Hello ! Here is my personal method : (i never remember your " tangent secant theorem ") I use your notations : points A,B,C,D,E,F ; o=angle (FAE) ; r is the radius of the circle Let's call O the center of the circle, G the mid-point of [BD] and H the point where the circle is tangent on [EF]. We can notice that (GO) is the mediator of chord segment [BD]. AO is the hypothenuse of 2 right-angled triangles : AGO and ACO AO^2=AG^2+GO^2=AC^2+CO^2 AG=AB+BG=AB+BD/2=6*sqrt(3)+9/2 BGO is a right-angled triangle then : GO^2=BO^2-BG^2=r^2-(9/2)^2 AC=AF-CF=AE*cos(o)-r=(12+6*sqrt(3))*cos(o)-r CO=r Finally, we have our first equation: AG^2+GO^2=AC^2+CO^2 (6*sqrt(3)+9/2)^2+r^2-(9/2)^2=((12+6*sqrt(3))*cos(o)-r)^2+r^2 (6*sqrt(3)+9/2)^2-(9/2)^2=((12+6*sqrt(3))*cos(o)-r)^2 108+54*sqrt(3)=((12+6*sqrt(3))*cos(o)-r)^2 (9+3*sqrt(3))^2=((12+6*sqrt(3))*cos(o)-r)^2 9+3*sqrt(3)=(12+6*sqrt(3))*cos(o)-r EO is the hypothenuse of 2 right-angled triangles : EGO and EHO EO^2=EG^2+GO^2=EH^2+HO^2 EG=ED+DG=ED+BD/2=3+9/2 DGO is a right-angled triangle then : GO^2=DO^2-DG^2=r^2-(9/2)^2 EH=EF-HF=AE*sin(o)-r=(12+6*sqrt(3))*sin(o)-r HO=r Finally, we have our second equation: EG^2+GO^2=EH^2+HO^2 (3+9/2)^2+r^2-(9/2)^2=((12+6*sqrt(3))*sin(o)-r)^2+r^2 (3+9/2)^2-(9/2)^2=((12+6*sqrt(3))*sin(o)-r)^2 9+27=((12+6*sqrt(3))*sin(o)-r)^2 6=(12+6*sqrt(3))*sin(o)-r Here is our system of 2 equations with 2 unknowns : radius r and angle o : 9+3*sqrt(3)=(12+6*sqrt(3))*cos(o)-r 6=(12+6*sqrt(3))*sin(o)-r 9+3*sqrt(3)-6=(12+6*sqrt(3))*(cos(o)-sin(o)) (sqrt(3)-1)/2=cos(o)-sin(o) ((sqrt(3)-1)/2)^2=(cos(o)-sin(o))^2 1-sqrt(3)/2=1-sin(2*o) sin(2*o)=sqrt(3)/2 Because 0
Ok friend, thanks for the comment. Please I don't know why exactly you said it's difficult to write, I mean I want to make it easier, but I need you to please explain what's making it difficult for you to write, I am 100% willing to improve the videos so that it becomes simpler. Like I always say I make the videos for you, waiting to hear from you, thanks
Hypotenuse of right triangle: c = 6√3+9+3 = 22,3923 km Tangent secant theorem : a²=3*(3+9) --> a=6 km b²=6√3*(6√3+9) --> b=14,196km Pytagorean theorem: (a+r)²+(b+r)² = c² (a²+2ar+r²)+(b²+2br+r²)=c² 2r²+(2a+2b)r+(a²+b²)=c² r² + (a+b)r+½(a²+b²-c²) = 0 r² + 20,196 r - 131,9443 = 0 r = 5,196244 = 3√3 km Area of circle: A = πr² = 27π km² ( Solved √)
Ok please I don't understand what you mean by "his instruction was this video" do you mean he instructed you to watch this video, or he had solve a Question similar to the one in the video, Which do you mean please?
Here is a second personal method with the same notations than my previous comment : I use again @MathandEngineering notations : points A,B,C,D,E,F ; o=angle (FAE) ; r is the radius of the circle Let's call O the center of the circle, G the mid-point of [BD] and H the point where the circle is tangent on [EF]. We can notice that (GO) is the mediator of chord segment [BD]. angle (GOC) = 180° - o ( because AGOC is a quadrilateral with angles : o, 90°, 180°-o , 90° ) Then : AG*sin(o)=OC+OG*cos(o) (AB+BD/2)*sin(o)=OC+sqrt(OB^2-BG^2)*cos(o) (6*sqrt(3)+9/2)*sin(o)=r+sqrt(r^2-(9/2)^2)*cos(o) angle (GOH) = 90° + o ( because EGOH is a quadrilateral with angles : 90°-o , 90°, 90°+o , 90° ) Then : EG*cos(o)=OH+OG*sin(o) (ED+BD/2)*cos(o)=OH+sqrt(OB^2-BG^2)*sin(o) (3+9/2)*cos(o)=r+sqrt(r^2-(9/2)^2)*sin(o) Here is our system of 2 equations with 2 unknowns : radius r and angle o : (6*sqrt(3)+9/2)*sin(o)=r+sqrt(r^2-(9/2)^2)*cos(o) (3+9/2)*cos(o)=r+sqrt(r^2-(9/2)^2)*sin(o) Compared to my first personal method, this method seems not to be a good one. For me, this system of equations seems hard to solve. I find : (y-3*sqrt(3)/2)*P(y)=0 P(y) is a polynomial expressed with variable y=OG=sqrt(r^2-(9/2)^2) For any r>0, we have y=OG>0 and P(y)>0 because, for information : P(y)=8*y^3+(108*sqrt(3)+192)*y^2+(2826+1440*sqrt(3))*y+(10368+6183*sqrt(3)) Then : y-3*sqrt(3)/2=0 y=sqrt(r^2-(9/2)^2)=3*sqrt(3)/2=sqrt(27/4) r^2-(9/2)^2=27/4 r^2=27 A(circle)=27*Pi km²
c = 9 km (given data) Tangent secant theorem : a²=3*(3+c) --> a=6 km Similarity of triangles: a / 2R = R/ c R² = ½a*c = ½*6*9 = 27 km² A = πR² = 27π km² ( Solved √ )
Very nice! And without tangent secant , setting H the point of tangency with EF considering similarity between EBH and BDH (right triangle inside the circle) you can write: (3+9): 2r=2r:9 4r^2= 9*12 R^2=27 The problem is that we don’t know if B is at the same level of H….😢
@@soli9mana-soli4953 Voglio dire, tu sei brillante e meravigliosa !! Come hai detto tu, non sai che B è allo stesso livello di H, questa è solo la conseguenza dell'analisi, non sono dati di input. I miei triangoli simili non sono BDH e BEH. I miei triangoli rettangoli sono BDD' e B'EH, dove DD' è il segmento DOD' del diametro del cerchio , e anche il segmento orizzontale B'H lo è. Il punto B' non corrisponde necessariamente con il punto B.
Hello !
Here is my personal method : (i never remember your " tangent secant theorem ")
I use your notations : points A,B,C,D,E,F ; o=angle (FAE) ; r is the radius of the circle
Let's call O the center of the circle, G the mid-point of [BD] and H the point where the circle is tangent on [EF].
We can notice that (GO) is the mediator of chord segment [BD].
AO is the hypothenuse of 2 right-angled triangles : AGO and ACO
AO^2=AG^2+GO^2=AC^2+CO^2
AG=AB+BG=AB+BD/2=6*sqrt(3)+9/2
BGO is a right-angled triangle then : GO^2=BO^2-BG^2=r^2-(9/2)^2
AC=AF-CF=AE*cos(o)-r=(12+6*sqrt(3))*cos(o)-r
CO=r
Finally, we have our first equation:
AG^2+GO^2=AC^2+CO^2
(6*sqrt(3)+9/2)^2+r^2-(9/2)^2=((12+6*sqrt(3))*cos(o)-r)^2+r^2
(6*sqrt(3)+9/2)^2-(9/2)^2=((12+6*sqrt(3))*cos(o)-r)^2
108+54*sqrt(3)=((12+6*sqrt(3))*cos(o)-r)^2
(9+3*sqrt(3))^2=((12+6*sqrt(3))*cos(o)-r)^2
9+3*sqrt(3)=(12+6*sqrt(3))*cos(o)-r
EO is the hypothenuse of 2 right-angled triangles : EGO and EHO
EO^2=EG^2+GO^2=EH^2+HO^2
EG=ED+DG=ED+BD/2=3+9/2
DGO is a right-angled triangle then : GO^2=DO^2-DG^2=r^2-(9/2)^2
EH=EF-HF=AE*sin(o)-r=(12+6*sqrt(3))*sin(o)-r
HO=r
Finally, we have our second equation:
EG^2+GO^2=EH^2+HO^2
(3+9/2)^2+r^2-(9/2)^2=((12+6*sqrt(3))*sin(o)-r)^2+r^2
(3+9/2)^2-(9/2)^2=((12+6*sqrt(3))*sin(o)-r)^2
9+27=((12+6*sqrt(3))*sin(o)-r)^2
6=(12+6*sqrt(3))*sin(o)-r
Here is our system of 2 equations with 2 unknowns : radius r and angle o :
9+3*sqrt(3)=(12+6*sqrt(3))*cos(o)-r
6=(12+6*sqrt(3))*sin(o)-r
9+3*sqrt(3)-6=(12+6*sqrt(3))*(cos(o)-sin(o))
(sqrt(3)-1)/2=cos(o)-sin(o)
((sqrt(3)-1)/2)^2=(cos(o)-sin(o))^2
1-sqrt(3)/2=1-sin(2*o)
sin(2*o)=sqrt(3)/2
Because 0
Your methods are perfect friend, I can imagine how interesting you find the Question, you shared two perfect methods, thanks. Best regards
Presentation is attractive but it is very difficult to write while viewing.
Ok friend, thanks for the comment.
Please I don't know why exactly you said it's difficult to write, I mean I want to make it easier, but I need you to please explain what's making it difficult for you to write, I am 100% willing to improve the videos so that it becomes simpler. Like I always say I make the videos for you, waiting to hear from you, thanks
Hypotenuse of right triangle:
c = 6√3+9+3 = 22,3923 km
Tangent secant theorem :
a²=3*(3+9) --> a=6 km
b²=6√3*(6√3+9) --> b=14,196km
Pytagorean theorem:
(a+r)²+(b+r)² = c²
(a²+2ar+r²)+(b²+2br+r²)=c²
2r²+(2a+2b)r+(a²+b²)=c²
r² + (a+b)r+½(a²+b²-c²) = 0
r² + 20,196 r - 131,9443 = 0
r = 5,196244 = 3√3 km
Area of circle:
A = πr² = 27π km² ( Solved √)
This is why I'm not able too understand , Math had a teacher Raj his instruction was this video !
Ok please I don't understand what you mean by "his instruction was this video" do you mean he instructed you to watch this video, or he had solve a Question similar to the one in the video,
Which do you mean please?
Here is a second personal method with the same notations than my previous comment :
I use again @MathandEngineering notations : points A,B,C,D,E,F ; o=angle (FAE) ; r is the radius of the circle
Let's call O the center of the circle, G the mid-point of [BD] and H the point where the circle is tangent on [EF].
We can notice that (GO) is the mediator of chord segment [BD].
angle (GOC) = 180° - o ( because AGOC is a quadrilateral with angles : o, 90°, 180°-o , 90° )
Then :
AG*sin(o)=OC+OG*cos(o)
(AB+BD/2)*sin(o)=OC+sqrt(OB^2-BG^2)*cos(o)
(6*sqrt(3)+9/2)*sin(o)=r+sqrt(r^2-(9/2)^2)*cos(o)
angle (GOH) = 90° + o ( because EGOH is a quadrilateral with angles : 90°-o , 90°, 90°+o , 90° )
Then :
EG*cos(o)=OH+OG*sin(o)
(ED+BD/2)*cos(o)=OH+sqrt(OB^2-BG^2)*sin(o)
(3+9/2)*cos(o)=r+sqrt(r^2-(9/2)^2)*sin(o)
Here is our system of 2 equations with 2 unknowns : radius r and angle o :
(6*sqrt(3)+9/2)*sin(o)=r+sqrt(r^2-(9/2)^2)*cos(o)
(3+9/2)*cos(o)=r+sqrt(r^2-(9/2)^2)*sin(o)
Compared to my first personal method, this method seems not to be a good one.
For me, this system of equations seems hard to solve.
I find :
(y-3*sqrt(3)/2)*P(y)=0
P(y) is a polynomial expressed with variable y=OG=sqrt(r^2-(9/2)^2)
For any r>0, we have y=OG>0 and P(y)>0 because, for information :
P(y)=8*y^3+(108*sqrt(3)+192)*y^2+(2826+1440*sqrt(3))*y+(10368+6183*sqrt(3))
Then :
y-3*sqrt(3)/2=0
y=sqrt(r^2-(9/2)^2)=3*sqrt(3)/2=sqrt(27/4)
r^2-(9/2)^2=27/4
r^2=27
A(circle)=27*Pi km²
c = 9 km (given data)
Tangent secant theorem :
a²=3*(3+c) --> a=6 km
Similarity of triangles:
a / 2R = R/ c
R² = ½a*c = ½*6*9 = 27 km²
A = πR² = 27π km² ( Solved √ )
Very nice! And without tangent secant , setting H the point of tangency with EF considering similarity between EBH and BDH (right triangle inside the circle) you can write:
(3+9): 2r=2r:9
4r^2= 9*12
R^2=27
The problem is that we don’t know if B is at the same level of H….😢
@@soli9mana-soli4953
Voglio dire, tu sei brillante e meravigliosa !!
Come hai detto tu, non sai che B è allo stesso livello di H, questa è solo la conseguenza dell'analisi, non sono dati di input.
I miei triangoli simili non sono BDH e BEH.
I miei triangoli rettangoli sono BDD' e B'EH, dove DD' è il segmento DOD' del diametro del cerchio , e anche il segmento orizzontale B'H lo è.
Il punto B' non corrisponde necessariamente con il punto B.