4 Sides CO,OD,DE,CE of Inscribable Quadrilateral CODE can be calculated.Also diagonals CD=4 and OE=R. We apply Ptolemy's Theorem saying ''The Product of diagonals is equal to the Sum of Products of the Opposite Sides''. CD•OE=CO•ED+CE•OD, so 4•R=(2√2)•(2)+(2√3)•(2√2) so 4•R=4•√2+4•√2•√3 so 4•R=4•(√2+√2•√3) Divide by 4 and R=√2+√2•√3=√2+√6. CALCULATION OF 4 Sides of Inscribable CODE,since sum of angles O+E=90+90=180 Degrees .1st DE is opposite of 30 degrees in right triangle. So is equal to the half of hypotenese CD=4 so DE=2. 2nd CE^2=CD^2-ED^2=4^2-2^2=16-4=12 so CE=√12=√4•√3=2•√3. 3rd and 4th CO^2+OD^2=CD^2 so x^2+x^2=4^2 so 2•x^2=16 so x^2=8 so x=√8=√4•√2=2•√2 hence CO=OD=2•√2.
Another scheme avoiding trig. Construct line from O to CD perpendicular to CD (point F). Do the same from E to CD (point G). OF = FD = Dc = 2. DG = 1 GE = sqrt(3). Path from O to E can be traced from O to F (2 long), F to G (1 long), & G to E (sqrt(3) long). These lines can be rearranged into right triangle, long leg 2+sqrt(3), short leg 1, hypotenuse R. From this triangle you get R^2 = 8 + 4 sqrt(3).
The center(F) of circle including C, O, D, E is on half of line CD. So, CF = FD = FO = FE = 2 triangle OFE has 2, 150 deg, 2 So, OE = sq[4 + 4 - 8×sq(3)/2 = sq6 + sq2
CD=4》DE=2》EC=2(sqrt3) 》》OE =r =Radio del círculo 》Mediante tres giros (de 90° y en el mismo sentido) de la figura propuesta obtenemos una circunferencia de radio r, con un cuadrado de lado (DE+EC) cuya diagonal es 2r 》(2+2sqrt3)(sqrt2)/2 =r=(sqrt2)(1+sqrt3) =3.8637 Gracias y un saludo cordial.
Difficult to prove AFC=OED (using properties of inscribable quadrilateral COED). Proving that the the extension DK of ED is equal to 2+2*sqrt3 is very difficult ( must prove that
There are two triangles inside the quarter circle. One is 45-45-90 and other is 30-60-90 triangle. The side length of 4 is common for both the triangles. For 45-45-90 triangle the side lenths are 2√2, 2√2, 4 and side lengths of 30-60-90 triangle are 2, 2√3, 4. Both the triangles taken together forms a quadrilateral which is cyclic bcoz sum of opposite angles are 180°. Side lengths of this cyclic quadrilateral are 2√2, 2√2, 2, 2√3 and one diagonal is 4. Now by using Ptolemy's theorem other diagonal can be calculated as √2+√6 which is also the radius of the given quarter circle.
मैंने इसे वैसे ही पहना है जैसे आपने किया था In 《CDE》= sin(30)=½=DE/CD ½=DE/4 DE=4/2=2 From Phythagorean thereom Radius CE=2*sqrt3 CO=2sqrt2 OD=2sqr2 Cos(105)=12-R²/8sqrt2 R=sqrt(8+4sqrt3)
Why Ptolemy theorem ???? It is a square !!! , and the square is inscribed in the circunference !!! Ptolemy theorem is not necessary Square: Side : Leg1 + Leg2 (of right triangle) Area = Side² = ½ Diameter² Area = ½ (2R)²= Here you have the Radius !!
@@marioalb9726 The big difficulty is to prove that if we extend EC and ED until the completed circle this is square.Its is rectangle and then what? I sent before 5 mininutes a complete proof why FEG is isosceles right triangle,prooving that OFE=45 degrees.You can take a look. Thanks for the comment.
@@sarantis40kalaitzis48 It is not a rectangle because the right triangle is "right" and its right vertex is on the cincunference, and the other 2 vertices are on the axis. Just could be a square and centered with the center of the circle !!! However, you were very close, and much better than the extremely complicated solution we see in this video Thanks
Take the quarter circle and copy it, rotate it 90 degrees ccw and connect it to the first quarter circle. The sides of the triangles connect at the same angle. Do that until you have a full circle in which a square consisting of 4 copies of the right triangle is made. Side of circle is 2+2(rt(3)) Multiply by rt(2) to get diameter or diagonal of square and then divide by 2 to get radius
It's a very good idea.I tried this and got 2•√2+2√2•√3 for diameter,so r=√2+√6, but i couldn't prove that this is square,or isosceles right triangle with the extended chords to be EQUAL Sides. Then i tried Ptolemy's Theorem,which works for every inscribable quadrilateral,since into CODE we have two opposite angles with sum 90+90=180 degrees and solved it immediately. Anyway we can prove that the extension of EC named EF=2 but is difficult comparing FCO and OED triangles.Also more difficult is to prove the other extension is 2+√3.
I forgot to say the formation C,O,D,E to CODE. Code Theory and Game Theory are very important branches of modern Mathematics. Ofcourse Trigonometry is the easiest way to OE=R, applying Rule of Cosines to triangle COE. OE^2=CO^2+CE^2-2•CO• CE•cos(OCE)=8+12-2• (2√2)•(2√3)•cos75. Calculate cos75=cos(45+30)=cos45•cos30-sin45•sin30= (√2/2)•(√3/2)-(√2/2)•(1/2) =(1/4)•√2•(√3-√2)...etc. The only difficullty is the same with the video, that we must find the perfect square form of the under square root answer.
Right triangle: Leg1 = 4.cos30° = 3,464 cm Leg2 = 4 sin30° = 2 cm Square inscribed in circle: Side= Leg1 + Leg2 S = 5,464 cm Area = S² = (2R)²/2 4R² = 2 Area R² = Area/2 = S²/2 R = 3,864 cm ( Solved √ )
4 Sides CO,OD,DE,CE of Inscribable Quadrilateral CODE can be calculated.Also diagonals CD=4 and OE=R. We apply Ptolemy's Theorem saying ''The Product of diagonals is equal to the Sum of Products of the Opposite Sides''. CD•OE=CO•ED+CE•OD, so 4•R=(2√2)•(2)+(2√3)•(2√2) so 4•R=4•√2+4•√2•√3 so 4•R=4•(√2+√2•√3) Divide by 4 and R=√2+√2•√3=√2+√6. CALCULATION OF 4 Sides of Inscribable CODE,since sum of angles O+E=90+90=180 Degrees .1st DE is opposite of 30 degrees in right triangle. So is equal to the half of hypotenese CD=4 so DE=2. 2nd CE^2=CD^2-ED^2=4^2-2^2=16-4=12 so CE=√12=√4•√3=2•√3. 3rd and 4th CO^2+OD^2=CD^2 so x^2+x^2=4^2 so 2•x^2=16 so x^2=8 so x=√8=√4•√2=2•√2 hence CO=OD=2•√2.
Another scheme avoiding trig. Construct line from O to CD perpendicular to CD (point F). Do the same from E to CD (point G). OF = FD = Dc = 2. DG = 1 GE = sqrt(3). Path from O to E can be traced from O to F (2 long), F to G (1 long), & G to E (sqrt(3) long). These lines can be rearranged into right triangle, long leg 2+sqrt(3), short leg 1, hypotenuse R. From this triangle you get R^2 = 8 + 4 sqrt(3).
The center(F) of circle including C, O, D, E is on half of line CD.
So, CF = FD = FO = FE = 2
triangle OFE has 2, 150 deg, 2
So, OE = sq[4 + 4 - 8×sq(3)/2 = sq6 + sq2
CD=4》DE=2》EC=2(sqrt3) 》》OE =r =Radio del círculo 》Mediante tres giros (de 90° y en el mismo sentido) de la figura propuesta obtenemos una circunferencia de radio r, con un cuadrado de lado (DE+EC) cuya diagonal es 2r 》(2+2sqrt3)(sqrt2)/2 =r=(sqrt2)(1+sqrt3) =3.8637
Gracias y un saludo cordial.
Another Geometrical aproach is to extend EC and ED intersecting the completed circle into points F and G.Then the inscribed angle
作△OFC全等△OED(利用旋轉即可,以O為定點將△OED逆時針轉90度,OD與OC重疊,E對應到F
則△OFE是等腰直角△
因為OF=OE=R且
FE=FC+CE=2+2sqrt3
(角ODE+角OCE=180度,F,C,E三點共線,角FOE=90度)
R=(2+2sqrt3)/sqrt2=sqrt2+sqrt6
Difficult to prove AFC=OED (using properties of inscribable quadrilateral COED). Proving that the the extension DK of ED is equal to 2+2*sqrt3 is very difficult ( must prove that
There are two triangles inside the quarter circle. One is 45-45-90 and other is 30-60-90 triangle. The side length of 4 is common for both the triangles. For 45-45-90 triangle the side lenths are 2√2, 2√2, 4 and side lengths of 30-60-90 triangle are 2, 2√3, 4. Both the triangles taken together forms a quadrilateral which is cyclic bcoz sum of opposite angles are 180°. Side lengths of this cyclic quadrilateral are 2√2, 2√2, 2, 2√3 and one diagonal is 4. Now by using Ptolemy's theorem other diagonal can be calculated as √2+√6 which is also the radius of the given quarter circle.
Note that
मैंने इसे वैसे ही पहना है जैसे आपने किया था
In 《CDE》= sin(30)=½=DE/CD
½=DE/4
DE=4/2=2
From Phythagorean thereom Radius
CE=2*sqrt3
CO=2sqrt2
OD=2sqr2
Cos(105)=12-R²/8sqrt2
R=sqrt(8+4sqrt3)
May I ask you what is your favourite number and why?
Right triangle:
Leg1 = 4.cos30° = 3,464 cm
Leg2 = 4 sin30° = 2 cm
Isosceles Right triangle:
Leg = 4 cos45° = 2√2 cm
Chord theorem:
(R + 2√2).(R-2√2) = 3.464 . 2
R² - 8 = 6.928
R² = 14,928
R = 3,864 cm ( Solved √ )
We cn also use plotemy theorem
I sent this solution before 18 hours.
@@sarantis40kalaitzis48 nice
Why Ptolemy theorem ????
It is a square !!! , and
the square is inscribed in the circunference !!!
Ptolemy theorem is not necessary
Square:
Side : Leg1 + Leg2 (of right triangle)
Area = Side² = ½ Diameter²
Area = ½ (2R)²=
Here you have the Radius !!
@@marioalb9726 The big difficulty is to prove that if we extend EC and ED until the completed circle this is square.Its is rectangle and then what? I sent before 5 mininutes a complete proof why FEG is isosceles right triangle,prooving that OFE=45 degrees.You can take a look. Thanks for the comment.
@@sarantis40kalaitzis48
It is not a rectangle because the right triangle is "right" and its right vertex is on the cincunference, and the other 2 vertices are on the axis.
Just could be a square and centered with the center of the circle !!!
However, you were very close, and much better than the extremely complicated solution we see in this video
Thanks
Take the quarter circle and copy it, rotate it 90 degrees ccw and connect it to the first quarter circle. The sides of the triangles connect at the same angle. Do that until you have a full circle in which a square consisting of 4 copies of the right triangle is made.
Side of circle is 2+2(rt(3))
Multiply by rt(2) to get diameter or diagonal of square and then divide by 2 to get radius
It's a very good idea.I tried this and got 2•√2+2√2•√3 for diameter,so r=√2+√6, but i couldn't prove that this is square,or isosceles right triangle with the extended chords to be EQUAL Sides. Then i tried Ptolemy's Theorem,which works for every inscribable quadrilateral,since into CODE we have two opposite angles with sum 90+90=180 degrees and solved it immediately. Anyway we can prove that the extension of EC named EF=2 but is difficult comparing FCO and OED triangles.Also more difficult is to prove the other extension is 2+√3.
I forgot to say the formation C,O,D,E to CODE. Code Theory and Game Theory are very important branches of modern Mathematics. Ofcourse Trigonometry is the easiest way to OE=R, applying Rule of Cosines to triangle COE. OE^2=CO^2+CE^2-2•CO• CE•cos(OCE)=8+12-2• (2√2)•(2√3)•cos75. Calculate cos75=cos(45+30)=cos45•cos30-sin45•sin30= (√2/2)•(√3/2)-(√2/2)•(1/2) =(1/4)•√2•(√3-√2)...etc. The only difficullty is the same with the video, that we must find the perfect square form of the under square root answer.
شكرآ
DE COUPE AC EN F
ON UTILISE LE TRIANGLE CEF
Täydennetään täydeksi ympyräksi, nähdään heti r=(2+sqrt3)/sqrt2 =sqrt2+sqrt6
you're too slow but that's ok
OB😂
Right triangle:
Leg1 = 4.cos30° = 3,464 cm
Leg2 = 4 sin30° = 2 cm
Square inscribed in circle:
Side= Leg1 + Leg2
S = 5,464 cm
Area = S² = (2R)²/2
4R² = 2 Area
R² = Area/2 = S²/2
R = 3,864 cm ( Solved √ )