How to find area of triangle from its medians The 2019 Stack Overflow Developer Survey Results Are InFinding the area of triangle if length of medians are givenProve that the area of the triangle formed by the medians is equal to $3/4$ the area of the original triangleTriangle MediansRatio of area of triangle to that formed by its mediansDrawing a triangle from mediansComputing area of triangle via equations of mediansCan we find inradius from medians of a triangle?Prove medians of a triangle can make a triangleGiven length of one side and its median and another median in a triangle. Find area of the triangleUsing values from vertexes to centroid to find area of triangleTriangle: Find area given 2 side lengths and that two medians are perpendicularSolving area of a triangle where medians are perpendicular.
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How to find area of triangle from its medians
The 2019 Stack Overflow Developer Survey Results Are InFinding the area of triangle if length of medians are givenProve that the area of the triangle formed by the medians is equal to $3/4$ the area of the original triangleTriangle MediansRatio of area of triangle to that formed by its mediansDrawing a triangle from mediansComputing area of triangle via equations of mediansCan we find inradius from medians of a triangle?Prove medians of a triangle can make a triangleGiven length of one side and its median and another median in a triangle. Find area of the triangleUsing values from vertexes to centroid to find area of triangleTriangle: Find area given 2 side lengths and that two medians are perpendicularSolving area of a triangle where medians are perpendicular.
$begingroup$
The length of three medians of a triangle are $9$,$12$ and $15$cm.The area (in sq. cm) of the triangle is
a) $48$
b) $144$
c) $24$
d) $72$
I don't want whole solution just give me the hint how can I solve it.Thanks.
geometry trigonometry triangles
asked May 19 '13 at 5:15
iostream007iostream007
3,71931439
$endgroup$
|
show 2 more comments
$begingroup$
The length of three medians of a triangle are $9$,$12$ and $15$cm.The area (in sq. cm) of the triangle is
a) $48$
b) $144$
c) $24$
d) $72$
I don't want whole solution just give me the hint how can I solve it.Thanks.
geometry trigonometry triangles
asked May 19 '13 at 5:15
iostream007iostream007
3,71931439
$endgroup$
2
$begingroup$
mathworld.wolfram.com/TriangleMedian.html
$endgroup$
– lab bhattacharjee
May 19 '13 at 5:17
$begingroup$
@labbhattacharjee Thanks I got a new formula.
$endgroup$
– iostream007
May 19 '13 at 5:40
$begingroup$
You can also use Appolonius theorem.
$endgroup$
– Aryabhata
May 19 '13 at 5:43
$begingroup$
@iostream007, welcome. Have you tried proving it?
$endgroup$
– lab bhattacharjee
May 19 '13 at 5:44
$begingroup$
i'm trying to solve it then i'll try how to prove it. Isn't any formula to the same because this question asked in SSC exam and there is not so much to spent on 1 question?
$endgroup$
– iostream007
May 19 '13 at 5:48
|
show 2 more comments
$begingroup$
The length of three medians of a triangle are $9$,$12$ and $15$cm.The area (in sq. cm) of the triangle is
a) $48$
b) $144$
c) $24$
d) $72$
I don't want whole solution just give me the hint how can I solve it.Thanks.
geometry trigonometry triangles
asked May 19 '13 at 5:15
iostream007iostream007
3,71931439
$endgroup$
The length of three medians of a triangle are $9$,$12$ and $15$cm.The area (in sq. cm) of the triangle is
a) $48$
b) $144$
c) $24$
d) $72$
I don't want whole solution just give me the hint how can I solve it.Thanks.
geometry trigonometry triangles
geometry trigonometry triangles
asked May 19 '13 at 5:15
iostream007iostream007
3,71931439
asked May 19 '13 at 5:15
iostream007iostream007
3,71931439
asked May 19 '13 at 5:15
iostream007iostream007
3,71931439
asked May 19 '13 at 5:15
iostream007iostream007
3,71931439
asked May 19 '13 at 5:15
iostream007iostream007
3,71931439
3,71931439
2
$begingroup$
mathworld.wolfram.com/TriangleMedian.html
$endgroup$
– lab bhattacharjee
May 19 '13 at 5:17
$begingroup$
@labbhattacharjee Thanks I got a new formula.
$endgroup$
– iostream007
May 19 '13 at 5:40
$begingroup$
You can also use Appolonius theorem.
$endgroup$
– Aryabhata
May 19 '13 at 5:43
$begingroup$
@iostream007, welcome. Have you tried proving it?
$endgroup$
– lab bhattacharjee
May 19 '13 at 5:44
$begingroup$
i'm trying to solve it then i'll try how to prove it. Isn't any formula to the same because this question asked in SSC exam and there is not so much to spent on 1 question?
$endgroup$
– iostream007
May 19 '13 at 5:48
|
show 2 more comments
2
$begingroup$
mathworld.wolfram.com/TriangleMedian.html
$endgroup$
– lab bhattacharjee
May 19 '13 at 5:17
$begingroup$
@labbhattacharjee Thanks I got a new formula.
$endgroup$
– iostream007
May 19 '13 at 5:40
$begingroup$
You can also use Appolonius theorem.
$endgroup$
– Aryabhata
May 19 '13 at 5:43
$begingroup$
@iostream007, welcome. Have you tried proving it?
$endgroup$
– lab bhattacharjee
May 19 '13 at 5:44
$begingroup$
i'm trying to solve it then i'll try how to prove it. Isn't any formula to the same because this question asked in SSC exam and there is not so much to spent on 1 question?
$endgroup$
– iostream007
May 19 '13 at 5:48
2
2
$begingroup$
mathworld.wolfram.com/TriangleMedian.html
$endgroup$
– lab bhattacharjee
May 19 '13 at 5:17
$begingroup$
mathworld.wolfram.com/TriangleMedian.html
$endgroup$
– lab bhattacharjee
May 19 '13 at 5:17
$begingroup$
@labbhattacharjee Thanks I got a new formula.
$endgroup$
– iostream007
May 19 '13 at 5:40
$begingroup$
@labbhattacharjee Thanks I got a new formula.
$endgroup$
– iostream007
May 19 '13 at 5:40
$begingroup$
You can also use Appolonius theorem.
$endgroup$
– Aryabhata
May 19 '13 at 5:43
$begingroup$
You can also use Appolonius theorem.
$endgroup$
– Aryabhata
May 19 '13 at 5:43
$begingroup$
@iostream007, welcome. Have you tried proving it?
$endgroup$
– lab bhattacharjee
May 19 '13 at 5:44
$begingroup$
@iostream007, welcome. Have you tried proving it?
$endgroup$
– lab bhattacharjee
May 19 '13 at 5:44
$begingroup$
i'm trying to solve it then i'll try how to prove it. Isn't any formula to the same because this question asked in SSC exam and there is not so much to spent on 1 question?
$endgroup$
– iostream007
May 19 '13 at 5:48
$begingroup$
i'm trying to solve it then i'll try how to prove it. Isn't any formula to the same because this question asked in SSC exam and there is not so much to spent on 1 question?
$endgroup$
– iostream007
May 19 '13 at 5:48
|
show 2 more comments
5 Answers
5
active
oldest
votes
$begingroup$
You know that medians divide a triangle to 6 equal areas. If you find one of them, multiplying with 6 give you the area of whole triangle. Let's denote one area as $S$, now see the figure:
I guess you saw the right triangle.
edited May 19 '13 at 6:49
DonAntonio
180k1494233
answered May 19 '13 at 5:57
newzadnewzad
3,2601346
$endgroup$
$begingroup$
What right triangle? Can you please elaborate?
$endgroup$
– Aryabhata
May 19 '13 at 6:05
$begingroup$
the triangle of 6-8-10. We get this by drawing paralels, 6 to 6, 10 to 10.
$endgroup$
– newzad
May 19 '13 at 6:06
$begingroup$
Thanks! I see it now :-) +1.
$endgroup$
– Aryabhata
May 19 '13 at 6:08
add a comment |
$begingroup$
Area of a Triangle from the Medians
A triangle is divided in to $6$ equal areas by its medians:
$hspace2cm$
In the case where the two blue triangles share a common side of the triangle, it is pretty simple to see they share a common altitude (dotted) and equal bases; therefore, equal areas.
In the case where the two red triangles share a common $frac23$ of a median, the altitudes (dotted) are equal since they are corresponding sides to two right triangles with equal hypotenuses and equal vertically opposite angles, and they share a common base; therefore, equal areas.
Now duplicate the original triangle (dark outline) by rotating it one-half a revolution on the middle of one of its sides:
$hspace3cm$
The triangle in green has sides $frac23a$, $frac23b$, and $frac23c$, and by Heron's formula has area
$$
frac49sqrts(s-a)(s-b)(s-c)tag1
$$
where $s=(a+b+c)/2$. Thus, each of the $6$ small, equal-area triangles in the original triangle has an area of half of that. Therefore, the area of the original triangle is $3$ times that given in $(1)$:
$$
frac43sqrts(s-a)(s-b)(s-c)tag2
$$
answered Sep 8 '14 at 1:19
robjohn♦robjohn
271k27313642
$endgroup$
add a comment |
$begingroup$
The area of a triangle made by the medians taken as sides is 75% of the triangle of which the medians are given. Now you can find the area by heron formula and the area thus you get will be 75% of the area of the triangle of which the medians are given.
answered Mar 21 '15 at 9:10
vaasievaasie
111
$endgroup$
$begingroup$
Are you sure about the $75%$? Connecting the medians of an equilateral triangle subdivides it into four congruent triangles.
$endgroup$
– N. F. Taussig
Mar 21 '15 at 9:47
add a comment |
$begingroup$
In this type of questions, given medians always make triplet (a right triangle). From these given triplet area of triangle can be find easily
A=4/3area of right triangle form by triplet
As according to your question:
A=4/30.5×(9×12)
=72
edited Aug 9 '17 at 8:21
Rafa Budría
5,9721825
answered Aug 9 '17 at 7:22
Miikash RainagMiikash Rainag
1
$endgroup$
add a comment |
$begingroup$
There is a direct formula:
Let
$$s = (m_1+m_2+m_3)/2,$$
Then
$$textarea = frac43sqrts(s-m_1)(s-m_2)(s-m_3).$$
This gives answer of above question as $72$.
edited Aug 26 '16 at 5:43
Frenzy Li
2,86522446
answered Oct 11 '13 at 18:20
Manuj KhullarManuj Khullar
1
$endgroup$
$begingroup$
There is a missing factor of $frac43$ missing from your formula.
$endgroup$
– robjohn♦
Sep 8 '14 at 2:17
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You know that medians divide a triangle to 6 equal areas. If you find one of them, multiplying with 6 give you the area of whole triangle. Let's denote one area as $S$, now see the figure:
I guess you saw the right triangle.
edited May 19 '13 at 6:49
DonAntonio
180k1494233
answered May 19 '13 at 5:57
newzadnewzad
3,2601346
$endgroup$
$begingroup$
What right triangle? Can you please elaborate?
$endgroup$
– Aryabhata
May 19 '13 at 6:05
$begingroup$
the triangle of 6-8-10. We get this by drawing paralels, 6 to 6, 10 to 10.
$endgroup$
– newzad
May 19 '13 at 6:06
$begingroup$
Thanks! I see it now :-) +1.
$endgroup$
– Aryabhata
May 19 '13 at 6:08
add a comment |
$begingroup$
You know that medians divide a triangle to 6 equal areas. If you find one of them, multiplying with 6 give you the area of whole triangle. Let's denote one area as $S$, now see the figure:
I guess you saw the right triangle.
edited May 19 '13 at 6:49
DonAntonio
180k1494233
answered May 19 '13 at 5:57
newzadnewzad
3,2601346
$endgroup$
$begingroup$
What right triangle? Can you please elaborate?
$endgroup$
– Aryabhata
May 19 '13 at 6:05
$begingroup$
the triangle of 6-8-10. We get this by drawing paralels, 6 to 6, 10 to 10.
$endgroup$
– newzad
May 19 '13 at 6:06
$begingroup$
Thanks! I see it now :-) +1.
$endgroup$
– Aryabhata
May 19 '13 at 6:08
add a comment |
$begingroup$
You know that medians divide a triangle to 6 equal areas. If you find one of them, multiplying with 6 give you the area of whole triangle. Let's denote one area as $S$, now see the figure:
I guess you saw the right triangle.
edited May 19 '13 at 6:49
DonAntonio
180k1494233
answered May 19 '13 at 5:57
newzadnewzad
3,2601346
$endgroup$
You know that medians divide a triangle to 6 equal areas. If you find one of them, multiplying with 6 give you the area of whole triangle. Let's denote one area as $S$, now see the figure:
I guess you saw the right triangle.
edited May 19 '13 at 6:49
DonAntonio
180k1494233
answered May 19 '13 at 5:57
newzadnewzad
3,2601346
edited May 19 '13 at 6:49
DonAntonio
180k1494233
edited May 19 '13 at 6:49
DonAntonio
180k1494233
edited May 19 '13 at 6:49
DonAntonio
180k1494233
180k1494233
answered May 19 '13 at 5:57
newzadnewzad
3,2601346
answered May 19 '13 at 5:57
newzadnewzad
3,2601346
answered May 19 '13 at 5:57
newzadnewzad
3,2601346
3,2601346
$begingroup$
What right triangle? Can you please elaborate?
$endgroup$
– Aryabhata
May 19 '13 at 6:05
$begingroup$
the triangle of 6-8-10. We get this by drawing paralels, 6 to 6, 10 to 10.
$endgroup$
– newzad
May 19 '13 at 6:06
$begingroup$
Thanks! I see it now :-) +1.
$endgroup$
– Aryabhata
May 19 '13 at 6:08
add a comment |
$begingroup$
What right triangle? Can you please elaborate?
$endgroup$
– Aryabhata
May 19 '13 at 6:05
$begingroup$
the triangle of 6-8-10. We get this by drawing paralels, 6 to 6, 10 to 10.
$endgroup$
– newzad
May 19 '13 at 6:06
$begingroup$
Thanks! I see it now :-) +1.
$endgroup$
– Aryabhata
May 19 '13 at 6:08
$begingroup$
What right triangle? Can you please elaborate?
$endgroup$
– Aryabhata
May 19 '13 at 6:05
$begingroup$
What right triangle? Can you please elaborate?
$endgroup$
– Aryabhata
May 19 '13 at 6:05
$begingroup$
the triangle of 6-8-10. We get this by drawing paralels, 6 to 6, 10 to 10.
$endgroup$
– newzad
May 19 '13 at 6:06
$begingroup$
the triangle of 6-8-10. We get this by drawing paralels, 6 to 6, 10 to 10.
$endgroup$
– newzad
May 19 '13 at 6:06
$begingroup$
Thanks! I see it now :-) +1.
$endgroup$
– Aryabhata
May 19 '13 at 6:08
$begingroup$
Thanks! I see it now :-) +1.
$endgroup$
– Aryabhata
May 19 '13 at 6:08
add a comment |
$begingroup$
Area of a Triangle from the Medians
A triangle is divided in to $6$ equal areas by its medians:
$hspace2cm$
In the case where the two blue triangles share a common side of the triangle, it is pretty simple to see they share a common altitude (dotted) and equal bases; therefore, equal areas.
In the case where the two red triangles share a common $frac23$ of a median, the altitudes (dotted) are equal since they are corresponding sides to two right triangles with equal hypotenuses and equal vertically opposite angles, and they share a common base; therefore, equal areas.
Now duplicate the original triangle (dark outline) by rotating it one-half a revolution on the middle of one of its sides:
$hspace3cm$
The triangle in green has sides $frac23a$, $frac23b$, and $frac23c$, and by Heron's formula has area
$$
frac49sqrts(s-a)(s-b)(s-c)tag1
$$
where $s=(a+b+c)/2$. Thus, each of the $6$ small, equal-area triangles in the original triangle has an area of half of that. Therefore, the area of the original triangle is $3$ times that given in $(1)$:
$$
frac43sqrts(s-a)(s-b)(s-c)tag2
$$
answered Sep 8 '14 at 1:19
robjohn♦robjohn
271k27313642
$endgroup$
add a comment |
$begingroup$
Area of a Triangle from the Medians
A triangle is divided in to $6$ equal areas by its medians:
$hspace2cm$
In the case where the two blue triangles share a common side of the triangle, it is pretty simple to see they share a common altitude (dotted) and equal bases; therefore, equal areas.
In the case where the two red triangles share a common $frac23$ of a median, the altitudes (dotted) are equal since they are corresponding sides to two right triangles with equal hypotenuses and equal vertically opposite angles, and they share a common base; therefore, equal areas.
Now duplicate the original triangle (dark outline) by rotating it one-half a revolution on the middle of one of its sides:
$hspace3cm$
The triangle in green has sides $frac23a$, $frac23b$, and $frac23c$, and by Heron's formula has area
$$
frac49sqrts(s-a)(s-b)(s-c)tag1
$$
where $s=(a+b+c)/2$. Thus, each of the $6$ small, equal-area triangles in the original triangle has an area of half of that. Therefore, the area of the original triangle is $3$ times that given in $(1)$:
$$
frac43sqrts(s-a)(s-b)(s-c)tag2
$$
answered Sep 8 '14 at 1:19
robjohn♦robjohn
271k27313642
$endgroup$
add a comment |
$begingroup$
Area of a Triangle from the Medians
A triangle is divided in to $6$ equal areas by its medians:
$hspace2cm$
In the case where the two blue triangles share a common side of the triangle, it is pretty simple to see they share a common altitude (dotted) and equal bases; therefore, equal areas.
In the case where the two red triangles share a common $frac23$ of a median, the altitudes (dotted) are equal since they are corresponding sides to two right triangles with equal hypotenuses and equal vertically opposite angles, and they share a common base; therefore, equal areas.
Now duplicate the original triangle (dark outline) by rotating it one-half a revolution on the middle of one of its sides:
$hspace3cm$
The triangle in green has sides $frac23a$, $frac23b$, and $frac23c$, and by Heron's formula has area
$$
frac49sqrts(s-a)(s-b)(s-c)tag1
$$
where $s=(a+b+c)/2$. Thus, each of the $6$ small, equal-area triangles in the original triangle has an area of half of that. Therefore, the area of the original triangle is $3$ times that given in $(1)$:
$$
frac43sqrts(s-a)(s-b)(s-c)tag2
$$
answered Sep 8 '14 at 1:19
robjohn♦robjohn
271k27313642
$endgroup$
Area of a Triangle from the Medians
A triangle is divided in to $6$ equal areas by its medians:
$hspace2cm$
In the case where the two blue triangles share a common side of the triangle, it is pretty simple to see they share a common altitude (dotted) and equal bases; therefore, equal areas.
In the case where the two red triangles share a common $frac23$ of a median, the altitudes (dotted) are equal since they are corresponding sides to two right triangles with equal hypotenuses and equal vertically opposite angles, and they share a common base; therefore, equal areas.
Now duplicate the original triangle (dark outline) by rotating it one-half a revolution on the middle of one of its sides:
$hspace3cm$
The triangle in green has sides $frac23a$, $frac23b$, and $frac23c$, and by Heron's formula has area
$$
frac49sqrts(s-a)(s-b)(s-c)tag1
$$
where $s=(a+b+c)/2$. Thus, each of the $6$ small, equal-area triangles in the original triangle has an area of half of that. Therefore, the area of the original triangle is $3$ times that given in $(1)$:
$$
frac43sqrts(s-a)(s-b)(s-c)tag2
$$
answered Sep 8 '14 at 1:19
robjohn♦robjohn
271k27313642
edited Sep 8 '14 at 5:40
answered Sep 8 '14 at 1:19
robjohn♦robjohn
271k27313642
answered Sep 8 '14 at 1:19
robjohn♦robjohn
271k27313642
answered Sep 8 '14 at 1:19
robjohn♦robjohn
271k27313642
271k27313642
add a comment |
add a comment |
$begingroup$
The area of a triangle made by the medians taken as sides is 75% of the triangle of which the medians are given. Now you can find the area by heron formula and the area thus you get will be 75% of the area of the triangle of which the medians are given.
answered Mar 21 '15 at 9:10
vaasievaasie
111
$endgroup$
$begingroup$
Are you sure about the $75%$? Connecting the medians of an equilateral triangle subdivides it into four congruent triangles.
$endgroup$
– N. F. Taussig
Mar 21 '15 at 9:47
add a comment |
$begingroup$
The area of a triangle made by the medians taken as sides is 75% of the triangle of which the medians are given. Now you can find the area by heron formula and the area thus you get will be 75% of the area of the triangle of which the medians are given.
answered Mar 21 '15 at 9:10
vaasievaasie
111
$endgroup$
$begingroup$
Are you sure about the $75%$? Connecting the medians of an equilateral triangle subdivides it into four congruent triangles.
$endgroup$
– N. F. Taussig
Mar 21 '15 at 9:47
add a comment |
$begingroup$
The area of a triangle made by the medians taken as sides is 75% of the triangle of which the medians are given. Now you can find the area by heron formula and the area thus you get will be 75% of the area of the triangle of which the medians are given.
answered Mar 21 '15 at 9:10
vaasievaasie
111
$endgroup$
The area of a triangle made by the medians taken as sides is 75% of the triangle of which the medians are given. Now you can find the area by heron formula and the area thus you get will be 75% of the area of the triangle of which the medians are given.
answered Mar 21 '15 at 9:10
vaasievaasie
111
answered Mar 21 '15 at 9:10
vaasievaasie
111
answered Mar 21 '15 at 9:10
vaasievaasie
111
answered Mar 21 '15 at 9:10
vaasievaasie
111
111
$begingroup$
Are you sure about the $75%$? Connecting the medians of an equilateral triangle subdivides it into four congruent triangles.
$endgroup$
– N. F. Taussig
Mar 21 '15 at 9:47
add a comment |
$begingroup$
Are you sure about the $75%$? Connecting the medians of an equilateral triangle subdivides it into four congruent triangles.
$endgroup$
– N. F. Taussig
Mar 21 '15 at 9:47
$begingroup$
Are you sure about the $75%$? Connecting the medians of an equilateral triangle subdivides it into four congruent triangles.
$endgroup$
– N. F. Taussig
Mar 21 '15 at 9:47
$begingroup$
Are you sure about the $75%$? Connecting the medians of an equilateral triangle subdivides it into four congruent triangles.
$endgroup$
– N. F. Taussig
Mar 21 '15 at 9:47
add a comment |
$begingroup$
In this type of questions, given medians always make triplet (a right triangle). From these given triplet area of triangle can be find easily
A=4/3area of right triangle form by triplet
As according to your question:
A=4/30.5×(9×12)
=72
edited Aug 9 '17 at 8:21
Rafa Budría
5,9721825
answered Aug 9 '17 at 7:22
Miikash RainagMiikash Rainag
1
$endgroup$
add a comment |
$begingroup$
In this type of questions, given medians always make triplet (a right triangle). From these given triplet area of triangle can be find easily
A=4/3area of right triangle form by triplet
As according to your question:
A=4/30.5×(9×12)
=72
edited Aug 9 '17 at 8:21
Rafa Budría
5,9721825
answered Aug 9 '17 at 7:22
Miikash RainagMiikash Rainag
1
$endgroup$
add a comment |
$begingroup$
In this type of questions, given medians always make triplet (a right triangle). From these given triplet area of triangle can be find easily
A=4/3area of right triangle form by triplet
As according to your question:
A=4/30.5×(9×12)
=72
edited Aug 9 '17 at 8:21
Rafa Budría
5,9721825
answered Aug 9 '17 at 7:22
Miikash RainagMiikash Rainag
1
$endgroup$
In this type of questions, given medians always make triplet (a right triangle). From these given triplet area of triangle can be find easily
A=4/3area of right triangle form by triplet
As according to your question:
A=4/30.5×(9×12)
=72
edited Aug 9 '17 at 8:21
Rafa Budría
5,9721825
answered Aug 9 '17 at 7:22
Miikash RainagMiikash Rainag
1
edited Aug 9 '17 at 8:21
Rafa Budría
5,9721825
edited Aug 9 '17 at 8:21
Rafa Budría
5,9721825
edited Aug 9 '17 at 8:21
Rafa Budría
5,9721825
5,9721825
answered Aug 9 '17 at 7:22
Miikash RainagMiikash Rainag
1
answered Aug 9 '17 at 7:22
Miikash RainagMiikash Rainag
1
answered Aug 9 '17 at 7:22
Miikash RainagMiikash Rainag
1
1
add a comment |
add a comment |
$begingroup$
There is a direct formula:
Let
$$s = (m_1+m_2+m_3)/2,$$
Then
$$textarea = frac43sqrts(s-m_1)(s-m_2)(s-m_3).$$
This gives answer of above question as $72$.
edited Aug 26 '16 at 5:43
Frenzy Li
2,86522446
answered Oct 11 '13 at 18:20
Manuj KhullarManuj Khullar
1
$endgroup$
$begingroup$
There is a missing factor of $frac43$ missing from your formula.
$endgroup$
– robjohn♦
Sep 8 '14 at 2:17
add a comment |
$begingroup$
There is a direct formula:
Let
$$s = (m_1+m_2+m_3)/2,$$
Then
$$textarea = frac43sqrts(s-m_1)(s-m_2)(s-m_3).$$
This gives answer of above question as $72$.
edited Aug 26 '16 at 5:43
Frenzy Li
2,86522446
answered Oct 11 '13 at 18:20
Manuj KhullarManuj Khullar
1
$endgroup$
$begingroup$
There is a missing factor of $frac43$ missing from your formula.
$endgroup$
– robjohn♦
Sep 8 '14 at 2:17
add a comment |
$begingroup$
There is a direct formula:
Let
$$s = (m_1+m_2+m_3)/2,$$
Then
$$textarea = frac43sqrts(s-m_1)(s-m_2)(s-m_3).$$
This gives answer of above question as $72$.
edited Aug 26 '16 at 5:43
Frenzy Li
2,86522446
answered Oct 11 '13 at 18:20
Manuj KhullarManuj Khullar
1
$endgroup$
There is a direct formula:
Let
$$s = (m_1+m_2+m_3)/2,$$
Then
$$textarea = frac43sqrts(s-m_1)(s-m_2)(s-m_3).$$
This gives answer of above question as $72$.
edited Aug 26 '16 at 5:43
Frenzy Li
2,86522446
answered Oct 11 '13 at 18:20
Manuj KhullarManuj Khullar
1
edited Aug 26 '16 at 5:43
Frenzy Li
2,86522446
edited Aug 26 '16 at 5:43
Frenzy Li
2,86522446
edited Aug 26 '16 at 5:43
Frenzy Li
2,86522446
2,86522446
answered Oct 11 '13 at 18:20
Manuj KhullarManuj Khullar
1
answered Oct 11 '13 at 18:20
Manuj KhullarManuj Khullar
1
answered Oct 11 '13 at 18:20
Manuj KhullarManuj Khullar
1
1
$begingroup$
There is a missing factor of $frac43$ missing from your formula.
$endgroup$
– robjohn♦
Sep 8 '14 at 2:17
add a comment |
$begingroup$
There is a missing factor of $frac43$ missing from your formula.
$endgroup$
– robjohn♦
Sep 8 '14 at 2:17
$begingroup$
There is a missing factor of $frac43$ missing from your formula.
$endgroup$
– robjohn♦
Sep 8 '14 at 2:17
$begingroup$
There is a missing factor of $frac43$ missing from your formula.
$endgroup$
– robjohn♦
Sep 8 '14 at 2:17
add a comment |
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2
$begingroup$
mathworld.wolfram.com/TriangleMedian.html
$endgroup$
– lab bhattacharjee
May 19 '13 at 5:17
$begingroup$
@labbhattacharjee Thanks I got a new formula.
$endgroup$
– iostream007
May 19 '13 at 5:40
$begingroup$
You can also use Appolonius theorem.
$endgroup$
– Aryabhata
May 19 '13 at 5:43
$begingroup$
@iostream007, welcome. Have you tried proving it?
$endgroup$
– lab bhattacharjee
May 19 '13 at 5:44
$begingroup$
i'm trying to solve it then i'll try how to prove it. Isn't any formula to the same because this question asked in SSC exam and there is not so much to spent on 1 question?
$endgroup$
– iostream007
May 19 '13 at 5:48