truss structure geometry - geometric induction?
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This might be a really simple Geometry rule that I'm missing, but I can't understand how in this problem they automatically knew to extend the truss to point X and knew that it was an additional 2 little triangles. I know that if I were to solve the entire truss using the typical tools like law of sines/cosines, and pythagorus, I could perhaps conclude the same thing, but it seems like there is a big shortcut I am missing to do this instantly.
attached are print screens from Meriam & Kraige Statics textbook of the problem and the solution:
geometry
asked May 16 '12 at 11:08
nofenofe
565513
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add a comment |
$begingroup$
This might be a really simple Geometry rule that I'm missing, but I can't understand how in this problem they automatically knew to extend the truss to point X and knew that it was an additional 2 little triangles. I know that if I were to solve the entire truss using the typical tools like law of sines/cosines, and pythagorus, I could perhaps conclude the same thing, but it seems like there is a big shortcut I am missing to do this instantly.
attached are print screens from Meriam & Kraige Statics textbook of the problem and the solution:
geometry
asked May 16 '12 at 11:08
nofenofe
565513
$endgroup$
add a comment |
$begingroup$
This might be a really simple Geometry rule that I'm missing, but I can't understand how in this problem they automatically knew to extend the truss to point X and knew that it was an additional 2 little triangles. I know that if I were to solve the entire truss using the typical tools like law of sines/cosines, and pythagorus, I could perhaps conclude the same thing, but it seems like there is a big shortcut I am missing to do this instantly.
attached are print screens from Meriam & Kraige Statics textbook of the problem and the solution:
geometry
asked May 16 '12 at 11:08
nofenofe
565513
$endgroup$
This might be a really simple Geometry rule that I'm missing, but I can't understand how in this problem they automatically knew to extend the truss to point X and knew that it was an additional 2 little triangles. I know that if I were to solve the entire truss using the typical tools like law of sines/cosines, and pythagorus, I could perhaps conclude the same thing, but it seems like there is a big shortcut I am missing to do this instantly.
attached are print screens from Meriam & Kraige Statics textbook of the problem and the solution:
geometry
geometry
asked May 16 '12 at 11:08
nofenofe
565513
asked May 16 '12 at 11:08
nofenofe
565513
asked May 16 '12 at 11:08
nofenofe
565513
asked May 16 '12 at 11:08
nofenofe
565513
asked May 16 '12 at 11:08
nofenofe
565513
565513
add a comment |
add a comment |
2 Answers
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I think that what they did here is a standard "trick" used a lot in geometry-trigonometry problems: they draw a a help-line (or help-circle or whatever) in order to make the problem easier and/or more manageable.
In the present case, they extend in a straight way the line FJ from the side of J all the way until it meets in point X the straight extension of line OL in order to form the straight-angle triangle $,Delta FOX,$, with $,angle FOX=90,$ , and now they can use trigonometric functions to do stuff...!
answered May 16 '12 at 11:32
DonAntonioDonAntonio
180k1494233
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Similar to DonAntonio's answer, it's often the case when working with (non-parallelogram) trapezoids that it's easier to consider the trapezoid as a truncated triangle—that is, extend the nonparallel sides of the trapezoid to their point of intersection, giving the triangle containing the trapezoid and the triangle outside the trapezoid, which are similar triangles. This is also particularly common in settings like yours, where you've got two nonparallel lines intersected by a number of parallel lines forming several trapezoids with parallel bases and nonparallel sides that lie on the same two lines.
edited Apr 13 '17 at 12:21
Community♦
1
answered May 17 '12 at 22:12
IsaacIsaac
30.1k1285128
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2 Answers
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active
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2 Answers
2
active
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$begingroup$
I think that what they did here is a standard "trick" used a lot in geometry-trigonometry problems: they draw a a help-line (or help-circle or whatever) in order to make the problem easier and/or more manageable.
In the present case, they extend in a straight way the line FJ from the side of J all the way until it meets in point X the straight extension of line OL in order to form the straight-angle triangle $,Delta FOX,$, with $,angle FOX=90,$ , and now they can use trigonometric functions to do stuff...!
answered May 16 '12 at 11:32
DonAntonioDonAntonio
180k1494233
$endgroup$
add a comment |
$begingroup$
I think that what they did here is a standard "trick" used a lot in geometry-trigonometry problems: they draw a a help-line (or help-circle or whatever) in order to make the problem easier and/or more manageable.
In the present case, they extend in a straight way the line FJ from the side of J all the way until it meets in point X the straight extension of line OL in order to form the straight-angle triangle $,Delta FOX,$, with $,angle FOX=90,$ , and now they can use trigonometric functions to do stuff...!
answered May 16 '12 at 11:32
DonAntonioDonAntonio
180k1494233
$endgroup$
add a comment |
$begingroup$
I think that what they did here is a standard "trick" used a lot in geometry-trigonometry problems: they draw a a help-line (or help-circle or whatever) in order to make the problem easier and/or more manageable.
In the present case, they extend in a straight way the line FJ from the side of J all the way until it meets in point X the straight extension of line OL in order to form the straight-angle triangle $,Delta FOX,$, with $,angle FOX=90,$ , and now they can use trigonometric functions to do stuff...!
answered May 16 '12 at 11:32
DonAntonioDonAntonio
180k1494233
$endgroup$
I think that what they did here is a standard "trick" used a lot in geometry-trigonometry problems: they draw a a help-line (or help-circle or whatever) in order to make the problem easier and/or more manageable.
In the present case, they extend in a straight way the line FJ from the side of J all the way until it meets in point X the straight extension of line OL in order to form the straight-angle triangle $,Delta FOX,$, with $,angle FOX=90,$ , and now they can use trigonometric functions to do stuff...!
answered May 16 '12 at 11:32
DonAntonioDonAntonio
180k1494233
answered May 16 '12 at 11:32
DonAntonioDonAntonio
180k1494233
answered May 16 '12 at 11:32
DonAntonioDonAntonio
180k1494233
answered May 16 '12 at 11:32
DonAntonioDonAntonio
180k1494233
180k1494233
add a comment |
add a comment |
$begingroup$
Similar to DonAntonio's answer, it's often the case when working with (non-parallelogram) trapezoids that it's easier to consider the trapezoid as a truncated triangle—that is, extend the nonparallel sides of the trapezoid to their point of intersection, giving the triangle containing the trapezoid and the triangle outside the trapezoid, which are similar triangles. This is also particularly common in settings like yours, where you've got two nonparallel lines intersected by a number of parallel lines forming several trapezoids with parallel bases and nonparallel sides that lie on the same two lines.
edited Apr 13 '17 at 12:21
Community♦
1
answered May 17 '12 at 22:12
IsaacIsaac
30.1k1285128
$endgroup$
add a comment |
$begingroup$
Similar to DonAntonio's answer, it's often the case when working with (non-parallelogram) trapezoids that it's easier to consider the trapezoid as a truncated triangle—that is, extend the nonparallel sides of the trapezoid to their point of intersection, giving the triangle containing the trapezoid and the triangle outside the trapezoid, which are similar triangles. This is also particularly common in settings like yours, where you've got two nonparallel lines intersected by a number of parallel lines forming several trapezoids with parallel bases and nonparallel sides that lie on the same two lines.
edited Apr 13 '17 at 12:21
Community♦
1
answered May 17 '12 at 22:12
IsaacIsaac
30.1k1285128
$endgroup$
add a comment |
$begingroup$
Similar to DonAntonio's answer, it's often the case when working with (non-parallelogram) trapezoids that it's easier to consider the trapezoid as a truncated triangle—that is, extend the nonparallel sides of the trapezoid to their point of intersection, giving the triangle containing the trapezoid and the triangle outside the trapezoid, which are similar triangles. This is also particularly common in settings like yours, where you've got two nonparallel lines intersected by a number of parallel lines forming several trapezoids with parallel bases and nonparallel sides that lie on the same two lines.
edited Apr 13 '17 at 12:21
Community♦
1
answered May 17 '12 at 22:12
IsaacIsaac
30.1k1285128
$endgroup$
Similar to DonAntonio's answer, it's often the case when working with (non-parallelogram) trapezoids that it's easier to consider the trapezoid as a truncated triangle—that is, extend the nonparallel sides of the trapezoid to their point of intersection, giving the triangle containing the trapezoid and the triangle outside the trapezoid, which are similar triangles. This is also particularly common in settings like yours, where you've got two nonparallel lines intersected by a number of parallel lines forming several trapezoids with parallel bases and nonparallel sides that lie on the same two lines.
edited Apr 13 '17 at 12:21
Community♦
1
answered May 17 '12 at 22:12
IsaacIsaac
30.1k1285128
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
answered May 17 '12 at 22:12
IsaacIsaac
30.1k1285128
answered May 17 '12 at 22:12
IsaacIsaac
30.1k1285128
answered May 17 '12 at 22:12
IsaacIsaac
30.1k1285128
30.1k1285128
add a comment |
add a comment |
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