Mixing
Perspective drawing of a train and parallel lines
$begingroup$
One of my students came in today with a textbook problem that ended up looking like this. It was a word problem based on the perspective drawing of a train. The vertical lines (which separated the carts in the picture) are given as parallel. The diagonals of the carts are also parallel.
The problem is to find the length of C2 based on similar figures.
I've tried comparing the nested triangles, but that doesn't yield anything useful. I've also tried setting up a system of linear equations, but that ends up folding back on itself. Does this have something to do with the properties of trapezoids?
Is there enough information to solve this problem? I've shown it to two other math people and they can't figure it out.
geometry
asked Jan 22 at 1:40
cygorxcygorx
79214
$endgroup$
add a comment |
$begingroup$
One of my students came in today with a textbook problem that ended up looking like this. It was a word problem based on the perspective drawing of a train. The vertical lines (which separated the carts in the picture) are given as parallel. The diagonals of the carts are also parallel.
The problem is to find the length of C2 based on similar figures.
I've tried comparing the nested triangles, but that doesn't yield anything useful. I've also tried setting up a system of linear equations, but that ends up folding back on itself. Does this have something to do with the properties of trapezoids?
Is there enough information to solve this problem? I've shown it to two other math people and they can't figure it out.
geometry
asked Jan 22 at 1:40
cygorxcygorx
79214
$endgroup$
add a comment |
$begingroup$
One of my students came in today with a textbook problem that ended up looking like this. It was a word problem based on the perspective drawing of a train. The vertical lines (which separated the carts in the picture) are given as parallel. The diagonals of the carts are also parallel.
The problem is to find the length of C2 based on similar figures.
I've tried comparing the nested triangles, but that doesn't yield anything useful. I've also tried setting up a system of linear equations, but that ends up folding back on itself. Does this have something to do with the properties of trapezoids?
Is there enough information to solve this problem? I've shown it to two other math people and they can't figure it out.
geometry
asked Jan 22 at 1:40
cygorxcygorx
79214
$endgroup$
One of my students came in today with a textbook problem that ended up looking like this. It was a word problem based on the perspective drawing of a train. The vertical lines (which separated the carts in the picture) are given as parallel. The diagonals of the carts are also parallel.
The problem is to find the length of C2 based on similar figures.
I've tried comparing the nested triangles, but that doesn't yield anything useful. I've also tried setting up a system of linear equations, but that ends up folding back on itself. Does this have something to do with the properties of trapezoids?
Is there enough information to solve this problem? I've shown it to two other math people and they can't figure it out.
geometry
geometry
asked Jan 22 at 1:40
cygorxcygorx
79214
asked Jan 22 at 1:40
cygorxcygorx
79214
asked Jan 22 at 1:40
cygorxcygorx
79214
asked Jan 22 at 1:40
cygorxcygorx
79214
asked Jan 22 at 1:40
cygorxcygorx
79214
79214
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The image above is a rough representation of the question's drawing, including just the pertinent information, with the $10.6$ length coming from $19 - 8.4$. Note that due to the common $angle ACE$, plus since $AE parallel BD$, we also have that $angle CBD = angle CAE$ and $angle CDB = angle CEA$, with this giving that $triangle BCD sim triangle ACE$. Thus, using the ratios of corresponding sides of a similar triangle, we get that
$$cfrac{leftlvert , CD , rightrvert}{leftlvert , CB , rightrvert} = cfrac{leftlvert , CE , rightrvert}{leftlvert , CA , rightrvert} tag{1}label{eq1}$$
Plugging in appropriate values from the diagram, we then get that
$$cfrac{5.4}{10.6} = cfrac{5.4 + text{C}2}{19} tag{2}label{eq2}$$
Cross-multiplying and simplifying gives that
$$102.6 = 57.24 + 10.6 ; text{C}2 Rightarrow text{C}2 = cfrac{left(102.6 - 57.24right)}{10.6} = cfrac{45.36}{10.6} tag{3}label{eq3}$$
The division creates a decimal starting with 4.279..., so you can decide yourself whether to use the exact value or an approximate decimal.
One interesting thing to note is that $text{C}2 + 5.4$ is less than $10.6$, so the drawing is misleading as it makes it seem that this length should be longer.
Also, as you can see, you don't need to use the other information provided, i.e., $text{C}1$ and the additional parallel lines. They were likely put in due to the context of a train & parallel lines, but possibly instead were used to try to confuse things, or perhaps to allow a different way to solve the problem (but I don't think any other method will be much, if any, simpler than what I've shown here).
answered Jan 22 at 2:08
John OmielanJohn Omielan
3,7801215
$endgroup$
$begingroup$
Maybe $C1$ is meant to be found too?
$endgroup$
– timtfj
Jan 22 at 3:05
$begingroup$
@timtfj You are quite possibly correct that perhaps $text{C}1$ is also meant to be found. I believe it would not be too difficult to do so, but the question only specifically asks for $text{C}2$. As such, I will leave it up to the OP to solve for $text{C}1$ if they wish, such as by using similar triangles as this is likely to be the simplest & easiest way.
$endgroup$
– John Omielan
Jan 22 at 3:09
add a comment |
$begingroup$
It is better to find all lengths so as to be later on available for reconstructing the full scenario.
Label horizontal segment above the slant segment marked c2 as segment of $x$ length.
A slant line is drawn at arbitrary inclination as shown. After rearrangement among three sets of similar triangles we have three variables $ (c1,c2,x). $
They can be solved from three simultaneous linear equations with pairwise common ratios between parallel lines by Cramer Rule:
( An error in data input corrected )
$$(19-x-8.4)/(19-8.4)=5.4/(5.4+c2);$$
$$(19-8.4)/19 =(c2+5.4)/(c1+c2+5.4);$$
$$(19-8.4)/19=5.4/(5.4 +c2)$$
which have the numeric solutions:
$${c1=7.67035;c2=4.27925;x=4.68632}.$$
Triangles and three parallel lines sets constructed with these segment lengths show similar triangles between parallel lines and confirming equal alternate angles after Geogebra construction.
answered Mar 2 at 17:24
NarasimhamNarasimham
20.9k62158
$endgroup$
add a comment |
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2 Answers
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$begingroup$
The image above is a rough representation of the question's drawing, including just the pertinent information, with the $10.6$ length coming from $19 - 8.4$. Note that due to the common $angle ACE$, plus since $AE parallel BD$, we also have that $angle CBD = angle CAE$ and $angle CDB = angle CEA$, with this giving that $triangle BCD sim triangle ACE$. Thus, using the ratios of corresponding sides of a similar triangle, we get that
$$cfrac{leftlvert , CD , rightrvert}{leftlvert , CB , rightrvert} = cfrac{leftlvert , CE , rightrvert}{leftlvert , CA , rightrvert} tag{1}label{eq1}$$
Plugging in appropriate values from the diagram, we then get that
$$cfrac{5.4}{10.6} = cfrac{5.4 + text{C}2}{19} tag{2}label{eq2}$$
Cross-multiplying and simplifying gives that
$$102.6 = 57.24 + 10.6 ; text{C}2 Rightarrow text{C}2 = cfrac{left(102.6 - 57.24right)}{10.6} = cfrac{45.36}{10.6} tag{3}label{eq3}$$
The division creates a decimal starting with 4.279..., so you can decide yourself whether to use the exact value or an approximate decimal.
One interesting thing to note is that $text{C}2 + 5.4$ is less than $10.6$, so the drawing is misleading as it makes it seem that this length should be longer.
Also, as you can see, you don't need to use the other information provided, i.e., $text{C}1$ and the additional parallel lines. They were likely put in due to the context of a train & parallel lines, but possibly instead were used to try to confuse things, or perhaps to allow a different way to solve the problem (but I don't think any other method will be much, if any, simpler than what I've shown here).
answered Jan 22 at 2:08
John OmielanJohn Omielan
3,7801215
$endgroup$
$begingroup$
Maybe $C1$ is meant to be found too?
$endgroup$
– timtfj
Jan 22 at 3:05
$begingroup$
@timtfj You are quite possibly correct that perhaps $text{C}1$ is also meant to be found. I believe it would not be too difficult to do so, but the question only specifically asks for $text{C}2$. As such, I will leave it up to the OP to solve for $text{C}1$ if they wish, such as by using similar triangles as this is likely to be the simplest & easiest way.
$endgroup$
– John Omielan
Jan 22 at 3:09
add a comment |
$begingroup$
The image above is a rough representation of the question's drawing, including just the pertinent information, with the $10.6$ length coming from $19 - 8.4$. Note that due to the common $angle ACE$, plus since $AE parallel BD$, we also have that $angle CBD = angle CAE$ and $angle CDB = angle CEA$, with this giving that $triangle BCD sim triangle ACE$. Thus, using the ratios of corresponding sides of a similar triangle, we get that
$$cfrac{leftlvert , CD , rightrvert}{leftlvert , CB , rightrvert} = cfrac{leftlvert , CE , rightrvert}{leftlvert , CA , rightrvert} tag{1}label{eq1}$$
Plugging in appropriate values from the diagram, we then get that
$$cfrac{5.4}{10.6} = cfrac{5.4 + text{C}2}{19} tag{2}label{eq2}$$
Cross-multiplying and simplifying gives that
$$102.6 = 57.24 + 10.6 ; text{C}2 Rightarrow text{C}2 = cfrac{left(102.6 - 57.24right)}{10.6} = cfrac{45.36}{10.6} tag{3}label{eq3}$$
The division creates a decimal starting with 4.279..., so you can decide yourself whether to use the exact value or an approximate decimal.
One interesting thing to note is that $text{C}2 + 5.4$ is less than $10.6$, so the drawing is misleading as it makes it seem that this length should be longer.
Also, as you can see, you don't need to use the other information provided, i.e., $text{C}1$ and the additional parallel lines. They were likely put in due to the context of a train & parallel lines, but possibly instead were used to try to confuse things, or perhaps to allow a different way to solve the problem (but I don't think any other method will be much, if any, simpler than what I've shown here).
answered Jan 22 at 2:08
John OmielanJohn Omielan
3,7801215
$endgroup$
$begingroup$
Maybe $C1$ is meant to be found too?
$endgroup$
– timtfj
Jan 22 at 3:05
$begingroup$
@timtfj You are quite possibly correct that perhaps $text{C}1$ is also meant to be found. I believe it would not be too difficult to do so, but the question only specifically asks for $text{C}2$. As such, I will leave it up to the OP to solve for $text{C}1$ if they wish, such as by using similar triangles as this is likely to be the simplest & easiest way.
$endgroup$
– John Omielan
Jan 22 at 3:09
add a comment |
$begingroup$
The image above is a rough representation of the question's drawing, including just the pertinent information, with the $10.6$ length coming from $19 - 8.4$. Note that due to the common $angle ACE$, plus since $AE parallel BD$, we also have that $angle CBD = angle CAE$ and $angle CDB = angle CEA$, with this giving that $triangle BCD sim triangle ACE$. Thus, using the ratios of corresponding sides of a similar triangle, we get that
$$cfrac{leftlvert , CD , rightrvert}{leftlvert , CB , rightrvert} = cfrac{leftlvert , CE , rightrvert}{leftlvert , CA , rightrvert} tag{1}label{eq1}$$
Plugging in appropriate values from the diagram, we then get that
$$cfrac{5.4}{10.6} = cfrac{5.4 + text{C}2}{19} tag{2}label{eq2}$$
Cross-multiplying and simplifying gives that
$$102.6 = 57.24 + 10.6 ; text{C}2 Rightarrow text{C}2 = cfrac{left(102.6 - 57.24right)}{10.6} = cfrac{45.36}{10.6} tag{3}label{eq3}$$
The division creates a decimal starting with 4.279..., so you can decide yourself whether to use the exact value or an approximate decimal.
One interesting thing to note is that $text{C}2 + 5.4$ is less than $10.6$, so the drawing is misleading as it makes it seem that this length should be longer.
Also, as you can see, you don't need to use the other information provided, i.e., $text{C}1$ and the additional parallel lines. They were likely put in due to the context of a train & parallel lines, but possibly instead were used to try to confuse things, or perhaps to allow a different way to solve the problem (but I don't think any other method will be much, if any, simpler than what I've shown here).
answered Jan 22 at 2:08
John OmielanJohn Omielan
3,7801215
$endgroup$
The image above is a rough representation of the question's drawing, including just the pertinent information, with the $10.6$ length coming from $19 - 8.4$. Note that due to the common $angle ACE$, plus since $AE parallel BD$, we also have that $angle CBD = angle CAE$ and $angle CDB = angle CEA$, with this giving that $triangle BCD sim triangle ACE$. Thus, using the ratios of corresponding sides of a similar triangle, we get that
$$cfrac{leftlvert , CD , rightrvert}{leftlvert , CB , rightrvert} = cfrac{leftlvert , CE , rightrvert}{leftlvert , CA , rightrvert} tag{1}label{eq1}$$
Plugging in appropriate values from the diagram, we then get that
$$cfrac{5.4}{10.6} = cfrac{5.4 + text{C}2}{19} tag{2}label{eq2}$$
Cross-multiplying and simplifying gives that
$$102.6 = 57.24 + 10.6 ; text{C}2 Rightarrow text{C}2 = cfrac{left(102.6 - 57.24right)}{10.6} = cfrac{45.36}{10.6} tag{3}label{eq3}$$
The division creates a decimal starting with 4.279..., so you can decide yourself whether to use the exact value or an approximate decimal.
One interesting thing to note is that $text{C}2 + 5.4$ is less than $10.6$, so the drawing is misleading as it makes it seem that this length should be longer.
Also, as you can see, you don't need to use the other information provided, i.e., $text{C}1$ and the additional parallel lines. They were likely put in due to the context of a train & parallel lines, but possibly instead were used to try to confuse things, or perhaps to allow a different way to solve the problem (but I don't think any other method will be much, if any, simpler than what I've shown here).
answered Jan 22 at 2:08
John OmielanJohn Omielan
3,7801215
edited Jan 23 at 0:12
answered Jan 22 at 2:08
John OmielanJohn Omielan
3,7801215
answered Jan 22 at 2:08
John OmielanJohn Omielan
3,7801215
answered Jan 22 at 2:08
John OmielanJohn Omielan
3,7801215
3,7801215
$begingroup$
Maybe $C1$ is meant to be found too?
$endgroup$
– timtfj
Jan 22 at 3:05
$begingroup$
@timtfj You are quite possibly correct that perhaps $text{C}1$ is also meant to be found. I believe it would not be too difficult to do so, but the question only specifically asks for $text{C}2$. As such, I will leave it up to the OP to solve for $text{C}1$ if they wish, such as by using similar triangles as this is likely to be the simplest & easiest way.
$endgroup$
– John Omielan
Jan 22 at 3:09
add a comment |
$begingroup$
Maybe $C1$ is meant to be found too?
$endgroup$
– timtfj
Jan 22 at 3:05
$begingroup$
@timtfj You are quite possibly correct that perhaps $text{C}1$ is also meant to be found. I believe it would not be too difficult to do so, but the question only specifically asks for $text{C}2$. As such, I will leave it up to the OP to solve for $text{C}1$ if they wish, such as by using similar triangles as this is likely to be the simplest & easiest way.
$endgroup$
– John Omielan
Jan 22 at 3:09
$begingroup$
Maybe $C1$ is meant to be found too?
$endgroup$
– timtfj
Jan 22 at 3:05
$begingroup$
Maybe $C1$ is meant to be found too?
$endgroup$
– timtfj
Jan 22 at 3:05
$begingroup$
@timtfj You are quite possibly correct that perhaps $text{C}1$ is also meant to be found. I believe it would not be too difficult to do so, but the question only specifically asks for $text{C}2$. As such, I will leave it up to the OP to solve for $text{C}1$ if they wish, such as by using similar triangles as this is likely to be the simplest & easiest way.
$endgroup$
– John Omielan
Jan 22 at 3:09
$begingroup$
@timtfj You are quite possibly correct that perhaps $text{C}1$ is also meant to be found. I believe it would not be too difficult to do so, but the question only specifically asks for $text{C}2$. As such, I will leave it up to the OP to solve for $text{C}1$ if they wish, such as by using similar triangles as this is likely to be the simplest & easiest way.
$endgroup$
– John Omielan
Jan 22 at 3:09
add a comment |
$begingroup$
It is better to find all lengths so as to be later on available for reconstructing the full scenario.
Label horizontal segment above the slant segment marked c2 as segment of $x$ length.
A slant line is drawn at arbitrary inclination as shown. After rearrangement among three sets of similar triangles we have three variables $ (c1,c2,x). $
They can be solved from three simultaneous linear equations with pairwise common ratios between parallel lines by Cramer Rule:
( An error in data input corrected )
$$(19-x-8.4)/(19-8.4)=5.4/(5.4+c2);$$
$$(19-8.4)/19 =(c2+5.4)/(c1+c2+5.4);$$
$$(19-8.4)/19=5.4/(5.4 +c2)$$
which have the numeric solutions:
$${c1=7.67035;c2=4.27925;x=4.68632}.$$
Triangles and three parallel lines sets constructed with these segment lengths show similar triangles between parallel lines and confirming equal alternate angles after Geogebra construction.
answered Mar 2 at 17:24
NarasimhamNarasimham
20.9k62158
$endgroup$
add a comment |
$begingroup$
It is better to find all lengths so as to be later on available for reconstructing the full scenario.
Label horizontal segment above the slant segment marked c2 as segment of $x$ length.
A slant line is drawn at arbitrary inclination as shown. After rearrangement among three sets of similar triangles we have three variables $ (c1,c2,x). $
They can be solved from three simultaneous linear equations with pairwise common ratios between parallel lines by Cramer Rule:
( An error in data input corrected )
$$(19-x-8.4)/(19-8.4)=5.4/(5.4+c2);$$
$$(19-8.4)/19 =(c2+5.4)/(c1+c2+5.4);$$
$$(19-8.4)/19=5.4/(5.4 +c2)$$
which have the numeric solutions:
$${c1=7.67035;c2=4.27925;x=4.68632}.$$
Triangles and three parallel lines sets constructed with these segment lengths show similar triangles between parallel lines and confirming equal alternate angles after Geogebra construction.
answered Mar 2 at 17:24
NarasimhamNarasimham
20.9k62158
$endgroup$
add a comment |
$begingroup$
It is better to find all lengths so as to be later on available for reconstructing the full scenario.
Label horizontal segment above the slant segment marked c2 as segment of $x$ length.
A slant line is drawn at arbitrary inclination as shown. After rearrangement among three sets of similar triangles we have three variables $ (c1,c2,x). $
They can be solved from three simultaneous linear equations with pairwise common ratios between parallel lines by Cramer Rule:
( An error in data input corrected )
$$(19-x-8.4)/(19-8.4)=5.4/(5.4+c2);$$
$$(19-8.4)/19 =(c2+5.4)/(c1+c2+5.4);$$
$$(19-8.4)/19=5.4/(5.4 +c2)$$
which have the numeric solutions:
$${c1=7.67035;c2=4.27925;x=4.68632}.$$
Triangles and three parallel lines sets constructed with these segment lengths show similar triangles between parallel lines and confirming equal alternate angles after Geogebra construction.
answered Mar 2 at 17:24
NarasimhamNarasimham
20.9k62158
$endgroup$
It is better to find all lengths so as to be later on available for reconstructing the full scenario.
Label horizontal segment above the slant segment marked c2 as segment of $x$ length.
A slant line is drawn at arbitrary inclination as shown. After rearrangement among three sets of similar triangles we have three variables $ (c1,c2,x). $
They can be solved from three simultaneous linear equations with pairwise common ratios between parallel lines by Cramer Rule:
( An error in data input corrected )
$$(19-x-8.4)/(19-8.4)=5.4/(5.4+c2);$$
$$(19-8.4)/19 =(c2+5.4)/(c1+c2+5.4);$$
$$(19-8.4)/19=5.4/(5.4 +c2)$$
which have the numeric solutions:
$${c1=7.67035;c2=4.27925;x=4.68632}.$$
Triangles and three parallel lines sets constructed with these segment lengths show similar triangles between parallel lines and confirming equal alternate angles after Geogebra construction.
answered Mar 2 at 17:24
NarasimhamNarasimham
20.9k62158
edited Mar 4 at 5:59
answered Mar 2 at 17:24
NarasimhamNarasimham
20.9k62158
answered Mar 2 at 17:24
NarasimhamNarasimham
20.9k62158
answered Mar 2 at 17:24
NarasimhamNarasimham
20.9k62158
20.9k62158
add a comment |
add a comment |
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