How can i Find the Area of a Traingle Formed inside a Triangle?
Here we have Co Ordinates of A ,B ,C and We have to Find the Area of the Triangle PQR Formed Inside the Triangle ABC.triangle ABC. Point D, E and F divides the sides BC, CA and AB into ratio 1:2 respectively. That is CD=2BD, AE=2CE and BF=2AF. A, D; B, E and C, F are connected. AD and BE intersects at P, BE and CF intersects at Q and CF and AD intersects at R. Please Check the Image in the link.
geometry triangle coordinate-systems
asked Nov 22 at 13:16
Raihan Rahman
11
|
show 5 more comments
Here we have Co Ordinates of A ,B ,C and We have to Find the Area of the Triangle PQR Formed Inside the Triangle ABC.triangle ABC. Point D, E and F divides the sides BC, CA and AB into ratio 1:2 respectively. That is CD=2BD, AE=2CE and BF=2AF. A, D; B, E and C, F are connected. AD and BE intersects at P, BE and CF intersects at Q and CF and AD intersects at R. Please Check the Image in the link.
geometry triangle coordinate-systems
asked Nov 22 at 13:16
Raihan Rahman
11
Angle bisectors are concurrent.
– Offlaw
Nov 22 at 13:22
Ok but how can I find Point P ,Q R and finally find the area of PQR .. Can you Explain...I don't Understand what You want to say.
– Raihan Rahman
Nov 22 at 13:26
As angle bisectors intersect at incenter of the triangle then you never get another triangle.
– Offlaw
Nov 22 at 13:33
Sorry...I think I made a mistake to ask the question...can you please check the image then you can understand what I ask for.
– Raihan Rahman
Nov 22 at 13:40
Give some more information.
– Offlaw
Nov 22 at 13:45
|
show 5 more comments
Here we have Co Ordinates of A ,B ,C and We have to Find the Area of the Triangle PQR Formed Inside the Triangle ABC.triangle ABC. Point D, E and F divides the sides BC, CA and AB into ratio 1:2 respectively. That is CD=2BD, AE=2CE and BF=2AF. A, D; B, E and C, F are connected. AD and BE intersects at P, BE and CF intersects at Q and CF and AD intersects at R. Please Check the Image in the link.
geometry triangle coordinate-systems
asked Nov 22 at 13:16
Raihan Rahman
11
Here we have Co Ordinates of A ,B ,C and We have to Find the Area of the Triangle PQR Formed Inside the Triangle ABC.triangle ABC. Point D, E and F divides the sides BC, CA and AB into ratio 1:2 respectively. That is CD=2BD, AE=2CE and BF=2AF. A, D; B, E and C, F are connected. AD and BE intersects at P, BE and CF intersects at Q and CF and AD intersects at R. Please Check the Image in the link.
geometry triangle coordinate-systems
geometry triangle coordinate-systems
asked Nov 22 at 13:16
Raihan Rahman
11
asked Nov 22 at 13:16
Raihan Rahman
11
edited Nov 22 at 13:58
asked Nov 22 at 13:16
Raihan Rahman
11
asked Nov 22 at 13:16
Raihan Rahman
11
asked Nov 22 at 13:16
Raihan Rahman
11
11
Angle bisectors are concurrent.
– Offlaw
Nov 22 at 13:22
Ok but how can I find Point P ,Q R and finally find the area of PQR .. Can you Explain...I don't Understand what You want to say.
– Raihan Rahman
Nov 22 at 13:26
As angle bisectors intersect at incenter of the triangle then you never get another triangle.
– Offlaw
Nov 22 at 13:33
Sorry...I think I made a mistake to ask the question...can you please check the image then you can understand what I ask for.
– Raihan Rahman
Nov 22 at 13:40
Give some more information.
– Offlaw
Nov 22 at 13:45
|
show 5 more comments
Angle bisectors are concurrent.
– Offlaw
Nov 22 at 13:22
Ok but how can I find Point P ,Q R and finally find the area of PQR .. Can you Explain...I don't Understand what You want to say.
– Raihan Rahman
Nov 22 at 13:26
As angle bisectors intersect at incenter of the triangle then you never get another triangle.
– Offlaw
Nov 22 at 13:33
Sorry...I think I made a mistake to ask the question...can you please check the image then you can understand what I ask for.
– Raihan Rahman
Nov 22 at 13:40
Give some more information.
– Offlaw
Nov 22 at 13:45
Angle bisectors are concurrent.
– Offlaw
Nov 22 at 13:22
Angle bisectors are concurrent.
– Offlaw
Nov 22 at 13:22
Ok but how can I find Point P ,Q R and finally find the area of PQR .. Can you Explain...I don't Understand what You want to say.
– Raihan Rahman
Nov 22 at 13:26
Ok but how can I find Point P ,Q R and finally find the area of PQR .. Can you Explain...I don't Understand what You want to say.
– Raihan Rahman
Nov 22 at 13:26
As angle bisectors intersect at incenter of the triangle then you never get another triangle.
– Offlaw
Nov 22 at 13:33
As angle bisectors intersect at incenter of the triangle then you never get another triangle.
– Offlaw
Nov 22 at 13:33
Sorry...I think I made a mistake to ask the question...can you please check the image then you can understand what I ask for.
– Raihan Rahman
Nov 22 at 13:40
Sorry...I think I made a mistake to ask the question...can you please check the image then you can understand what I ask for.
– Raihan Rahman
Nov 22 at 13:40
Give some more information.
– Offlaw
Nov 22 at 13:45
Give some more information.
– Offlaw
Nov 22 at 13:45
|
show 5 more comments
1 Answer
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Denote by $[ABC]$ the area of the triangle $Delta ABC$.
First note that $$frac{[ADC]}{[ABC]}=frac{overline{DC}}{overline {BC}}=frac{2}{3}Rightarrow [ADC]=frac{2[ABC]}{3}quad (1)$$
Now, by Menelaus's Theorem regarding the triangle $Delta ADB$ and the line through $F,R$ and $C$:
$$frac{overline {AF}}{overline {FB}}*frac{overline {BC}}{overline {DC}}*frac{overline {RD}}{overline {AR}}=1 Rightarrow frac{overline {RD}}{overline {AR}}=frac{overline {FB}}{overline {AF}}*frac{overline {DC}}{overline {BC}}=2*frac{2}{3}=frac{4}{3}$$
Note now that (recall $(1)$) $$frac{overline {RD}}{overline {AR}}=frac{[ADC]-[ARC]}{[ARC]}=frac{[ADC]}{[ARC]}-1=frac{2[ABC]}{3[ARC]}-1=frac{4}{3}$$ $$Rightarrow [ARC]=frac{2[ABC]}{7}$$
Analogously $$[ARC]=[CQB]=[BPA]=frac{2[ABC]}{7}$$
Since $$[PQR]=[ABC]-Bigl([ARC]+[CQB]+[BPA]Bigr)$$
Your have $$[PQR]=[ABC]-3Biggl(frac{2[ABC]}{7}Biggr)=frac{[ABC]}{7}$$
$mathbf {Remark:}$
The problem (configuration) you've suggested is known as the "One-seventh area triangle" and is a special case of "Routh's theorem" which gives a general solution in order to calculate the area of the inner triangle for any three cevians forming this one.
answered Nov 22 at 20:39
Dr. Mathva
931315
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Denote by $[ABC]$ the area of the triangle $Delta ABC$.
First note that $$frac{[ADC]}{[ABC]}=frac{overline{DC}}{overline {BC}}=frac{2}{3}Rightarrow [ADC]=frac{2[ABC]}{3}quad (1)$$
Now, by Menelaus's Theorem regarding the triangle $Delta ADB$ and the line through $F,R$ and $C$:
$$frac{overline {AF}}{overline {FB}}*frac{overline {BC}}{overline {DC}}*frac{overline {RD}}{overline {AR}}=1 Rightarrow frac{overline {RD}}{overline {AR}}=frac{overline {FB}}{overline {AF}}*frac{overline {DC}}{overline {BC}}=2*frac{2}{3}=frac{4}{3}$$
Note now that (recall $(1)$) $$frac{overline {RD}}{overline {AR}}=frac{[ADC]-[ARC]}{[ARC]}=frac{[ADC]}{[ARC]}-1=frac{2[ABC]}{3[ARC]}-1=frac{4}{3}$$ $$Rightarrow [ARC]=frac{2[ABC]}{7}$$
Analogously $$[ARC]=[CQB]=[BPA]=frac{2[ABC]}{7}$$
Since $$[PQR]=[ABC]-Bigl([ARC]+[CQB]+[BPA]Bigr)$$
Your have $$[PQR]=[ABC]-3Biggl(frac{2[ABC]}{7}Biggr)=frac{[ABC]}{7}$$
$mathbf {Remark:}$
The problem (configuration) you've suggested is known as the "One-seventh area triangle" and is a special case of "Routh's theorem" which gives a general solution in order to calculate the area of the inner triangle for any three cevians forming this one.
answered Nov 22 at 20:39
Dr. Mathva
931315
add a comment |
Denote by $[ABC]$ the area of the triangle $Delta ABC$.
First note that $$frac{[ADC]}{[ABC]}=frac{overline{DC}}{overline {BC}}=frac{2}{3}Rightarrow [ADC]=frac{2[ABC]}{3}quad (1)$$
Now, by Menelaus's Theorem regarding the triangle $Delta ADB$ and the line through $F,R$ and $C$:
$$frac{overline {AF}}{overline {FB}}*frac{overline {BC}}{overline {DC}}*frac{overline {RD}}{overline {AR}}=1 Rightarrow frac{overline {RD}}{overline {AR}}=frac{overline {FB}}{overline {AF}}*frac{overline {DC}}{overline {BC}}=2*frac{2}{3}=frac{4}{3}$$
Note now that (recall $(1)$) $$frac{overline {RD}}{overline {AR}}=frac{[ADC]-[ARC]}{[ARC]}=frac{[ADC]}{[ARC]}-1=frac{2[ABC]}{3[ARC]}-1=frac{4}{3}$$ $$Rightarrow [ARC]=frac{2[ABC]}{7}$$
Analogously $$[ARC]=[CQB]=[BPA]=frac{2[ABC]}{7}$$
Since $$[PQR]=[ABC]-Bigl([ARC]+[CQB]+[BPA]Bigr)$$
Your have $$[PQR]=[ABC]-3Biggl(frac{2[ABC]}{7}Biggr)=frac{[ABC]}{7}$$
$mathbf {Remark:}$
The problem (configuration) you've suggested is known as the "One-seventh area triangle" and is a special case of "Routh's theorem" which gives a general solution in order to calculate the area of the inner triangle for any three cevians forming this one.
answered Nov 22 at 20:39
Dr. Mathva
931315
add a comment |
Denote by $[ABC]$ the area of the triangle $Delta ABC$.
First note that $$frac{[ADC]}{[ABC]}=frac{overline{DC}}{overline {BC}}=frac{2}{3}Rightarrow [ADC]=frac{2[ABC]}{3}quad (1)$$
Now, by Menelaus's Theorem regarding the triangle $Delta ADB$ and the line through $F,R$ and $C$:
$$frac{overline {AF}}{overline {FB}}*frac{overline {BC}}{overline {DC}}*frac{overline {RD}}{overline {AR}}=1 Rightarrow frac{overline {RD}}{overline {AR}}=frac{overline {FB}}{overline {AF}}*frac{overline {DC}}{overline {BC}}=2*frac{2}{3}=frac{4}{3}$$
Note now that (recall $(1)$) $$frac{overline {RD}}{overline {AR}}=frac{[ADC]-[ARC]}{[ARC]}=frac{[ADC]}{[ARC]}-1=frac{2[ABC]}{3[ARC]}-1=frac{4}{3}$$ $$Rightarrow [ARC]=frac{2[ABC]}{7}$$
Analogously $$[ARC]=[CQB]=[BPA]=frac{2[ABC]}{7}$$
Since $$[PQR]=[ABC]-Bigl([ARC]+[CQB]+[BPA]Bigr)$$
Your have $$[PQR]=[ABC]-3Biggl(frac{2[ABC]}{7}Biggr)=frac{[ABC]}{7}$$
$mathbf {Remark:}$
The problem (configuration) you've suggested is known as the "One-seventh area triangle" and is a special case of "Routh's theorem" which gives a general solution in order to calculate the area of the inner triangle for any three cevians forming this one.
answered Nov 22 at 20:39
Dr. Mathva
931315
Denote by $[ABC]$ the area of the triangle $Delta ABC$.
First note that $$frac{[ADC]}{[ABC]}=frac{overline{DC}}{overline {BC}}=frac{2}{3}Rightarrow [ADC]=frac{2[ABC]}{3}quad (1)$$
Now, by Menelaus's Theorem regarding the triangle $Delta ADB$ and the line through $F,R$ and $C$:
$$frac{overline {AF}}{overline {FB}}*frac{overline {BC}}{overline {DC}}*frac{overline {RD}}{overline {AR}}=1 Rightarrow frac{overline {RD}}{overline {AR}}=frac{overline {FB}}{overline {AF}}*frac{overline {DC}}{overline {BC}}=2*frac{2}{3}=frac{4}{3}$$
Note now that (recall $(1)$) $$frac{overline {RD}}{overline {AR}}=frac{[ADC]-[ARC]}{[ARC]}=frac{[ADC]}{[ARC]}-1=frac{2[ABC]}{3[ARC]}-1=frac{4}{3}$$ $$Rightarrow [ARC]=frac{2[ABC]}{7}$$
Analogously $$[ARC]=[CQB]=[BPA]=frac{2[ABC]}{7}$$
Since $$[PQR]=[ABC]-Bigl([ARC]+[CQB]+[BPA]Bigr)$$
Your have $$[PQR]=[ABC]-3Biggl(frac{2[ABC]}{7}Biggr)=frac{[ABC]}{7}$$
$mathbf {Remark:}$
The problem (configuration) you've suggested is known as the "One-seventh area triangle" and is a special case of "Routh's theorem" which gives a general solution in order to calculate the area of the inner triangle for any three cevians forming this one.
answered Nov 22 at 20:39
Dr. Mathva
931315
answered Nov 22 at 20:39
Dr. Mathva
931315
answered Nov 22 at 20:39
Dr. Mathva
931315
answered Nov 22 at 20:39
Dr. Mathva
931315
931315
add a comment |
add a comment |
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Angle bisectors are concurrent.
– Offlaw
Nov 22 at 13:22
Ok but how can I find Point P ,Q R and finally find the area of PQR .. Can you Explain...I don't Understand what You want to say.
– Raihan Rahman
Nov 22 at 13:26
As angle bisectors intersect at incenter of the triangle then you never get another triangle.
– Offlaw
Nov 22 at 13:33
Sorry...I think I made a mistake to ask the question...can you please check the image then you can understand what I ask for.
– Raihan Rahman
Nov 22 at 13:40
Give some more information.
– Offlaw
Nov 22 at 13:45