A line segment is drawn from one centroid of a triangle to the centroid of another triangle. Is the line...
$begingroup$
Given that $G$ and $G'$ are the centroids of triangles $ABC$ and $ACD$, respectively, and that $|GG'|=12$ cm, what is the value of $|BD|$?
Figure is not to scale, so $C$ may or may not be the middle point of $BD$.
I can find the answer ($36$ cm) by assuming that $GG'$ is parallel to $BD$. After that, it is a matter of setting up the similarity ratios to find $BD$.
I wanted to know if I'm correct in my assumption that these two lines are parallel? If yes, can you show me how?
geometry
asked Dec 3 '18 at 8:41
Eldar RahimliEldar Rahimli
1269
$endgroup$
|
show 1 more comment
$begingroup$
Given that $G$ and $G'$ are the centroids of triangles $ABC$ and $ACD$, respectively, and that $|GG'|=12$ cm, what is the value of $|BD|$?
Figure is not to scale, so $C$ may or may not be the middle point of $BD$.
I can find the answer ($36$ cm) by assuming that $GG'$ is parallel to $BD$. After that, it is a matter of setting up the similarity ratios to find $BD$.
I wanted to know if I'm correct in my assumption that these two lines are parallel? If yes, can you show me how?
geometry
asked Dec 3 '18 at 8:41
Eldar RahimliEldar Rahimli
1269
$endgroup$
$begingroup$
Which center of the triangle are you referring to? There are several. See en.wikipedia.org/wiki/Triangle_center
$endgroup$
– 1123581321
Dec 3 '18 at 8:45
$begingroup$
I didn't know. I'll clarify and edit my question.Thanks
$endgroup$
– Eldar Rahimli
Dec 3 '18 at 8:46
1
$begingroup$
For centroids, $G' - G = (A+D+C)/3 - (A+C+B)/3 = (D-B)/3$
$endgroup$
– achille hui
Dec 3 '18 at 8:47
$begingroup$
Yes, It is centroid, where the 3 medians intersect
$endgroup$
– Eldar Rahimli
Dec 3 '18 at 8:48
$begingroup$
As suggested by @achillehui's comment, since centroid of a triangle XYZ in vector notation is given by $frac{vec{X} + vec{Y} + vec{Z}}{3}, $ you can find $vec{GG'} = vec{G'} - vec{G} = (frac{vec{D} - vec{B}}{3})$, so $GG'$ is parallel to $DB$ and one-third of it in length.
$endgroup$
– ab123
Dec 3 '18 at 9:00
|
show 1 more comment
$begingroup$
Given that $G$ and $G'$ are the centroids of triangles $ABC$ and $ACD$, respectively, and that $|GG'|=12$ cm, what is the value of $|BD|$?
Figure is not to scale, so $C$ may or may not be the middle point of $BD$.
I can find the answer ($36$ cm) by assuming that $GG'$ is parallel to $BD$. After that, it is a matter of setting up the similarity ratios to find $BD$.
I wanted to know if I'm correct in my assumption that these two lines are parallel? If yes, can you show me how?
geometry
asked Dec 3 '18 at 8:41
Eldar RahimliEldar Rahimli
1269
$endgroup$
Given that $G$ and $G'$ are the centroids of triangles $ABC$ and $ACD$, respectively, and that $|GG'|=12$ cm, what is the value of $|BD|$?
Figure is not to scale, so $C$ may or may not be the middle point of $BD$.
I can find the answer ($36$ cm) by assuming that $GG'$ is parallel to $BD$. After that, it is a matter of setting up the similarity ratios to find $BD$.
I wanted to know if I'm correct in my assumption that these two lines are parallel? If yes, can you show me how?
geometry
geometry
asked Dec 3 '18 at 8:41
Eldar RahimliEldar Rahimli
1269
asked Dec 3 '18 at 8:41
Eldar RahimliEldar Rahimli
1269
edited Dec 28 '18 at 7:32
Eldar Rahimli
asked Dec 3 '18 at 8:41
Eldar RahimliEldar Rahimli
1269
asked Dec 3 '18 at 8:41
Eldar RahimliEldar Rahimli
1269
asked Dec 3 '18 at 8:41
Eldar RahimliEldar Rahimli
1269
1269
$begingroup$
Which center of the triangle are you referring to? There are several. See en.wikipedia.org/wiki/Triangle_center
$endgroup$
– 1123581321
Dec 3 '18 at 8:45
$begingroup$
I didn't know. I'll clarify and edit my question.Thanks
$endgroup$
– Eldar Rahimli
Dec 3 '18 at 8:46
1
$begingroup$
For centroids, $G' - G = (A+D+C)/3 - (A+C+B)/3 = (D-B)/3$
$endgroup$
– achille hui
Dec 3 '18 at 8:47
$begingroup$
Yes, It is centroid, where the 3 medians intersect
$endgroup$
– Eldar Rahimli
Dec 3 '18 at 8:48
$begingroup$
As suggested by @achillehui's comment, since centroid of a triangle XYZ in vector notation is given by $frac{vec{X} + vec{Y} + vec{Z}}{3}, $ you can find $vec{GG'} = vec{G'} - vec{G} = (frac{vec{D} - vec{B}}{3})$, so $GG'$ is parallel to $DB$ and one-third of it in length.
$endgroup$
– ab123
Dec 3 '18 at 9:00
|
show 1 more comment
$begingroup$
Which center of the triangle are you referring to? There are several. See en.wikipedia.org/wiki/Triangle_center
$endgroup$
– 1123581321
Dec 3 '18 at 8:45
$begingroup$
I didn't know. I'll clarify and edit my question.Thanks
$endgroup$
– Eldar Rahimli
Dec 3 '18 at 8:46
1
$begingroup$
For centroids, $G' - G = (A+D+C)/3 - (A+C+B)/3 = (D-B)/3$
$endgroup$
– achille hui
Dec 3 '18 at 8:47
$begingroup$
Yes, It is centroid, where the 3 medians intersect
$endgroup$
– Eldar Rahimli
Dec 3 '18 at 8:48
$begingroup$
As suggested by @achillehui's comment, since centroid of a triangle XYZ in vector notation is given by $frac{vec{X} + vec{Y} + vec{Z}}{3}, $ you can find $vec{GG'} = vec{G'} - vec{G} = (frac{vec{D} - vec{B}}{3})$, so $GG'$ is parallel to $DB$ and one-third of it in length.
$endgroup$
– ab123
Dec 3 '18 at 9:00
$begingroup$
Which center of the triangle are you referring to? There are several. See en.wikipedia.org/wiki/Triangle_center
$endgroup$
– 1123581321
Dec 3 '18 at 8:45
$begingroup$
Which center of the triangle are you referring to? There are several. See en.wikipedia.org/wiki/Triangle_center
$endgroup$
– 1123581321
Dec 3 '18 at 8:45
$begingroup$
I didn't know. I'll clarify and edit my question.Thanks
$endgroup$
– Eldar Rahimli
Dec 3 '18 at 8:46
$begingroup$
I didn't know. I'll clarify and edit my question.Thanks
$endgroup$
– Eldar Rahimli
Dec 3 '18 at 8:46
1
1
$begingroup$
For centroids, $G' - G = (A+D+C)/3 - (A+C+B)/3 = (D-B)/3$
$endgroup$
– achille hui
Dec 3 '18 at 8:47
$begingroup$
For centroids, $G' - G = (A+D+C)/3 - (A+C+B)/3 = (D-B)/3$
$endgroup$
– achille hui
Dec 3 '18 at 8:47
$begingroup$
Yes, It is centroid, where the 3 medians intersect
$endgroup$
– Eldar Rahimli
Dec 3 '18 at 8:48
$begingroup$
Yes, It is centroid, where the 3 medians intersect
$endgroup$
– Eldar Rahimli
Dec 3 '18 at 8:48
$begingroup$
As suggested by @achillehui's comment, since centroid of a triangle XYZ in vector notation is given by $frac{vec{X} + vec{Y} + vec{Z}}{3}, $ you can find $vec{GG'} = vec{G'} - vec{G} = (frac{vec{D} - vec{B}}{3})$, so $GG'$ is parallel to $DB$ and one-third of it in length.
$endgroup$
– ab123
Dec 3 '18 at 9:00
$begingroup$
As suggested by @achillehui's comment, since centroid of a triangle XYZ in vector notation is given by $frac{vec{X} + vec{Y} + vec{Z}}{3}, $ you can find $vec{GG'} = vec{G'} - vec{G} = (frac{vec{D} - vec{B}}{3})$, so $GG'$ is parallel to $DB$ and one-third of it in length.
$endgroup$
– ab123
Dec 3 '18 at 9:00
|
show 1 more comment
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$begingroup$
Which center of the triangle are you referring to? There are several. See en.wikipedia.org/wiki/Triangle_center
$endgroup$
– 1123581321
Dec 3 '18 at 8:45
$begingroup$
I didn't know. I'll clarify and edit my question.Thanks
$endgroup$
– Eldar Rahimli
Dec 3 '18 at 8:46
1
$begingroup$
For centroids, $G' - G = (A+D+C)/3 - (A+C+B)/3 = (D-B)/3$
$endgroup$
– achille hui
Dec 3 '18 at 8:47
$begingroup$
Yes, It is centroid, where the 3 medians intersect
$endgroup$
– Eldar Rahimli
Dec 3 '18 at 8:48
$begingroup$
As suggested by @achillehui's comment, since centroid of a triangle XYZ in vector notation is given by $frac{vec{X} + vec{Y} + vec{Z}}{3}, $ you can find $vec{GG'} = vec{G'} - vec{G} = (frac{vec{D} - vec{B}}{3})$, so $GG'$ is parallel to $DB$ and one-third of it in length.
$endgroup$
– ab123
Dec 3 '18 at 9:00