Basic similarity of triangles problem
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The question is to find the length of AB. I'm currently studying civil engineering, but I'm trying to refresh some old knowledge as I got a part time job this semester as a math assistant at a middle school. You're not supposed to use trigonometry.
My attempt involves similarity of triangles of course, since all the angles are obviously identical.
$fracABAD = frac36$
$fracABAB+BD = frac12$
In the solution they somehow know that the length of BD is 4. I figured out that they're at least equal using trigonometry, but I don't see where the 4 comes from.
geometry euclidean-geometry triangle
asked Sep 7 at 7:16
novo
34918
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up vote
0
down vote
favorite
The question is to find the length of AB. I'm currently studying civil engineering, but I'm trying to refresh some old knowledge as I got a part time job this semester as a math assistant at a middle school. You're not supposed to use trigonometry.
My attempt involves similarity of triangles of course, since all the angles are obviously identical.
$fracABAD = frac36$
$fracABAB+BD = frac12$
In the solution they somehow know that the length of BD is 4. I figured out that they're at least equal using trigonometry, but I don't see where the 4 comes from.
geometry euclidean-geometry triangle
asked Sep 7 at 7:16
novo
34918
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The question is to find the length of AB. I'm currently studying civil engineering, but I'm trying to refresh some old knowledge as I got a part time job this semester as a math assistant at a middle school. You're not supposed to use trigonometry.
My attempt involves similarity of triangles of course, since all the angles are obviously identical.
$fracABAD = frac36$
$fracABAB+BD = frac12$
In the solution they somehow know that the length of BD is 4. I figured out that they're at least equal using trigonometry, but I don't see where the 4 comes from.
geometry euclidean-geometry triangle
asked Sep 7 at 7:16
novo
34918
The question is to find the length of AB. I'm currently studying civil engineering, but I'm trying to refresh some old knowledge as I got a part time job this semester as a math assistant at a middle school. You're not supposed to use trigonometry.
My attempt involves similarity of triangles of course, since all the angles are obviously identical.
$fracABAD = frac36$
$fracABAB+BD = frac12$
In the solution they somehow know that the length of BD is 4. I figured out that they're at least equal using trigonometry, but I don't see where the 4 comes from.
geometry euclidean-geometry triangle
geometry euclidean-geometry triangle
asked Sep 7 at 7:16
novo
34918
asked Sep 7 at 7:16
novo
34918
asked Sep 7 at 7:16
novo
34918
asked Sep 7 at 7:16
novo
34918
asked Sep 7 at 7:16
novo
34918
34918
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
0
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Updated answer:
The picture in the question is missing $BD = 4$.
From $fracABAB+4 = frac12$, we have $2 AB = AB + 4$, so $AB=4$.
answered Sep 7 at 7:30
Toby Mak
2,8751925
I don't think it makes sense to guess. You're supposed to find the length of AB. If you make that guess, you basically guessed the solution.
– novo
Sep 7 at 7:35
@novo That's true, but the problem doesn't really make sense. Problems similar to this have another side given, so that you can find the side using similarity ratios.
– Toby Mak
Sep 7 at 7:37
Ok, thank you for the assistance. The problem is taken from math10.com/en/geometry/similar-triangles/… example 3
– novo
Sep 7 at 7:46
@novo The picture is probably missing $BD = 4$. If this is added in, all their steps make sense.
– Toby Mak
Sep 7 at 7:52
add a comment |Â
up vote
0
down vote
There may be more information here than the problem requires.
Since they are right triangles, you know by Pythagoras that$$AB^2+BC^2=AC^2$$
And since $BC= 3$, then if we grant that $5^2$ and $4^2$ are the only two square numbers that differ by $3^2$, it will follow that $AC=5$ and $AB=4$. No trigonometry needed, nor similar triangles.
On the other hand, allowing for irrational numbers, supposing $AC=sqrt24$, $sqrt23$, or $sqrt22$, then by Pythagoras $AB=sqrt15$, $sqrt14$, or $sqrt13$, respectively. We know in every case that $AB=BD$, but beyond that? The problem seems indeterminate, and again similarity of triangles is no help.
answered Sep 8 at 6:32
Edward Porcella
1,1831411
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Updated answer:
The picture in the question is missing $BD = 4$.
From $fracABAB+4 = frac12$, we have $2 AB = AB + 4$, so $AB=4$.
answered Sep 7 at 7:30
Toby Mak
2,8751925
I don't think it makes sense to guess. You're supposed to find the length of AB. If you make that guess, you basically guessed the solution.
– novo
Sep 7 at 7:35
@novo That's true, but the problem doesn't really make sense. Problems similar to this have another side given, so that you can find the side using similarity ratios.
– Toby Mak
Sep 7 at 7:37
Ok, thank you for the assistance. The problem is taken from math10.com/en/geometry/similar-triangles/… example 3
– novo
Sep 7 at 7:46
@novo The picture is probably missing $BD = 4$. If this is added in, all their steps make sense.
– Toby Mak
Sep 7 at 7:52
add a comment |Â
up vote
0
down vote
Updated answer:
The picture in the question is missing $BD = 4$.
From $fracABAB+4 = frac12$, we have $2 AB = AB + 4$, so $AB=4$.
answered Sep 7 at 7:30
Toby Mak
2,8751925
I don't think it makes sense to guess. You're supposed to find the length of AB. If you make that guess, you basically guessed the solution.
– novo
Sep 7 at 7:35
@novo That's true, but the problem doesn't really make sense. Problems similar to this have another side given, so that you can find the side using similarity ratios.
– Toby Mak
Sep 7 at 7:37
Ok, thank you for the assistance. The problem is taken from math10.com/en/geometry/similar-triangles/… example 3
– novo
Sep 7 at 7:46
@novo The picture is probably missing $BD = 4$. If this is added in, all their steps make sense.
– Toby Mak
Sep 7 at 7:52
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Updated answer:
The picture in the question is missing $BD = 4$.
From $fracABAB+4 = frac12$, we have $2 AB = AB + 4$, so $AB=4$.
answered Sep 7 at 7:30
Toby Mak
2,8751925
Updated answer:
The picture in the question is missing $BD = 4$.
From $fracABAB+4 = frac12$, we have $2 AB = AB + 4$, so $AB=4$.
answered Sep 7 at 7:30
Toby Mak
2,8751925
edited Sep 8 at 0:02
answered Sep 7 at 7:30
Toby Mak
2,8751925
answered Sep 7 at 7:30
Toby Mak
2,8751925
answered Sep 7 at 7:30
Toby Mak
2,8751925
2,8751925
I don't think it makes sense to guess. You're supposed to find the length of AB. If you make that guess, you basically guessed the solution.
– novo
Sep 7 at 7:35
@novo That's true, but the problem doesn't really make sense. Problems similar to this have another side given, so that you can find the side using similarity ratios.
– Toby Mak
Sep 7 at 7:37
Ok, thank you for the assistance. The problem is taken from math10.com/en/geometry/similar-triangles/… example 3
– novo
Sep 7 at 7:46
@novo The picture is probably missing $BD = 4$. If this is added in, all their steps make sense.
– Toby Mak
Sep 7 at 7:52
add a comment |Â
I don't think it makes sense to guess. You're supposed to find the length of AB. If you make that guess, you basically guessed the solution.
– novo
Sep 7 at 7:35
@novo That's true, but the problem doesn't really make sense. Problems similar to this have another side given, so that you can find the side using similarity ratios.
– Toby Mak
Sep 7 at 7:37
Ok, thank you for the assistance. The problem is taken from math10.com/en/geometry/similar-triangles/… example 3
– novo
Sep 7 at 7:46
@novo The picture is probably missing $BD = 4$. If this is added in, all their steps make sense.
– Toby Mak
Sep 7 at 7:52
I don't think it makes sense to guess. You're supposed to find the length of AB. If you make that guess, you basically guessed the solution.
– novo
Sep 7 at 7:35
I don't think it makes sense to guess. You're supposed to find the length of AB. If you make that guess, you basically guessed the solution.
– novo
Sep 7 at 7:35
@novo That's true, but the problem doesn't really make sense. Problems similar to this have another side given, so that you can find the side using similarity ratios.
– Toby Mak
Sep 7 at 7:37
@novo That's true, but the problem doesn't really make sense. Problems similar to this have another side given, so that you can find the side using similarity ratios.
– Toby Mak
Sep 7 at 7:37
Ok, thank you for the assistance. The problem is taken from math10.com/en/geometry/similar-triangles/… example 3
– novo
Sep 7 at 7:46
Ok, thank you for the assistance. The problem is taken from math10.com/en/geometry/similar-triangles/… example 3
– novo
Sep 7 at 7:46
@novo The picture is probably missing $BD = 4$. If this is added in, all their steps make sense.
– Toby Mak
Sep 7 at 7:52
@novo The picture is probably missing $BD = 4$. If this is added in, all their steps make sense.
– Toby Mak
Sep 7 at 7:52
add a comment |Â
up vote
0
down vote
There may be more information here than the problem requires.
Since they are right triangles, you know by Pythagoras that$$AB^2+BC^2=AC^2$$
And since $BC= 3$, then if we grant that $5^2$ and $4^2$ are the only two square numbers that differ by $3^2$, it will follow that $AC=5$ and $AB=4$. No trigonometry needed, nor similar triangles.
On the other hand, allowing for irrational numbers, supposing $AC=sqrt24$, $sqrt23$, or $sqrt22$, then by Pythagoras $AB=sqrt15$, $sqrt14$, or $sqrt13$, respectively. We know in every case that $AB=BD$, but beyond that? The problem seems indeterminate, and again similarity of triangles is no help.
answered Sep 8 at 6:32
Edward Porcella
1,1831411
add a comment |Â
up vote
0
down vote
There may be more information here than the problem requires.
Since they are right triangles, you know by Pythagoras that$$AB^2+BC^2=AC^2$$
And since $BC= 3$, then if we grant that $5^2$ and $4^2$ are the only two square numbers that differ by $3^2$, it will follow that $AC=5$ and $AB=4$. No trigonometry needed, nor similar triangles.
On the other hand, allowing for irrational numbers, supposing $AC=sqrt24$, $sqrt23$, or $sqrt22$, then by Pythagoras $AB=sqrt15$, $sqrt14$, or $sqrt13$, respectively. We know in every case that $AB=BD$, but beyond that? The problem seems indeterminate, and again similarity of triangles is no help.
answered Sep 8 at 6:32
Edward Porcella
1,1831411
add a comment |Â
up vote
0
down vote
up vote
0
down vote
There may be more information here than the problem requires.
Since they are right triangles, you know by Pythagoras that$$AB^2+BC^2=AC^2$$
And since $BC= 3$, then if we grant that $5^2$ and $4^2$ are the only two square numbers that differ by $3^2$, it will follow that $AC=5$ and $AB=4$. No trigonometry needed, nor similar triangles.
On the other hand, allowing for irrational numbers, supposing $AC=sqrt24$, $sqrt23$, or $sqrt22$, then by Pythagoras $AB=sqrt15$, $sqrt14$, or $sqrt13$, respectively. We know in every case that $AB=BD$, but beyond that? The problem seems indeterminate, and again similarity of triangles is no help.
answered Sep 8 at 6:32
Edward Porcella
1,1831411
There may be more information here than the problem requires.
Since they are right triangles, you know by Pythagoras that$$AB^2+BC^2=AC^2$$
And since $BC= 3$, then if we grant that $5^2$ and $4^2$ are the only two square numbers that differ by $3^2$, it will follow that $AC=5$ and $AB=4$. No trigonometry needed, nor similar triangles.
On the other hand, allowing for irrational numbers, supposing $AC=sqrt24$, $sqrt23$, or $sqrt22$, then by Pythagoras $AB=sqrt15$, $sqrt14$, or $sqrt13$, respectively. We know in every case that $AB=BD$, but beyond that? The problem seems indeterminate, and again similarity of triangles is no help.
answered Sep 8 at 6:32
Edward Porcella
1,1831411
edited Sep 8 at 15:48
answered Sep 8 at 6:32
Edward Porcella
1,1831411
answered Sep 8 at 6:32
Edward Porcella
1,1831411
answered Sep 8 at 6:32
Edward Porcella
1,1831411
1,1831411
add a comment |Â
add a comment |Â
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