x3*y1 - 3*x4*y1 - 2*x1*y2 + x3*y2 + x4*y2 - x1*y3 - x2*y3 + 2*x4*y3 + 3*x1*y4 - x2*y4 - 2*x3*y4)
Area_BCR = TriangleShoeLaceArea(B,C,R)
Area_BCR = -1/2*(x2*y1 - x3*y1 - x1*y2 + x3*y2 + x1*y3 - x2*y3)*(x3*y2 - x4*y2 - x2*y3 + x4*y3 + x2*y4 - x3*y4)/(2*x2*y1 - x3*y1 - x4*y1 - 2*x1*y2 + 3*x3*y2 - x4*y2 + x1*y3 - 3*x2*y3 + 2*x4*y3 + x1*y4 + x2*y4 - 2*x3*y4)
answered Nov 17 at 18:16
Dominique Laurain
472
It works for any signed quadrilateral (a signed quadrilateral is a quadrilateral where you take care of clockwise orientation for computing area with the shoe lace method) ... convex or not, self intersecting or not.
– Dominique Laurain
Nov 17 at 18:19
add a comment |
2 Answers
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2 Answers
2
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active
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active
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I could not draw the diagram but here is the solution. Draw a dotted line from A parallel to DC. Draw dotted perpendiculars from A to DC, M to DC and also from B to DC, they intersect DC at K, G and H respectively (say). Now MG and BH also intersect the line drawn from A parallel to DC say at point E and F respectively. Now as AM = MB so AE = EF and ME = (1/2) BF, let ME = x, BF = 2x. Now say AN and MD intersect at S1 and MC and BN intersect at S2. Now
$Delta ADN = (1/2){overline{DN}}.{overline{AK}}$,
$Delta MDN = (1/2){overline{DN}}.{overline{MG}} = (1/2){overline{DN}}.({overline{AK}} + x) $,
$Delta MNC = (1/2){overline{NC}}.({overline{AK}} + x) = (1/2){overline{DN}}.({overline{AK}} + x)$,
$Delta BNC = (1/2){overline{DN}}.({overline{AK}} + 2x)$,
Now $Delta ADN = Delta ADS_1+ Delta S_1DN$,
$Delta MDN = Delta S_1DN + Delta MS_1N$,
$Delta MNC = Delta MS_2N + Delta S_2NC$,
$Delta BNC = Delta S_2NC + Delta BS_2C$
So, $Delta MDN - Delta ADN = Delta MS_1N - Delta ADS_1 = (1/2){overline{DN}}x$
$Delta BNC - Delta MNC = Delta BS_2C - Delta MS_2N = (1/2){overline{DN}}x$
So $Delta MS_1N + Delta MS_2N = Delta BS_2C + Delta ADS_1$
Note: I assumed BH > AK , but anyway it does not matter.
answered Jun 12 at 20:19
AMITAVA
713
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up vote
1
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I could not draw the diagram but here is the solution. Draw a dotted line from A parallel to DC. Draw dotted perpendiculars from A to DC, M to DC and also from B to DC, they intersect DC at K, G and H respectively (say). Now MG and BH also intersect the line drawn from A parallel to DC say at point E and F respectively. Now as AM = MB so AE = EF and ME = (1/2) BF, let ME = x, BF = 2x. Now say AN and MD intersect at S1 and MC and BN intersect at S2. Now
$Delta ADN = (1/2){overline{DN}}.{overline{AK}}$,
$Delta MDN = (1/2){overline{DN}}.{overline{MG}} = (1/2){overline{DN}}.({overline{AK}} + x) $,
$Delta MNC = (1/2){overline{NC}}.({overline{AK}} + x) = (1/2){overline{DN}}.({overline{AK}} + x)$,
$Delta BNC = (1/2){overline{DN}}.({overline{AK}} + 2x)$,
Now $Delta ADN = Delta ADS_1+ Delta S_1DN$,
$Delta MDN = Delta S_1DN + Delta MS_1N$,
$Delta MNC = Delta MS_2N + Delta S_2NC$,
$Delta BNC = Delta S_2NC + Delta BS_2C$
So, $Delta MDN - Delta ADN = Delta MS_1N - Delta ADS_1 = (1/2){overline{DN}}x$
$Delta BNC - Delta MNC = Delta BS_2C - Delta MS_2N = (1/2){overline{DN}}x$
So $Delta MS_1N + Delta MS_2N = Delta BS_2C + Delta ADS_1$
Note: I assumed BH > AK , but anyway it does not matter.
answered Jun 12 at 20:19
AMITAVA
713
add a comment |
up vote
1
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up vote
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I could not draw the diagram but here is the solution. Draw a dotted line from A parallel to DC. Draw dotted perpendiculars from A to DC, M to DC and also from B to DC, they intersect DC at K, G and H respectively (say). Now MG and BH also intersect the line drawn from A parallel to DC say at point E and F respectively. Now as AM = MB so AE = EF and ME = (1/2) BF, let ME = x, BF = 2x. Now say AN and MD intersect at S1 and MC and BN intersect at S2. Now
$Delta ADN = (1/2){overline{DN}}.{overline{AK}}$,
$Delta MDN = (1/2){overline{DN}}.{overline{MG}} = (1/2){overline{DN}}.({overline{AK}} + x) $,
$Delta MNC = (1/2){overline{NC}}.({overline{AK}} + x) = (1/2){overline{DN}}.({overline{AK}} + x)$,
$Delta BNC = (1/2){overline{DN}}.({overline{AK}} + 2x)$,
Now $Delta ADN = Delta ADS_1+ Delta S_1DN$,
$Delta MDN = Delta S_1DN + Delta MS_1N$,
$Delta MNC = Delta MS_2N + Delta S_2NC$,
$Delta BNC = Delta S_2NC + Delta BS_2C$
So, $Delta MDN - Delta ADN = Delta MS_1N - Delta ADS_1 = (1/2){overline{DN}}x$
$Delta BNC - Delta MNC = Delta BS_2C - Delta MS_2N = (1/2){overline{DN}}x$
So $Delta MS_1N + Delta MS_2N = Delta BS_2C + Delta ADS_1$
Note: I assumed BH > AK , but anyway it does not matter.
answered Jun 12 at 20:19
AMITAVA
713
I could not draw the diagram but here is the solution. Draw a dotted line from A parallel to DC. Draw dotted perpendiculars from A to DC, M to DC and also from B to DC, they intersect DC at K, G and H respectively (say). Now MG and BH also intersect the line drawn from A parallel to DC say at point E and F respectively. Now as AM = MB so AE = EF and ME = (1/2) BF, let ME = x, BF = 2x. Now say AN and MD intersect at S1 and MC and BN intersect at S2. Now
$Delta ADN = (1/2){overline{DN}}.{overline{AK}}$,
$Delta MDN = (1/2){overline{DN}}.{overline{MG}} = (1/2){overline{DN}}.({overline{AK}} + x) $,
$Delta MNC = (1/2){overline{NC}}.({overline{AK}} + x) = (1/2){overline{DN}}.({overline{AK}} + x)$,
$Delta BNC = (1/2){overline{DN}}.({overline{AK}} + 2x)$,
Now $Delta ADN = Delta ADS_1+ Delta S_1DN$,
$Delta MDN = Delta S_1DN + Delta MS_1N$,
$Delta MNC = Delta MS_2N + Delta S_2NC$,
$Delta BNC = Delta S_2NC + Delta BS_2C$
So, $Delta MDN - Delta ADN = Delta MS_1N - Delta ADS_1 = (1/2){overline{DN}}x$
$Delta BNC - Delta MNC = Delta BS_2C - Delta MS_2N = (1/2){overline{DN}}x$
So $Delta MS_1N + Delta MS_2N = Delta BS_2C + Delta ADS_1$
Note: I assumed BH > AK , but anyway it does not matter.
answered Jun 12 at 20:19
AMITAVA
713
edited Jun 13 at 6:54
answered Jun 12 at 20:19
AMITAVA
713
answered Jun 12 at 20:19
AMITAVA
713
answered Jun 12 at 20:19
AMITAVA
713
713
add a comment |
add a comment |
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0
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Using Shoe Lace formulas for quadrilateral and triangle, you can verify algebrically the sum of areas of triangle is area of the quadrilateral.
A = [x1,y1]; B = [x2,y2]; C = [x3,y3]; D = [x4,y4]
M = (1/2*x1 + 1/2*x2, 1/2*y1 + 1/2*y2)
N = (1/2*x3 + 1/2*x4, 1/2*y3 + 1/2*y4)
Q = [(x1*x3*y1 + x2*x3*y1 - x1*x4*y1 + x2*x4*y1 - x3*x4*y1 - x4^2*y1 - 2*x1*x4*y2 + x3*x4*y2 + x4^2*y2 - x1^2*y3 - x1*x2*y3 + 2*x1*x4*y3 + x1^2*y4 + x1*x2*y4 - x1*x3*y4 - x2*x3*y4 + x1*x4*y4 - x2*x4*y4)/(2*x2*y1 + x3*y1 - 3*x4*y1 - 2*x1*y2 + x3*y2 + x4*y2 - x1*y3 - x2*y3 + 2*x4*y3 + 3*x1*y4 - x2*y4 - 2*x3*y4), (x3*y1^2 - x4*y1^2 + x3*y1*y2 - x4*y1*y2 - x1*y1*y3 + x4*y1*y3 - x1*y2*y3 + x4*y2*y3 + x1*y1*y4 + 2*x2*y1*y4 - 2*x3*y1*y4 - x4*y1*y4 - x1*y2*y4 + x4*y2*y4 + x1*y3*y4 - x2*y3*y4 + x1*y4^2 - x2*y4^2)/(2*x2*y1 + x3*y1 - 3*x4*y1 - 2*x1*y2 + x3*y2 + x4*y2 - x1*y3 - x2*y3 + 2*x4*y3 + 3*x1*y4 - x2*y4 - 2*x3*y4)]
R = [(2*x2*x3*y1 - x3^2*y1 - x3*x4*y1 - x1*x3*y2 + x2*x3*y2 + x3^2*y2 - x1*x4*y2 - x2*x4*y2 + x3*x4*y2 - x1*x2*y3 - x2^2*y3 + x1*x3*y3 - x2*x3*y3 + x1*x4*y3 + x2*x4*y3 + x1*x2*y4 + x2^2*y4 - 2*x2*x3*y4)/(2*x2*y1 - x3*y1 - x4*y1 - 2*x1*y2 + 3*x3*y2 - x4*y2 + x1*y3 - 3*x2*y3 + 2*x4*y3 + x1*y4 + x2*y4 - 2*x3*y4), (x3*y1*y2 - x4*y1*y2 + x3*y2^2 - x4*y2^2 + x2*y1*y3 - x3*y1*y3 - 2*x1*y2*y3 - x2*y2*y3 + x3*y2*y3 + 2*x4*y2*y3 + x1*y3^2 - x2*y3^2 + x2*y1*y4 - x3*y1*y4 + x2*y2*y4 - x3*y2*y4 + x1*y3*y4 - x2*y3*y4)/(2*x2*y1 - x3*y1 - x4*y1 - 2*x1*y2 + 3*x3*y2 - x4*y2 + x1*y3 - 3*x2*y3 + 2*x4*y3 + x1*y4 + x2*y4 - 2*x3*y4)]
Area_MRNQ = QuadrilateralShoeLaceArea(M,R,N,Q)
Area_MRNQ = -1/2*(2*x2*y1 - x3*y1 - x4*y1 - 2*x1*y2 + x3*y2 + x4*y2 + x1*y3 - x2*y3 + x1*y4 - x2*y4)*(x2*y1 - x4*y1 - x1*y2 + x3*y2 - x2*y3 + x4*y3 + x1*y4 - x3*y4)*(x3*y1 - x4*y1 + x3*y2 - x4*y2 - x1*y3 - x2*y3 + 2*x4*y3 + x1*y4 + x2*y4 - 2*x3*y4)/((2*x2*y1 + x3*y1 - 3*x4*y1 - 2*x1*y2 + x3*y2 + x4*y2 - x1*y3 - x2*y3 + 2*x4*y3 + 3*x1*y4 - x2*y4 - 2*x3*y4)*(2*x2*y1 - x3*y1 - x4*y1 - 2*x1*y2 + 3*x3*y2 - x4*y2 + x1*y3 - 3*x2*y3 + 2*x4*y3 + x1*y4 + x2*y4 - 2*x3*y4))
Area_AQD = TriangleShoeLaceArea(A,Q,D)
Area_AQD = -1/2*(x2*y1 - x4*y1 - x1*y2 + x4*y2 + x1*y4 - x2*y4)*(x3*y1 - x4*y1 - x1*y3 + x4*y3 + x1*y4 - x3*y4)/(2*x2*y1 + x3*y1 - 3*x4*y1 - 2*x1*y2 + x3*y2 + x4*y2 - x1*y3 - x2*y3 + 2*x4*y3 + 3*x1*y4 - x2*y4 - 2*x3*y4)
Area_BCR = TriangleShoeLaceArea(B,C,R)
Area_BCR = -1/2*(x2*y1 - x3*y1 - x1*y2 + x3*y2 + x1*y3 - x2*y3)*(x3*y2 - x4*y2 - x2*y3 + x4*y3 + x2*y4 - x3*y4)/(2*x2*y1 - x3*y1 - x4*y1 - 2*x1*y2 + 3*x3*y2 - x4*y2 + x1*y3 - 3*x2*y3 + 2*x4*y3 + x1*y4 + x2*y4 - 2*x3*y4)
answered Nov 17 at 18:16
Dominique Laurain
472
It works for any signed quadrilateral (a signed quadrilateral is a quadrilateral where you take care of clockwise orientation for computing area with the shoe lace method) ... convex or not, self intersecting or not.
– Dominique Laurain
Nov 17 at 18:19
add a comment |
up vote
0
down vote
Using Shoe Lace formulas for quadrilateral and triangle, you can verify algebrically the sum of areas of triangle is area of the quadrilateral.
A = [x1,y1]; B = [x2,y2]; C = [x3,y3]; D = [x4,y4]
M = (1/2*x1 + 1/2*x2, 1/2*y1 + 1/2*y2)
N = (1/2*x3 + 1/2*x4, 1/2*y3 + 1/2*y4)
Q = [(x1*x3*y1 + x2*x3*y1 - x1*x4*y1 + x2*x4*y1 - x3*x4*y1 - x4^2*y1 - 2*x1*x4*y2 + x3*x4*y2 + x4^2*y2 - x1^2*y3 - x1*x2*y3 + 2*x1*x4*y3 + x1^2*y4 + x1*x2*y4 - x1*x3*y4 - x2*x3*y4 + x1*x4*y4 - x2*x4*y4)/(2*x2*y1 + x3*y1 - 3*x4*y1 - 2*x1*y2 + x3*y2 + x4*y2 - x1*y3 - x2*y3 + 2*x4*y3 + 3*x1*y4 - x2*y4 - 2*x3*y4), (x3*y1^2 - x4*y1^2 + x3*y1*y2 - x4*y1*y2 - x1*y1*y3 + x4*y1*y3 - x1*y2*y3 + x4*y2*y3 + x1*y1*y4 + 2*x2*y1*y4 - 2*x3*y1*y4 - x4*y1*y4 - x1*y2*y4 + x4*y2*y4 + x1*y3*y4 - x2*y3*y4 + x1*y4^2 - x2*y4^2)/(2*x2*y1 + x3*y1 - 3*x4*y1 - 2*x1*y2 + x3*y2 + x4*y2 - x1*y3 - x2*y3 + 2*x4*y3 + 3*x1*y4 - x2*y4 - 2*x3*y4)]
R = [(2*x2*x3*y1 - x3^2*y1 - x3*x4*y1 - x1*x3*y2 + x2*x3*y2 + x3^2*y2 - x1*x4*y2 - x2*x4*y2 + x3*x4*y2 - x1*x2*y3 - x2^2*y3 + x1*x3*y3 - x2*x3*y3 + x1*x4*y3 + x2*x4*y3 + x1*x2*y4 + x2^2*y4 - 2*x2*x3*y4)/(2*x2*y1 - x3*y1 - x4*y1 - 2*x1*y2 + 3*x3*y2 - x4*y2 + x1*y3 - 3*x2*y3 + 2*x4*y3 + x1*y4 + x2*y4 - 2*x3*y4), (x3*y1*y2 - x4*y1*y2 + x3*y2^2 - x4*y2^2 + x2*y1*y3 - x3*y1*y3 - 2*x1*y2*y3 - x2*y2*y3 + x3*y2*y3 + 2*x4*y2*y3 + x1*y3^2 - x2*y3^2 + x2*y1*y4 - x3*y1*y4 + x2*y2*y4 - x3*y2*y4 + x1*y3*y4 - x2*y3*y4)/(2*x2*y1 - x3*y1 - x4*y1 - 2*x1*y2 + 3*x3*y2 - x4*y2 + x1*y3 - 3*x2*y3 + 2*x4*y3 + x1*y4 + x2*y4 - 2*x3*y4)]
Area_MRNQ = QuadrilateralShoeLaceArea(M,R,N,Q)
Area_MRNQ = -1/2*(2*x2*y1 - x3*y1 - x4*y1 - 2*x1*y2 + x3*y2 + x4*y2 + x1*y3 - x2*y3 + x1*y4 - x2*y4)*(x2*y1 - x4*y1 - x1*y2 + x3*y2 - x2*y3 + x4*y3 + x1*y4 - x3*y4)*(x3*y1 - x4*y1 + x3*y2 - x4*y2 - x1*y3 - x2*y3 + 2*x4*y3 + x1*y4 + x2*y4 - 2*x3*y4)/((2*x2*y1 + x3*y1 - 3*x4*y1 - 2*x1*y2 + x3*y2 + x4*y2 - x1*y3 - x2*y3 + 2*x4*y3 + 3*x1*y4 - x2*y4 - 2*x3*y4)*(2*x2*y1 - x3*y1 - x4*y1 - 2*x1*y2 + 3*x3*y2 - x4*y2 + x1*y3 - 3*x2*y3 + 2*x4*y3 + x1*y4 + x2*y4 - 2*x3*y4))
Area_AQD = TriangleShoeLaceArea(A,Q,D)
Area_AQD = -1/2*(x2*y1 - x4*y1 - x1*y2 + x4*y2 + x1*y4 - x2*y4)*(x3*y1 - x4*y1 - x1*y3 + x4*y3 + x1*y4 - x3*y4)/(2*x2*y1 + x3*y1 - 3*x4*y1 - 2*x1*y2 + x3*y2 + x4*y2 - x1*y3 - x2*y3 + 2*x4*y3 + 3*x1*y4 - x2*y4 - 2*x3*y4)
Area_BCR = TriangleShoeLaceArea(B,C,R)
Area_BCR = -1/2*(x2*y1 - x3*y1 - x1*y2 + x3*y2 + x1*y3 - x2*y3)*(x3*y2 - x4*y2 - x2*y3 + x4*y3 + x2*y4 - x3*y4)/(2*x2*y1 - x3*y1 - x4*y1 - 2*x1*y2 + 3*x3*y2 - x4*y2 + x1*y3 - 3*x2*y3 + 2*x4*y3 + x1*y4 + x2*y4 - 2*x3*y4)
answered Nov 17 at 18:16
Dominique Laurain
472
It works for any signed quadrilateral (a signed quadrilateral is a quadrilateral where you take care of clockwise orientation for computing area with the shoe lace method) ... convex or not, self intersecting or not.
– Dominique Laurain
Nov 17 at 18:19
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Using Shoe Lace formulas for quadrilateral and triangle, you can verify algebrically the sum of areas of triangle is area of the quadrilateral.
A = [x1,y1]; B = [x2,y2]; C = [x3,y3]; D = [x4,y4]
M = (1/2*x1 + 1/2*x2, 1/2*y1 + 1/2*y2)
N = (1/2*x3 + 1/2*x4, 1/2*y3 + 1/2*y4)
Q = [(x1*x3*y1 + x2*x3*y1 - x1*x4*y1 + x2*x4*y1 - x3*x4*y1 - x4^2*y1 - 2*x1*x4*y2 + x3*x4*y2 + x4^2*y2 - x1^2*y3 - x1*x2*y3 + 2*x1*x4*y3 + x1^2*y4 + x1*x2*y4 - x1*x3*y4 - x2*x3*y4 + x1*x4*y4 - x2*x4*y4)/(2*x2*y1 + x3*y1 - 3*x4*y1 - 2*x1*y2 + x3*y2 + x4*y2 - x1*y3 - x2*y3 + 2*x4*y3 + 3*x1*y4 - x2*y4 - 2*x3*y4), (x3*y1^2 - x4*y1^2 + x3*y1*y2 - x4*y1*y2 - x1*y1*y3 + x4*y1*y3 - x1*y2*y3 + x4*y2*y3 + x1*y1*y4 + 2*x2*y1*y4 - 2*x3*y1*y4 - x4*y1*y4 - x1*y2*y4 + x4*y2*y4 + x1*y3*y4 - x2*y3*y4 + x1*y4^2 - x2*y4^2)/(2*x2*y1 + x3*y1 - 3*x4*y1 - 2*x1*y2 + x3*y2 + x4*y2 - x1*y3 - x2*y3 + 2*x4*y3 + 3*x1*y4 - x2*y4 - 2*x3*y4)]
R = [(2*x2*x3*y1 - x3^2*y1 - x3*x4*y1 - x1*x3*y2 + x2*x3*y2 + x3^2*y2 - x1*x4*y2 - x2*x4*y2 + x3*x4*y2 - x1*x2*y3 - x2^2*y3 + x1*x3*y3 - x2*x3*y3 + x1*x4*y3 + x2*x4*y3 + x1*x2*y4 + x2^2*y4 - 2*x2*x3*y4)/(2*x2*y1 - x3*y1 - x4*y1 - 2*x1*y2 + 3*x3*y2 - x4*y2 + x1*y3 - 3*x2*y3 + 2*x4*y3 + x1*y4 + x2*y4 - 2*x3*y4), (x3*y1*y2 - x4*y1*y2 + x3*y2^2 - x4*y2^2 + x2*y1*y3 - x3*y1*y3 - 2*x1*y2*y3 - x2*y2*y3 + x3*y2*y3 + 2*x4*y2*y3 + x1*y3^2 - x2*y3^2 + x2*y1*y4 - x3*y1*y4 + x2*y2*y4 - x3*y2*y4 + x1*y3*y4 - x2*y3*y4)/(2*x2*y1 - x3*y1 - x4*y1 - 2*x1*y2 + 3*x3*y2 - x4*y2 + x1*y3 - 3*x2*y3 + 2*x4*y3 + x1*y4 + x2*y4 - 2*x3*y4)]
Area_MRNQ = QuadrilateralShoeLaceArea(M,R,N,Q)
Area_MRNQ = -1/2*(2*x2*y1 - x3*y1 - x4*y1 - 2*x1*y2 + x3*y2 + x4*y2 + x1*y3 - x2*y3 + x1*y4 - x2*y4)*(x2*y1 - x4*y1 - x1*y2 + x3*y2 - x2*y3 + x4*y3 + x1*y4 - x3*y4)*(x3*y1 - x4*y1 + x3*y2 - x4*y2 - x1*y3 - x2*y3 + 2*x4*y3 + x1*y4 + x2*y4 - 2*x3*y4)/((2*x2*y1 + x3*y1 - 3*x4*y1 - 2*x1*y2 + x3*y2 + x4*y2 - x1*y3 - x2*y3 + 2*x4*y3 + 3*x1*y4 - x2*y4 - 2*x3*y4)*(2*x2*y1 - x3*y1 - x4*y1 - 2*x1*y2 + 3*x3*y2 - x4*y2 + x1*y3 - 3*x2*y3 + 2*x4*y3 + x1*y4 + x2*y4 - 2*x3*y4))
Area_AQD = TriangleShoeLaceArea(A,Q,D)
Area_AQD = -1/2*(x2*y1 - x4*y1 - x1*y2 + x4*y2 + x1*y4 - x2*y4)*(x3*y1 - x4*y1 - x1*y3 + x4*y3 + x1*y4 - x3*y4)/(2*x2*y1 + x3*y1 - 3*x4*y1 - 2*x1*y2 + x3*y2 + x4*y2 - x1*y3 - x2*y3 + 2*x4*y3 + 3*x1*y4 - x2*y4 - 2*x3*y4)
Area_BCR = TriangleShoeLaceArea(B,C,R)
Area_BCR = -1/2*(x2*y1 - x3*y1 - x1*y2 + x3*y2 + x1*y3 - x2*y3)*(x3*y2 - x4*y2 - x2*y3 + x4*y3 + x2*y4 - x3*y4)/(2*x2*y1 - x3*y1 - x4*y1 - 2*x1*y2 + 3*x3*y2 - x4*y2 + x1*y3 - 3*x2*y3 + 2*x4*y3 + x1*y4 + x2*y4 - 2*x3*y4)
answered Nov 17 at 18:16
Dominique Laurain
472
Using Shoe Lace formulas for quadrilateral and triangle, you can verify algebrically the sum of areas of triangle is area of the quadrilateral.
A = [x1,y1]; B = [x2,y2]; C = [x3,y3]; D = [x4,y4]
M = (1/2*x1 + 1/2*x2, 1/2*y1 + 1/2*y2)
N = (1/2*x3 + 1/2*x4, 1/2*y3 + 1/2*y4)
Q = [(x1*x3*y1 + x2*x3*y1 - x1*x4*y1 + x2*x4*y1 - x3*x4*y1 - x4^2*y1 - 2*x1*x4*y2 + x3*x4*y2 + x4^2*y2 - x1^2*y3 - x1*x2*y3 + 2*x1*x4*y3 + x1^2*y4 + x1*x2*y4 - x1*x3*y4 - x2*x3*y4 + x1*x4*y4 - x2*x4*y4)/(2*x2*y1 + x3*y1 - 3*x4*y1 - 2*x1*y2 + x3*y2 + x4*y2 - x1*y3 - x2*y3 + 2*x4*y3 + 3*x1*y4 - x2*y4 - 2*x3*y4), (x3*y1^2 - x4*y1^2 + x3*y1*y2 - x4*y1*y2 - x1*y1*y3 + x4*y1*y3 - x1*y2*y3 + x4*y2*y3 + x1*y1*y4 + 2*x2*y1*y4 - 2*x3*y1*y4 - x4*y1*y4 - x1*y2*y4 + x4*y2*y4 + x1*y3*y4 - x2*y3*y4 + x1*y4^2 - x2*y4^2)/(2*x2*y1 + x3*y1 - 3*x4*y1 - 2*x1*y2 + x3*y2 + x4*y2 - x1*y3 - x2*y3 + 2*x4*y3 + 3*x1*y4 - x2*y4 - 2*x3*y4)]
R = [(2*x2*x3*y1 - x3^2*y1 - x3*x4*y1 - x1*x3*y2 + x2*x3*y2 + x3^2*y2 - x1*x4*y2 - x2*x4*y2 + x3*x4*y2 - x1*x2*y3 - x2^2*y3 + x1*x3*y3 - x2*x3*y3 + x1*x4*y3 + x2*x4*y3 + x1*x2*y4 + x2^2*y4 - 2*x2*x3*y4)/(2*x2*y1 - x3*y1 - x4*y1 - 2*x1*y2 + 3*x3*y2 - x4*y2 + x1*y3 - 3*x2*y3 + 2*x4*y3 + x1*y4 + x2*y4 - 2*x3*y4), (x3*y1*y2 - x4*y1*y2 + x3*y2^2 - x4*y2^2 + x2*y1*y3 - x3*y1*y3 - 2*x1*y2*y3 - x2*y2*y3 + x3*y2*y3 + 2*x4*y2*y3 + x1*y3^2 - x2*y3^2 + x2*y1*y4 - x3*y1*y4 + x2*y2*y4 - x3*y2*y4 + x1*y3*y4 - x2*y3*y4)/(2*x2*y1 - x3*y1 - x4*y1 - 2*x1*y2 + 3*x3*y2 - x4*y2 + x1*y3 - 3*x2*y3 + 2*x4*y3 + x1*y4 + x2*y4 - 2*x3*y4)]
Area_MRNQ = QuadrilateralShoeLaceArea(M,R,N,Q)
Area_MRNQ = -1/2*(2*x2*y1 - x3*y1 - x4*y1 - 2*x1*y2 + x3*y2 + x4*y2 + x1*y3 - x2*y3 + x1*y4 - x2*y4)*(x2*y1 - x4*y1 - x1*y2 + x3*y2 - x2*y3 + x4*y3 + x1*y4 - x3*y4)*(x3*y1 - x4*y1 + x3*y2 - x4*y2 - x1*y3 - x2*y3 + 2*x4*y3 + x1*y4 + x2*y4 - 2*x3*y4)/((2*x2*y1 + x3*y1 - 3*x4*y1 - 2*x1*y2 + x3*y2 + x4*y2 - x1*y3 - x2*y3 + 2*x4*y3 + 3*x1*y4 - x2*y4 - 2*x3*y4)*(2*x2*y1 - x3*y1 - x4*y1 - 2*x1*y2 + 3*x3*y2 - x4*y2 + x1*y3 - 3*x2*y3 + 2*x4*y3 + x1*y4 + x2*y4 - 2*x3*y4))
Area_AQD = TriangleShoeLaceArea(A,Q,D)
Area_AQD = -1/2*(x2*y1 - x4*y1 - x1*y2 + x4*y2 + x1*y4 - x2*y4)*(x3*y1 - x4*y1 - x1*y3 + x4*y3 + x1*y4 - x3*y4)/(2*x2*y1 + x3*y1 - 3*x4*y1 - 2*x1*y2 + x3*y2 + x4*y2 - x1*y3 - x2*y3 + 2*x4*y3 + 3*x1*y4 - x2*y4 - 2*x3*y4)
Area_BCR = TriangleShoeLaceArea(B,C,R)
Area_BCR = -1/2*(x2*y1 - x3*y1 - x1*y2 + x3*y2 + x1*y3 - x2*y3)*(x3*y2 - x4*y2 - x2*y3 + x4*y3 + x2*y4 - x3*y4)/(2*x2*y1 - x3*y1 - x4*y1 - 2*x1*y2 + 3*x3*y2 - x4*y2 + x1*y3 - 3*x2*y3 + 2*x4*y3 + x1*y4 + x2*y4 - 2*x3*y4)
answered Nov 17 at 18:16
Dominique Laurain
472
answered Nov 17 at 18:16
Dominique Laurain
472
answered Nov 17 at 18:16
Dominique Laurain
472
answered Nov 17 at 18:16
Dominique Laurain
472
472
It works for any signed quadrilateral (a signed quadrilateral is a quadrilateral where you take care of clockwise orientation for computing area with the shoe lace method) ... convex or not, self intersecting or not.
– Dominique Laurain
Nov 17 at 18:19
add a comment |
It works for any signed quadrilateral (a signed quadrilateral is a quadrilateral where you take care of clockwise orientation for computing area with the shoe lace method) ... convex or not, self intersecting or not.
– Dominique Laurain
Nov 17 at 18:19
It works for any signed quadrilateral (a signed quadrilateral is a quadrilateral where you take care of clockwise orientation for computing area with the shoe lace method) ... convex or not, self intersecting or not.
– Dominique Laurain
Nov 17 at 18:19
It works for any signed quadrilateral (a signed quadrilateral is a quadrilateral where you take care of clockwise orientation for computing area with the shoe lace method) ... convex or not, self intersecting or not.
– Dominique Laurain
Nov 17 at 18:19
add a comment |
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Do you mean "an arbitrary convex quadrilateral"?
– ajotatxe
Jun 12 at 18:51
Yes, It should be a convex quadrilateral. I will change the statement.
– DiegoMath
Jun 12 at 18:57