R is the midpoint of PS. Extend QR to touch PT at a new point M. MT = 7. TV = 6. MV = MQ = 13 (isosceles). STV is also isosceles. STV is similar to QMV ST = 6 so RM = 3. As QM = 13, QR = 10 Alright, you called it A and I called it M> Same difference :)
I got an alternative solution with trigonometry to share with. Applying sine law to triangle PQV sin( 2*sin(a)*cos(a))/20=sin(a)/PQ ->PQ=10/cos(a) or PQ*cos(a)=10 cos(a)=QR/PQ ->QR=PQ*cos(a)=10 that’s our answer.
Wanting to do something different, here it is my much longer solution: As shown QA = AV = 6+7=13 Applying the bisector theorem on QPV with QA bisector we have: QP/QV=7/13 then we can write QP = 7x QV = 13x now applying the stewart th. we can write: QA² = QP*QV - PA*AV 13² = 7x*13x - 7*13 x = √ 20/7 QP = 7x = 7 √ 20/7 Finally to find h we can apply pythagorean theorem on PQR and PRA in this way: 1) PR² = QP² - QR² => PR² = ( 7 √ 20/7)² - h² 2) PR² = PA² - AR² => PR² = 7² - (13 - h)² comparing 1) = 2) we have: 140 - h² = 26h - h² - 120 26h = 260 h = 10 needless to say that the methods of the Prof and Xualain are much better...
No friend don't say that, yours is Perfect, I agree Mr Xualain method is also perfect, do not underate your methodology. its very clear, If someone who doesn't know the angle bisector theorem or the Stewart's theorem should read your solution, they'll end up understanding the method, get the opportunity to know the angle bisector theorem and Stewart's theorem, Hence you've educated them, I don't know if you have noticed that most school do not teach students angle bisector theorem and Stewart's Theorem, I don't maybe the do it at you place but most of the places I have gone to, schools ignore angle bisector theorems and Stewarts theorem
@@MathandEngineering I don't know, Prof, I'm not a teacher. I can tell you that I wasn't taught this, but that was over 40 years ago, haha. However, I had the chance to review a textbook used by students participating in the Math Olympiads, and the Stewart's Theorem wasn't included! Nonetheless, for teaching purposes, I want to warn the student who might take a look at the solution I proposed that Stewart's Theorem is a bit more complicated. What I presented is a version used when there's a bisector involved.
QR || ST. So, RQS=TSV=a and TS=6. Thus, in ∆PTS, PS= √160 and /_PTS = a+a = 2a We calculate cos 2a and then cos a and sin a and apply in ∆PQR and the value of h. BTW, where i come from, lines and angles can be equal to each other, but the word congruent is applicable only for shapes, triangle and upwards.
Ok friend thank you, you said where you come from lines and angles can be equal with each other not Congruent, ok though my understanding of the word "Congruent" It means "exactly the same", If me calling lines Congruent is stopping you from using my videos the way you want, you tell me please I will try and make some adjustments so that it fits in the category you want it, like I always say, I make them videos for you, so if there is any adjustments you want which is not in contradiction with established fact, I am willing to do it thanks, I will be expecting your reply thanks
Great ! @MathandEngineering regularly deletes his old videos to re-upload them later with even better demonstrations. Here is my method, a little less brilliant than @MathandEngineering and @xualain3129 : angle(TSV)=180°-angle(TSP)-angle(RSQ)=180°-90°-(90°-a)=a angle(TSV)=angle(TVS)=a then TSV is an isoscele triangle and : ST=VT=6 angle(STP)=180°-angle(STV)=angle(TSV)+angle(TVS)=a+a=2*a ST=PT*cos(2*a) 6=14*cos(2*a) cos(2*a)=2*(cos(a))^2-1=6/14=3/7 (cos(a))^2=5/7 PS=PT*sin(2*a) PR+RS=PT*sin(2*a) QR*tan(a)+QR*tan(a)=PT*sin(2*a) 2*QR*sin(a)/cos(a)=2*PT*sin(a)*cos(a) QR=PT*(cos(a))^2 QR=14*5/7 QR=10
R is the midpoint of PS.
Extend QR to touch PT at a new point M.
MT = 7.
TV = 6.
MV = MQ = 13 (isosceles).
STV is also isosceles.
STV is similar to QMV
ST = 6 so RM = 3.
As QM = 13, QR = 10
Alright, you called it A and I called it M> Same difference :)
I got an alternative solution with trigonometry to share with.
Applying sine law to triangle PQV
sin(
2*sin(a)*cos(a))/20=sin(a)/PQ ->PQ=10/cos(a) or PQ*cos(a)=10
cos(a)=QR/PQ ->QR=PQ*cos(a)=10 that’s our answer.
demonstration as smart as @MathandEngineering
Wanting to do something different, here it is my much longer solution:
As shown QA = AV = 6+7=13
Applying the bisector theorem on QPV with QA bisector we have:
QP/QV=7/13 then we can write
QP = 7x
QV = 13x
now applying the stewart th. we can write:
QA² = QP*QV - PA*AV
13² = 7x*13x - 7*13
x = √ 20/7
QP = 7x = 7 √ 20/7
Finally to find h we can apply pythagorean theorem on PQR and PRA in this way:
1) PR² = QP² - QR² => PR² = ( 7 √ 20/7)² - h²
2) PR² = PA² - AR² => PR² = 7² - (13 - h)²
comparing 1) = 2) we have:
140 - h² = 26h - h² - 120
26h = 260
h = 10
needless to say that the methods of the Prof and Xualain are much better...
No friend don't say that, yours is Perfect, I agree Mr Xualain method is also perfect, do not underate your methodology. its very clear,
If someone who doesn't know the angle bisector theorem or the Stewart's theorem should read your solution, they'll end up understanding the method, get the opportunity to know the angle bisector theorem and Stewart's theorem,
Hence you've educated them,
I don't know if you have noticed that most school do not teach students angle bisector theorem and Stewart's Theorem,
I don't maybe the do it at you place but most of the places I have gone to, schools ignore angle bisector theorems and Stewarts theorem
@@MathandEngineering I don't know, Prof, I'm not a teacher. I can tell you that I wasn't taught this, but that was over 40 years ago, haha. However, I had the chance to review a textbook used by students participating in the Math Olympiads, and the Stewart's Theorem wasn't included! Nonetheless, for teaching purposes, I want to warn the student who might take a look at the solution I proposed that Stewart's Theorem is a bit more complicated. What I presented is a version used when there's a bisector involved.
QR || ST. So, RQS=TSV=a and TS=6. Thus, in ∆PTS, PS= √160 and /_PTS = a+a = 2a
We calculate cos 2a and then cos a and sin a and apply in ∆PQR and the value of h.
BTW, where i come from, lines and angles can be equal to each other, but the word congruent is applicable only for shapes, triangle and upwards.
Ok friend thank you, you said where you come from lines and angles can be equal with each other not Congruent, ok though my understanding of the word "Congruent" It means "exactly the same",
If me calling lines Congruent is stopping you from using my videos the way you want, you tell me please I will try and make some adjustments so that it fits in the category you want it, like I always say, I make them videos for you, so if there is any adjustments you want which is not in contradiction with established fact, I am willing to do it thanks, I will be expecting your reply thanks
Triangle QAV is isósceles
QA = 6+14/2 = 13 cm
Triangle STV is isósceles
ST = 6 cm
Segment QR:
h = 13 - 6/2 = 10 cm ( Solved √ )
This is exactly the demonstration of @MathandEngineering
@matthieudutriaux
Yes, there was a coincidence, same thought. I always try to solve without watching the video.
@@marioalb9726
Okay
Great !
@MathandEngineering regularly deletes his old videos to re-upload them later with even better demonstrations.
Here is my method, a little less brilliant than @MathandEngineering and @xualain3129 :
angle(TSV)=180°-angle(TSP)-angle(RSQ)=180°-90°-(90°-a)=a
angle(TSV)=angle(TVS)=a then TSV is an isoscele triangle and : ST=VT=6
angle(STP)=180°-angle(STV)=angle(TSV)+angle(TVS)=a+a=2*a
ST=PT*cos(2*a)
6=14*cos(2*a)
cos(2*a)=2*(cos(a))^2-1=6/14=3/7
(cos(a))^2=5/7
PS=PT*sin(2*a)
PR+RS=PT*sin(2*a)
QR*tan(a)+QR*tan(a)=PT*sin(2*a)
2*QR*sin(a)/cos(a)=2*PT*sin(a)*cos(a)
QR=PT*(cos(a))^2
QR=14*5/7
QR=10