triangle relation equation of a pentagon [closed]
$begingroup$
ABCDE is an irregular-shaped pentagon. No angle is more than 180.
prove that,
area of (ABD)(AEC)-area of (ABC)(ADE) = area of (ABE)*(ACD)
geometry
asked Dec 19 '18 at 21:19
epicsaadepicsaad
62
$endgroup$
closed as off-topic by Saad, Lord Shark the Unknown, Shailesh, Karn Watcharasupat, Leucippus Dec 20 '18 at 3:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
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$begingroup$
ABCDE is an irregular-shaped pentagon. No angle is more than 180.
prove that,
area of (ABD)(AEC)-area of (ABC)(ADE) = area of (ABE)*(ACD)
geometry
asked Dec 19 '18 at 21:19
epicsaadepicsaad
62
$endgroup$
closed as off-topic by Saad, Lord Shark the Unknown, Shailesh, Karn Watcharasupat, Leucippus Dec 20 '18 at 3:19
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." – Saad, Lord Shark the Unknown, Shailesh, Karn Watcharasupat, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
ABCDE is an irregular-shaped pentagon. No angle is more than 180.
prove that,
area of (ABD)(AEC)-area of (ABC)(ADE) = area of (ABE)*(ACD)
geometry
asked Dec 19 '18 at 21:19
epicsaadepicsaad
62
$endgroup$
ABCDE is an irregular-shaped pentagon. No angle is more than 180.
prove that,
area of (ABD)(AEC)-area of (ABC)(ADE) = area of (ABE)*(ACD)
geometry
geometry
asked Dec 19 '18 at 21:19
epicsaadepicsaad
62
asked Dec 19 '18 at 21:19
epicsaadepicsaad
62
asked Dec 19 '18 at 21:19
epicsaadepicsaad
62
asked Dec 19 '18 at 21:19
epicsaadepicsaad
62
asked Dec 19 '18 at 21:19
epicsaadepicsaad
62
62
closed as off-topic by Saad, Lord Shark the Unknown, Shailesh, Karn Watcharasupat, Leucippus Dec 20 '18 at 3:19
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." – Saad, Lord Shark the Unknown, Shailesh, Karn Watcharasupat, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Lord Shark the Unknown, Shailesh, Karn Watcharasupat, Leucippus Dec 20 '18 at 3:19
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." – Saad, Lord Shark the Unknown, Shailesh, Karn Watcharasupat, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
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1 Answer
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$begingroup$
We can use coordinate geometry to solve this problem. Let side AE lie on the x-axis and point A is $(0,0)$, B is $(x_1, y_1)$, C is $(x_2, y_2)$, D is $(x_3, y_3)$, E is $(x_4, 0)$.
In coordinate geometry we have a way to find the area of any polygon using its vertices.
Area ABD is $frac{1}{2}(x_1y_3 - x_3y_1)$
Area AEC is $frac{1}{2}(x_4y_2)$
Area ABC is $frac{1}{2}(x_1y_2 - x_2y_1)$
Area ADE is $frac{1}{2}(x_4y_3)$
Area ABE is $frac{1}{2}(x_4y_1)$
Area ACD is $frac{1}{2}(x_2y_3 - x_3y_2)$
Hence Area ABD x Area AEC - Area ABC x Area ADE is
$frac{1}{4}(x_2x_4y_1y_3 - x_3x_4y_1y_2)$
For the right hand side Area ABE x Area ACD is the same. Hence proved.
answered Dec 20 '18 at 2:58
KY TangKY Tang
37435
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We can use coordinate geometry to solve this problem. Let side AE lie on the x-axis and point A is $(0,0)$, B is $(x_1, y_1)$, C is $(x_2, y_2)$, D is $(x_3, y_3)$, E is $(x_4, 0)$.
In coordinate geometry we have a way to find the area of any polygon using its vertices.
Area ABD is $frac{1}{2}(x_1y_3 - x_3y_1)$
Area AEC is $frac{1}{2}(x_4y_2)$
Area ABC is $frac{1}{2}(x_1y_2 - x_2y_1)$
Area ADE is $frac{1}{2}(x_4y_3)$
Area ABE is $frac{1}{2}(x_4y_1)$
Area ACD is $frac{1}{2}(x_2y_3 - x_3y_2)$
Hence Area ABD x Area AEC - Area ABC x Area ADE is
$frac{1}{4}(x_2x_4y_1y_3 - x_3x_4y_1y_2)$
For the right hand side Area ABE x Area ACD is the same. Hence proved.
answered Dec 20 '18 at 2:58
KY TangKY Tang
37435
$endgroup$
add a comment |
$begingroup$
We can use coordinate geometry to solve this problem. Let side AE lie on the x-axis and point A is $(0,0)$, B is $(x_1, y_1)$, C is $(x_2, y_2)$, D is $(x_3, y_3)$, E is $(x_4, 0)$.
In coordinate geometry we have a way to find the area of any polygon using its vertices.
Area ABD is $frac{1}{2}(x_1y_3 - x_3y_1)$
Area AEC is $frac{1}{2}(x_4y_2)$
Area ABC is $frac{1}{2}(x_1y_2 - x_2y_1)$
Area ADE is $frac{1}{2}(x_4y_3)$
Area ABE is $frac{1}{2}(x_4y_1)$
Area ACD is $frac{1}{2}(x_2y_3 - x_3y_2)$
Hence Area ABD x Area AEC - Area ABC x Area ADE is
$frac{1}{4}(x_2x_4y_1y_3 - x_3x_4y_1y_2)$
For the right hand side Area ABE x Area ACD is the same. Hence proved.
answered Dec 20 '18 at 2:58
KY TangKY Tang
37435
$endgroup$
add a comment |
$begingroup$
We can use coordinate geometry to solve this problem. Let side AE lie on the x-axis and point A is $(0,0)$, B is $(x_1, y_1)$, C is $(x_2, y_2)$, D is $(x_3, y_3)$, E is $(x_4, 0)$.
In coordinate geometry we have a way to find the area of any polygon using its vertices.
Area ABD is $frac{1}{2}(x_1y_3 - x_3y_1)$
Area AEC is $frac{1}{2}(x_4y_2)$
Area ABC is $frac{1}{2}(x_1y_2 - x_2y_1)$
Area ADE is $frac{1}{2}(x_4y_3)$
Area ABE is $frac{1}{2}(x_4y_1)$
Area ACD is $frac{1}{2}(x_2y_3 - x_3y_2)$
Hence Area ABD x Area AEC - Area ABC x Area ADE is
$frac{1}{4}(x_2x_4y_1y_3 - x_3x_4y_1y_2)$
For the right hand side Area ABE x Area ACD is the same. Hence proved.
answered Dec 20 '18 at 2:58
KY TangKY Tang
37435
$endgroup$
We can use coordinate geometry to solve this problem. Let side AE lie on the x-axis and point A is $(0,0)$, B is $(x_1, y_1)$, C is $(x_2, y_2)$, D is $(x_3, y_3)$, E is $(x_4, 0)$.
In coordinate geometry we have a way to find the area of any polygon using its vertices.
Area ABD is $frac{1}{2}(x_1y_3 - x_3y_1)$
Area AEC is $frac{1}{2}(x_4y_2)$
Area ABC is $frac{1}{2}(x_1y_2 - x_2y_1)$
Area ADE is $frac{1}{2}(x_4y_3)$
Area ABE is $frac{1}{2}(x_4y_1)$
Area ACD is $frac{1}{2}(x_2y_3 - x_3y_2)$
Hence Area ABD x Area AEC - Area ABC x Area ADE is
$frac{1}{4}(x_2x_4y_1y_3 - x_3x_4y_1y_2)$
For the right hand side Area ABE x Area ACD is the same. Hence proved.
answered Dec 20 '18 at 2:58
KY TangKY Tang
37435
answered Dec 20 '18 at 2:58
KY TangKY Tang
37435
answered Dec 20 '18 at 2:58
KY TangKY Tang
37435
answered Dec 20 '18 at 2:58
KY TangKY Tang
37435
37435
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