Actually that's exactly what I was thinking, since I was looking up lines tangent on 2 circles recently and found their construction methods, AB is an internal tangent of the 2 circles, which makes it equal to a segment tangent on a circle of center P and radius R1+R2=8, from point Q
Solution: If you move the line AB parallel upwards so that B coincides with Q, you get a right-angled triangle with the hypotenuse PQ = 16 and the leg 5+3 = 8, so that you can use the Pythagorean theorem to calculate the other leg AB: AB = √(16²-8²) = √192 = √(64*3) = 8*√3 = 13.8564
Sir, I am astonished. You did not use your favourite 30-60-90 special triangle theory in working out the length of AC & BC when hypotenuse and leg are 10 and 5 & 6 and 3. Thanks a lot.
Помню, что по какой-то там теореме APǁBQ, тогда ▲АСР~▲BCQ, пропорцию видим - 3 к 5. Тогда РС=5х, CQ=3х, 5х+3х=8х=16⇔х=2. Отсюда РС=10, CQ=6. BC:=у. По теореме о касательной и секущей 9*3=у²=27⇔y=3√3. Возвращаясь к первоначальной пропорции, видно, что АС в таком случае будет 5√3, общая длина тоже становится очевидной - 8√3.
Another solution is to complete the rectangle ABQD. This makes a right triangle PDQ whose hypotenuse PQ is 16 and whose shorter leg PD is 8. This is a 30-60-90 right triangle, so the other leg DQ=8*sqrt(3), and AB is the opposite side of the rectangle, hence also 8*sqrt(3).
Another method, pls pin Apc and cbq are similar triangles. the ratio b/w PC and cq is 5:3 , and pc+cq=16, so pc=10cm and cq=6cm, if we consider triangle PAC we get AC=5×root 3 and in triangle cbq we get cq=3root3 so the length AB=8 root 3
Woohoo! I did it in my head just from the thumbnail! I figured it was gonna get messy with the sqrt75 but simplified to 5sqrt3 then got 3sqrt3 being that they're proportional, so the total was 8sqrt3. ✨Magic!✨
Triangles PAC and CBQ are similar (same angles). So PC/PA = QC/QB, or PC/5 = QC/3 = (PC+QC)/(5+3) = PQ/8 = 16/8 = 2, (*) so we have: PC = 10 and QC = 6. PAC and QBC are then 30°-60°-90° triangles, so AC = sqrt(3).PA = 5.sqrt(3) and BC = sqrt(3).QB = 3.qqrt(3), and finally AB = AC + BC = 8.sqrt(3). (*) If a/b = c/d, then a/b = c/d = (a+c)/(b+d), naturally with b, d and b+d non equal to 0.
Once you know the two right triangles are similar, the ratio is 5:3, so the line of centers is split into segments of length 5c and 3c, That makes 8c=16, so c=2 and the hypotenuses are 10 and 6. Since the hypotenuse of each triangle is twice a leg, the triangles are 30-60-90 triangles, so the common internal tangent is 5*sqrt(3)+3*sqrt(3), which is 8*sqrt(3).
Let's find the length: . .. ... .... ..... Since AB is a tangent to both circles, we know that ∠PAC=∠QBC=90°. Additionally we have ∠ACP=∠BCQ. Therefore, ACP and BCQ are similar triangles and we can conclude: CP/CQ = AP/BQ = 5/3 ⇒ CP = (5/3)CQ CP + CQ = PQ (5/3)CQ + CQ = 16 (5/3)CQ + (3/3)CQ = 16 (8/3)CQ = 16 ⇒ CQ = 16*3/8 = 6 ⇒ CP = (5/3)CQ = (5/3)*6 = 10 By applying the Pythagorean theorem to the right triangles ACP and BCQ we obtain: AP² + AC² = CP² ⇒ AC² = CP² − AP² = 10² − 5² = 100 − 25 = 75 ⇒ AC = √75 = 5√3 BQ² + BC² = CQ² ⇒ BC² = CQ² − BQ² = 6² − 3² = 36 − 9 = 27 ⇒ BC = √27 = 3√3 Now we are able to calculate the length of AB: AB = AC + BC = 5√3 + 3√3 = 8√3 Best regards from Germany
We are looking at two similar right triangles whose linear ratio = 5 : 3 The sum of their hypotenuses = 16, so their respective hypotenuse = 16(5/8) & 16(3/8) = 10 & 6 We can then calculate AB = √(10^2 - 5^2) + √(6^2 - 3^2) = √75 + √27 = 8√3
As AB is tangent to circle P at A and circle Q at B, then ∠CAP = ∠CBQ = 90°. As ∠PCA and ∠QCB are vertical angles and thus equal, then ∆CAP and ∆CBQ are similar triangles. QC/QB = PC/PA QC/3 = PC/5 3PC = 5QC = 5(16-PC) 3PC = 80 - 5PC 8PC = 80 PC = 80/8 = 10 CA² + PA² = PC² CA² + 5² = 10² CA² = 100 - 25 = 75 CA = √75 = 5√3 CB/QB = CA/PA CB/3 = 5√3/5 = √3 CB = 3√3 AB = CA+CB = 5√3 + 3√3 [ AB = 8√3 ≈ 13.86 units ]
Shift AB upward towards the center of small circle. You will end up having right triangle with 8 unit length, AB and hyp=16 unit length. It is 30,60,90 triangle with AB= 8 sqrt(3) DONE
👍 Length of direct conmon tangent = sqrt { PQ^2 - ( difference of radii)^2 } Length of transverse common tangent = sqrt { PQ^2 - ( sum of radii)^2 } where PQ = distance between centres of circles.
From the original image, you didn't say that A and B are tangential points, you just added then that AB is tangent to both circles. From the screenshot image, why would the 2 circles need to be lined up like that? By not stating that AB is a tangent line, it could have several solutions. It is only when you say that it's a tangent... I mean, from the original image, you can just move A and B anywhere, and you can have circles of radius 5 and 3, PQ equal 16, and AB has many values, but in those cases AB is not a tangent line.
1/The two triangles PAC and QBC are similar so, CP/AP=CQ/QB =(CP+CQ)/(AP+QB)=16/8=2 --> CP= 2AP= 10 and CQ= 2QB=6 --> the two triangles mentioned above are special 30-60-90 ones -> AC= 5sqrt3 and BC= 3sqrt3 --> AB = 8sqrt3😅😅😅
PAC~CBQ===> 5/3=PC/(16-PC)===> PC=10 Ambos triángulos son notables (K, 2K y K raíz de 5, por tanto: AC = 5 raíz de 5, CB = 3 raíz de 5 AB = 8 raíz de 5 Saludos
@ 9:12 X²-3=0 ...🤔 Irrational numbers bore me half to death. They just don't have any quality. Take π for example, ...now that number has brilliance! 😊
Thanks . I got the exact result 13.8, but the length of a in my way is 9. (5+3=8-16=8/2=4 so a =9). Tri.pac= 25+x^2 =81 x=7.48 and tri. Qbc y^2+9=49==y=6.32, so 7.48+6.32=13.8. Isn't it so strange?
MY RESOLUTION PROPOSAL : 01) AP = 5 lin un 02) BQ = 3 lin un 03) 5 + 3 = 8 ; R = 5 :: 3 or R = 3 :: 5 03) We must keep this Linear Proportional Ratio in every Linear Measures, between AP and BQ. 04) Dividing PQ = 16, in 8 Equal Parts we get : (16/8) = 2 05) CP = 5 * 2 = 10 lin 06) CQ = 3 * 2 = 6 lin un 07) Checking : 10 + 6 = 16 : Fact Checked!! 08) BC^2 = QC^2 - 9 ; BC^2 = 36 - 9 ; BC^2 = 27 ; BC = sqrt(27) lin un ; BC = [3*sqrt(3)] lin un 09) AP^2 = PC^2 - 25 ; AP^2 = 100 - 25 ; AP^2 = 75 ; AP = sqrt(75) lin un ; AP = [5*sqrt(3)] lin un 10) AB = [3*sqrt(3) + 5*sqrt(3)] lin un 11) AB = 8*sqrt(3) lin un Therefore, MY BEST ANSWER IS : Line AB Length is equal to approx. 13,86 Linear Units. Exact Form = (8*sqrt(3)) Linear Units.
Just move AB 3 points where B meets Q, and you have a right triangle 16,8,AB
@yenerkorkmaz:
Very nice solution! Thank you so much!❤
А ведь и правда. Про прямой угол я и забыл.
Yes, this is a much simpler solution, thinking outside the box!
Actually that's exactly what I was thinking, since I was looking up lines tangent on 2 circles recently and found their construction methods, AB is an internal tangent of the 2 circles, which makes it equal to a segment tangent on a circle of center P and radius R1+R2=8, from point Q
Exactly my approach! Simple and to the point.
Solution:
If you move the line AB parallel upwards so that B coincides with Q, you get a right-angled triangle with the hypotenuse PQ = 16 and the leg 5+3 = 8, so that you can use the Pythagorean theorem to calculate the other leg AB:
AB = √(16²-8²) = √192 = √(64*3) = 8*√3 = 13.8564
AB²(TCT)= PQ²-(PA+BQ)²
i.e AB²=16²-(5+3)²
=256-64
AB² =192
Or AB =√192
=8√3 ans.
Sir,
I am astonished. You did not use your favourite
30-60-90 special triangle theory in working out the length of
AC & BC
when hypotenuse and leg are 10 and 5
& 6 and 3.
Thanks a lot.
Nice! PQ = 16 = PC + QC = (16 - k) + k; sin(CAP) = sin(CBQ) = 1 → PCA = QCB = δ →
sin(δ) = 5/(16 - k) = 3/k → k = 6 → sin(δ) = 1/2 → tan(δ) = √3/3 → BC = 3√3 → AC = 5√3 → AB = 8√3
CQ=c---> PQ=5c/3 +c=16---> c=6---> CP=16-6=10 ---> CA=√(10²-5²)=5√3---> CB=(3/5)5√3 =3√3 ---> AB=8√3.
Gracias y saludos
Помню, что по какой-то там теореме APǁBQ, тогда ▲АСР~▲BCQ, пропорцию видим - 3 к 5. Тогда РС=5х, CQ=3х, 5х+3х=8х=16⇔х=2. Отсюда РС=10, CQ=6. BC:=у. По теореме о касательной и секущей 9*3=у²=27⇔y=3√3. Возвращаясь к первоначальной пропорции, видно, что АС в таком случае будет 5√3, общая длина тоже становится очевидной - 8√3.
Another solution is to complete the rectangle ABQD. This makes a right triangle PDQ whose hypotenuse PQ is 16 and whose shorter leg PD is 8. This is a 30-60-90 right triangle, so the other leg DQ=8*sqrt(3), and AB is the opposite side of the rectangle, hence also 8*sqrt(3).
PC=5x QC=3x 8x=16 x=2
PC=10 QC=6
AC=√[10²-5²]=5√3 BC=√[6²-5²]=3√3
AB=AB+BC=8√3
Here I'm ready and prepared for this.
Very great question professor ☺️
Another method, pls pin
Apc and cbq are similar triangles. the ratio b/w PC and cq is 5:3 , and pc+cq=16, so pc=10cm and cq=6cm, if we consider triangle PAC we get AC=5×root 3 and in triangle cbq we get cq=3root3 so the length AB=8 root 3
Pytagorean theorem:
d² = 16² - (5+3)²
d = 8√3 = 13,856 cm ( Solved √ )
Extend QB to QR such that RP is perpendicular to AP.
∆QRP is a Right triangle with QR=3+5=8 and hypotenuse PQ = 16
PR= AB =8√3
Thank you!
Woohoo! I did it in my head just from the thumbnail!
I figured it was gonna get messy with the sqrt75 but simplified to 5sqrt3 then got 3sqrt3 being that they're proportional, so the total was 8sqrt3.
✨Magic!✨
Premath is a great teacher. 😎
@@michaelgarrow3239 He's wonderful.
Triangles PAC and CBQ are similar (same angles). So PC/PA = QC/QB, or PC/5 = QC/3 = (PC+QC)/(5+3) = PQ/8 = 16/8 = 2, (*)
so we have: PC = 10 and QC = 6.
PAC and QBC are then 30°-60°-90° triangles, so AC = sqrt(3).PA = 5.sqrt(3) and BC = sqrt(3).QB = 3.qqrt(3), and finally AB = AC + BC = 8.sqrt(3).
(*) If a/b = c/d, then a/b = c/d = (a+c)/(b+d), naturally with b, d and b+d non equal to 0.
Boa tarde Mestre
Obrigado pela aula
AB=8√3≈13,856≈13,86
Teorema del coseno 16^2=5^2+(AB^2+3^2)-2*5*√...cos(90+arctg3/AB)...AB^2=256-25-9-30=192
Once you know the two right triangles are similar, the ratio is 5:3, so the line of centers is split into segments of length 5c and 3c, That makes 8c=16, so c=2 and the hypotenuses are 10 and 6. Since the hypotenuse of each triangle is twice a leg, the triangles are 30-60-90 triangles, so the common internal tangent is 5*sqrt(3)+3*sqrt(3), which is 8*sqrt(3).
Let's find the length:
.
..
...
....
.....
Since AB is a tangent to both circles, we know that ∠PAC=∠QBC=90°. Additionally we have ∠ACP=∠BCQ. Therefore, ACP and BCQ are similar triangles and we can conclude:
CP/CQ = AP/BQ = 5/3 ⇒ CP = (5/3)CQ
CP + CQ = PQ
(5/3)CQ + CQ = 16
(5/3)CQ + (3/3)CQ = 16
(8/3)CQ = 16
⇒ CQ = 16*3/8 = 6
⇒ CP = (5/3)CQ = (5/3)*6 = 10
By applying the Pythagorean theorem to the right triangles ACP and BCQ we obtain:
AP² + AC² = CP² ⇒ AC² = CP² − AP² = 10² − 5² = 100 − 25 = 75 ⇒ AC = √75 = 5√3
BQ² + BC² = CQ² ⇒ BC² = CQ² − BQ² = 6² − 3² = 36 − 9 = 27 ⇒ BC = √27 = 3√3
Now we are able to calculate the length of AB:
AB = AC + BC = 5√3 + 3√3 = 8√3
Best regards from Germany
We are looking at two similar right triangles whose linear ratio = 5 : 3
The sum of their hypotenuses = 16, so their respective hypotenuse = 16(5/8) & 16(3/8) = 10 & 6
We can then calculate AB = √(10^2 - 5^2) + √(6^2 - 3^2) = √75 + √27 = 8√3
I realized 5/8 of 16 would be part of one triangle and 3/8 part of the other one. Then you got 5-10 and 3-6 which are 30-60-90 numbers.
As AB is tangent to circle P at A and circle Q at B, then ∠CAP = ∠CBQ = 90°. As ∠PCA and ∠QCB are vertical angles and thus equal, then ∆CAP and ∆CBQ are similar triangles.
QC/QB = PC/PA
QC/3 = PC/5
3PC = 5QC = 5(16-PC)
3PC = 80 - 5PC
8PC = 80
PC = 80/8 = 10
CA² + PA² = PC²
CA² + 5² = 10²
CA² = 100 - 25 = 75
CA = √75 = 5√3
CB/QB = CA/PA
CB/3 = 5√3/5 = √3
CB = 3√3
AB = CA+CB = 5√3 + 3√3
[ AB = 8√3 ≈ 13.86 units ]
(AB) (AB) = 16*16 - (5 + 3) (5 + 3)
(AB) (AB) = 256 - 64 = 192
AB = √192
AB = √(64 * 3) = 8√3
Shift AB upward towards the center of small circle. You will end up having right triangle with 8 unit length, AB and hyp=16 unit length. It is 30,60,90 triangle with AB= 8 sqrt(3)
DONE
👍
Length of direct conmon tangent = sqrt { PQ^2 - ( difference of radii)^2 }
Length of transverse common tangent = sqrt { PQ^2 - ( sum of radii)^2 }
where PQ = distance between centres of circles.
From point Q, parallel to AB, to the intersection with the continuation of PA.
AB= √(16^2-8^2)= 8√3
From the original image, you didn't say that A and B are tangential points, you just added then that AB is tangent to both circles. From the screenshot image, why would the 2 circles need to be lined up like that? By not stating that AB is a tangent line, it could have several solutions. It is only when you say that it's a tangent... I mean, from the original image, you can just move A and B anywhere, and you can have circles of radius 5 and 3, PQ equal 16, and AB has many values, but in those cases AB is not a tangent line.
1/The two triangles PAC and QBC are similar so,
CP/AP=CQ/QB =(CP+CQ)/(AP+QB)=16/8=2
--> CP= 2AP= 10 and CQ= 2QB=6
--> the two triangles mentioned above are special 30-60-90 ones
-> AC= 5sqrt3 and BC= 3sqrt3
--> AB = 8sqrt3😅😅😅
Nice video sir
8sqrt3
Good question sir
I am happy that I solved it 😊😊😊
In which class you are
@zupitoxyt 9th
@@jtechnicalgaming7831 me to. , your school!
PAC~CBQ===>
5/3=PC/(16-PC)===>
PC=10
Ambos triángulos son notables (K, 2K y K raíz de 5, por tanto:
AC = 5 raíz de 5,
CB = 3 raíz de 5
AB = 8 raíz de 5
Saludos
@ 9:12 X²-3=0 ...🤔 Irrational numbers bore me half to death. They just don't have any quality. Take π for example, ...now that number has brilliance! 😊
Thanks . I got the exact result 13.8, but the length of a in my way is 9. (5+3=8-16=8/2=4 so a =9). Tri.pac= 25+x^2 =81 x=7.48 and tri. Qbc y^2+9=49==y=6.32, so 7.48+6.32=13.8. Isn't it so strange?
Solution:
AB = AC + BC ... ¹
So, we must find AC and BC length
Since ∆ ACP and ∆ BCQ are similar by AA, we have proportions
Hence:
CP/AP = CQ/BQ
CP/CQ = AP/BQ
CP/CQ = 5/3
CP = 5k
CQ = 3k
PQ = 16
PQ = CP + CQ
16 = 5k + 3k
8k = 16
k = 2
Applying Pythagorean Theorem in ∆ ACP:
AP² + AC² = CP²
5² + AC² = (5k)²
25 + AC² = 25k²
AC² = 25k² - 25
AC² = 25 (2)² - 25
AC² = 75
AC = 5√3 ... ²
Once again, applying Pythagorean Theorem, now in ∆ BCQ:
BQ² + BC² = CQ²
3² + BC² = (3k)²
9 + BC² = 9k²
BC² = 9k² - 9
BC² = 9 (2)² - 9
BC² = 27
BC = 3√3 ... ³
Final Step
Substituting ² and ³ in Equation ¹, we have:
AB = 5√3 + 3√3
AB = 8√3 Units ✅
AB ≈ 13.8564 Units ✅
Is that a Bicycle?
MY RESOLUTION PROPOSAL :
01) AP = 5 lin un
02) BQ = 3 lin un
03) 5 + 3 = 8 ; R = 5 :: 3 or R = 3 :: 5
03) We must keep this Linear Proportional Ratio in every Linear Measures, between AP and BQ.
04) Dividing PQ = 16, in 8 Equal Parts we get : (16/8) = 2
05) CP = 5 * 2 = 10 lin
06) CQ = 3 * 2 = 6 lin un
07) Checking : 10 + 6 = 16 : Fact Checked!!
08) BC^2 = QC^2 - 9 ; BC^2 = 36 - 9 ; BC^2 = 27 ; BC = sqrt(27) lin un ; BC = [3*sqrt(3)] lin un
09) AP^2 = PC^2 - 25 ; AP^2 = 100 - 25 ; AP^2 = 75 ; AP = sqrt(75) lin un ; AP = [5*sqrt(3)] lin un
10) AB = [3*sqrt(3) + 5*sqrt(3)] lin un
11) AB = 8*sqrt(3) lin un
Therefore,
MY BEST ANSWER IS :
Line AB Length is equal to approx. 13,86 Linear Units. Exact Form = (8*sqrt(3)) Linear Units.