That would be great if he can teach kids at a lower level to prepare them for college level math but Mr Pre Math is a professor at a university. Isn’t that right, Mr Pre Math?
Call G the opposite of C along the diameter in the semicircle, then I noticed a similitude between GEC and DEC triangles (because they are rectangles and have an angle in common) GC : EC = EC : DC 2r : 12sqrt(3) = 12sqrt(3) : DC and we get DC = 216/r Area = r*216/r = 216
you can construct a perpendicular to the chord of EC and OC would be length of the rectangle since its the radius of the circle and CD would be breadth and then use similarities We get, L/6√3=12√3/B LB=6√3*12√3=216 (I have taken BC as length and AB as breadth)
Interestingly, point E can be moved to coincide with point F and we then have a square FBCO with the diagonal of 12 root 3. The area of that square is given by the formula 1/2 x diagonal squared. 1/2 x (12 root 3 )^2 = 216.
@ EC is a given dimension which remains constant, but the radius of the circle is not constant and will increase in value as point E moves towards point F.
Draw a perpendicular OG to EC. Join OE. By calculations of angles. OD = OG. So DE = 6*sqrt3. Angle OCE = 30. So DC = 18, OC = 12. So Area ABCD = 12*18 = 216
The area of the rectangle is(1/2)*semicircle chord square. We can solve it by shortly if CF is the radius of your drawn figure of semicircle, then by tangent triangle are ∆ EDC & ∆ FEC are congruent triangles, so we have EC/FC=DC/EC, EC^2=FC*DC, 12√3 * 12√3= 2r * (x+r), 432= 2r * (x+r), 216= r * (x+r) = area of rectangle
I dare say there is some room left for optimization: if we leave aside EA and use only DE for our workout we will have a slight simplified process: 1) Let h= DE; 2) r^2= = h^2+x^2 => h^2= r^2-x^2; 3) (12*sqrt(3))^2= (r+x)^2+ h^2; 4) lets substitute h^2: 432= (r+x)^2+ (r^2-x^2); 432= r^2+x^2+2rx+r^2-x^2; 432=2r^2+2rx AreaBlue = r^2+rx= 432/2=216.
Lets call the other end of the semicircle's diameter point G. Now draw EG. Triangle EGC is a right triangle from angle GEC subtending the diameter GC. Triangle EDC is given as a right triangle. Triangles EDC and GEC share common angle ECD/GCD. Then triangles EDC and GEC are similar triangles. From similar triangles we have GC/CE = CE/ED. CE = 12*sqrt(3) is given. Let the radius be r, then diameter GC equals 2*r and the height of the rectangle is BC=r. CD is the length of the rectangle, call that x. The area of the rectangle can be expressed as CD*BC = r*x. Now just substitute and simplify: GC/CE = CE/ED -> (2*r)/(12*sqrt(3)) = (12*sqrt(3))/x -> r*x = 216. The area of the rectangle is 216.
Let r be the radius of the semicircle. And OD be a. DE^2= r^2 -a^2 By Pythagorean theorem, DE^2 + CD^2= 144•3 r^2 -a^2 +(r+a)^2=144•3 2r^2 + 2ar=144•3 r(r+a)=216 Area of blue rectangle = r(r+a)=216
On peut librement faire varier l'angle DCE. Prenons 45°. OCBF (ou ABCD, c'est le même) est alors un carré de coté c et dont la diagonale est 12.3^0,5 . Ce qui donne : 2c^2=(12.3^0,5)^2 c^2=216
I used a very similar approach, but instead of letting AE = y and DE = r−y, I used DE = y (this makes it slightly simpler when squaring binomials, since we are eliminating binomial r−y So we have OD = x and DE = y. Area of rectangle = BC × CD = r(r+x) (1) Using Pythagorean theorem in △ODE, we get: OD² + DE² = OE² x² + y² = r² (2) Using Pythagorean theorem in △CDE, we get: CD² + DE² = CE² (r+x)² + y² = (12√3)² r² + 2rx + x² + y² = 432 → From (2) we replace x² + y² with r² r² + 2rx + r² = 432 2r² + 2rx = 432 2r (r + x) = 432 r (r + x) = 216 But from (1), the left side of expression above [r(r+x)] is area of rectangle Area of rectangle = 216
Llamamos “G” al extremo izquierdo del diámetro horizontal → ∠GEC=90º por estar inscrito en un semicírculo → La longitud de EC sugiere que se trata de la altura de la mitad de un triángulo equilátero (GEC) → GE=12 → GC=2*12=24 → Radio del semicírculo, r=12=OE=GE → ∆OGE es equilátero → ∠GCE=30º y su ángulo central ∠GOE=60º → La hipótesis inicial es correcta → OD=DG=12/2=6 → Área azul ABCD=(2r-6)r=(2*12 -6)12=18*12=216 Gracias y saludos.
Let G be the unmarked end of the diameter of semicircle O. Draw EG. As G and C are ends of a diameter and E is a point on the circumference, ∠GEC = 90°. As ∠CDE = 90° as well and ∠DCE is common, ∆EDC and ∆GEC are similar triangles. Let OD = x. CD/EC = EC/CG (r+x)/12√3 = 12√3/2r 2r(r+x) = 144(3) = 432 r(r+x) = 216 As ABCD is a rectangle and OF, as a radius intersecting a tangent, is perpendicular to AB, the height of ABCD is DA = BC = OF = r, and the width of ABCD is OC+OD = AB = r+x. So the area of ABCD is r(r+x) = 216 units.
Congratulations for your videos. But this one I'd solve using another way. The rectangle width (w) is equal to the radius (r). The rectangle length (l) is a cathetus from CDE triangle. So l^2 + (DE)^2 = (CE)^2=432. DE is the geometric mean of l and (2r-l). So (DE)^2 = l(2r-l). If we use the first equation, we have l^2 + l(2r-l) = 432, so l^2 + 2lr - l^2 = 432. So, 2lr=432. But r=w, so wl = 216. And that's the answer.
Интересно, что к этому времени только три пользователя догадались, что задачу можно решить простым и быстрым способом, основанном на том, что результат не зависит от положения точки E на стороне AD (угол DCE может изменятся от 0 до 45°, соотношение сторон ABCD - от 1:1 до 2:1). Все они рассмотрели один крайний случай: E совпадает с A (угол DCE=45°, ABCD - квадрат). Мне остаётся только рассмотреть другой крайний случай: E совпадает с D (угол DCE=0, ABCD - прямоугольник 2:1). [ABCD]=12sqrt3×6sqrt3=216 Что и требовалось доказать!
Let the point where Diameter CD ends be called P. Let length and breadth of rectangle be l and r respectively. Now ED is perpendicular to CP and Triangle PEC is right angled at P. (Angle inside semicircle) Hence DE^2 = PD * DC (Property of similar triangles) => EC^2 - DC^2 = PD * DC => 432-l^2 = l* (2r-l) Hence rl = 216. (after canceling out l^2)
Another way: Set the point G to have OG symmetric OC. CG is the diameter of the circle. If DG =x , we have ED*2=DG×DC=x(2r_x) ED*2+DC*2=EC*2=(12×3^2)*2 x(2r_x)+(2r_x)*2=144×3 2rx_x*2+4r*2_4rx+x*2=432 4r*2_2rx=432 2r*2_rx =216 r(2r_x) =216 AD×CD=216 Blue area=216
[ rectangle ABCD] = BC x CD = CO x CD = CF x CD /2 Herein F is other end of diameter joining C and O. = CE^2 , ∆CEF is right angular, & CD perpendicular to hypotenuse CF Hereby [ rectangle ABCD] = CE^2/2
c = 12 √3 = 20,785 cm Similarly of triangles: (c/2) / R = (R+x) / c (R+x). R = (c/2). c = Area (R+x). R = c² / 2 (R+x). R = 20,785² / 2 Area = 216 cm² ( Solved √ ) The area of this rectangle with base R+x and height R is equal to: The area of a square with diagonal 12√3 !!!!!!
Fantastic. What a seemingly unsolvable problem and what an amazing solution
Thanks PreMath Guru Ji
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So nice of you, dear
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Wow! Amaze!
So nice of you.
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Wish you were my school teacher. Thank you for the way the problem was solved. Very methodical.
Wow, thanks!
So nice of you, dear 😀
That would be great if he can teach kids at a lower level to prepare them for college level math but Mr Pre Math is a professor at a university. Isn’t that right, Mr Pre Math?
تمرين جميل جيد . رسم واضح مرتب . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا . تحياتنا لكم من غزة فلسطين .
So nice of you, dear
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That one was much juicier. Thanks. Remembering to break it down into identities looks paramount.
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Your approach to these minimal information projects is fascinating!🥂😀❤
Glad you think so.
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Call G the opposite of C along the diameter in the semicircle, then I noticed a similitude between GEC and DEC triangles (because they are rectangles and have an angle in common)
GC : EC = EC : DC
2r : 12sqrt(3) = 12sqrt(3) : DC
and we get
DC = 216/r
Area = r*216/r = 216
This is the simplest way to do it. Nice!
you can construct a perpendicular to the chord of EC and OC would be length of the rectangle since its the radius of the circle and CD would be breadth and then use similarities
We get,
L/6√3=12√3/B
LB=6√3*12√3=216
(I have taken BC as length and AB as breadth)
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What a explanation premath
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Awesome! I think I'm ready now too tackle a Millineum Prize Prize Problem . I need the money!🙂
Good luck!
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Thanks for video.Good luck sir!!!!!!!!!!!!
So nice of you
Interestingly, point E can be moved to coincide with point F and we then have a square FBCO with the diagonal of 12 root 3.
The area of that square is given by the formula 1/2 x diagonal squared.
1/2 x (12 root 3 )^2 = 216.
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If I move E to coincide with F the resulting square doesn't have. diagonal of 12 root 3 thus I don't understand your reasoning
@ Agree. If CE = k → CF = r√2 ≠ k
@ EC is a given dimension which remains constant, but the radius of the circle is not constant and will increase in value as point E moves towards point F.
@@georgebliss964 Ah, clever.
Draw a perpendicular OG to EC. Join OE. By calculations of angles. OD = OG. So DE = 6*sqrt3. Angle OCE = 30.
So DC = 18, OC = 12. So Area ABCD = 12*18 = 216
This particular case is just as remarkable as when ∠OCE = 45°😉
Oh, that's a lovely little trick.
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Awesome
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The area of the rectangle is(1/2)*semicircle chord square. We can solve it by shortly if CF is the radius of your drawn figure of semicircle, then by tangent triangle are ∆ EDC & ∆ FEC are congruent triangles, so we have EC/FC=DC/EC, EC^2=FC*DC, 12√3 * 12√3= 2r * (x+r), 432= 2r * (x+r), 216= r * (x+r) = area of rectangle
I dare say there is some room left for optimization: if we leave aside EA and use only DE for our workout we will have a slight simplified process:
1) Let h= DE;
2) r^2= = h^2+x^2 => h^2= r^2-x^2;
3) (12*sqrt(3))^2= (r+x)^2+ h^2;
4) lets substitute h^2:
432= (r+x)^2+ (r^2-x^2);
432= r^2+x^2+2rx+r^2-x^2;
432=2r^2+2rx
AreaBlue = r^2+rx= 432/2=216.
Thank you! Cheers! 😀
Lets call the other end of the semicircle's diameter point G. Now draw EG.
Triangle EGC is a right triangle from angle GEC subtending the diameter GC.
Triangle EDC is given as a right triangle.
Triangles EDC and GEC share common angle ECD/GCD.
Then triangles EDC and GEC are similar triangles.
From similar triangles we have GC/CE = CE/ED.
CE = 12*sqrt(3) is given.
Let the radius be r, then diameter GC equals 2*r and the height of the rectangle is BC=r.
CD is the length of the rectangle, call that x.
The area of the rectangle can be expressed as CD*BC = r*x.
Now just substitute and simplify: GC/CE = CE/ED -> (2*r)/(12*sqrt(3)) = (12*sqrt(3))/x -> r*x = 216.
The area of the rectangle is 216.
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Let r be the radius of the semicircle. And OD be a.
DE^2= r^2 -a^2
By Pythagorean theorem,
DE^2 + CD^2= 144•3
r^2 -a^2 +(r+a)^2=144•3
2r^2 + 2ar=144•3
r(r+a)=216
Area of blue rectangle = r(r+a)=216
otetaan taas yksinkertaisin ehdot täyttävä tapaus. Piste E siirretään puoliympyrän vasempaan nurkkaan . Ala on 12*sqrt3 * 6*sqrt3.
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Yeah 😊 for the first time I say wow, I didn't see it wow
So nice of you.
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On peut librement faire varier l'angle DCE. Prenons 45°. OCBF (ou ABCD, c'est le même) est alors un carré de coté c et dont la diagonale est 12.3^0,5 . Ce qui donne : 2c^2=(12.3^0,5)^2 c^2=216
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Wow
So nice of you.
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Can you show me how to calculate covariance correlation coefficient please
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Please send an email to:
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Nice problem, square root of 3 was my homing beacon, then a spoonful of Thales and alakazam 🤓
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Amazing.. And very elegant explanations as always. Thank you teacher 🙏.
Glad you liked it!
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I used a very similar approach, but instead of letting AE = y and DE = r−y, I used DE = y (this makes it slightly simpler when squaring binomials, since we are eliminating binomial r−y
So we have OD = x and DE = y.
Area of rectangle = BC × CD = r(r+x) (1)
Using Pythagorean theorem in △ODE, we get:
OD² + DE² = OE²
x² + y² = r² (2)
Using Pythagorean theorem in △CDE, we get:
CD² + DE² = CE²
(r+x)² + y² = (12√3)²
r² + 2rx + x² + y² = 432 → From (2) we replace x² + y² with r²
r² + 2rx + r² = 432
2r² + 2rx = 432
2r (r + x) = 432
r (r + x) = 216
But from (1), the left side of expression above [r(r+x)] is area of rectangle
Area of rectangle = 216
This sum seems to so much difficult
Your solution is excellent.
So nice of you.
Thank you! Cheers! 😀
Llamamos “G” al extremo izquierdo del diámetro horizontal → ∠GEC=90º por estar inscrito en un semicírculo → La longitud de EC sugiere que se trata de la altura de la mitad de un triángulo equilátero (GEC) → GE=12 → GC=2*12=24 → Radio del semicírculo, r=12=OE=GE → ∆OGE es equilátero → ∠GCE=30º y su ángulo central ∠GOE=60º → La hipótesis inicial es correcta → OD=DG=12/2=6 → Área azul ABCD=(2r-6)r=(2*12 -6)12=18*12=216
Gracias y saludos.
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Generalization: the area of the blue rectangle is CE² / 2
Trivia: the area of this blue rectangle is 3³ + 4³ + 5³ = 6³
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Shortcut Method:
Name the diametre as CP
Join EP
∆ CEP~∆CED
CP/CE=CE/CD
CP.CD=CE.CE
2r(r+x)=(12√3)(12√3)
2A=12×12×3
A=12×12×3/2=216 (Answer)
A = DC x OC
12v3/DC=OC/6v3
DC x OC = 3 x 72 = 216
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Let G be the unmarked end of the diameter of semicircle O. Draw EG. As G and C are ends of a diameter and E is a point on the circumference, ∠GEC = 90°. As ∠CDE = 90° as well and ∠DCE is common, ∆EDC and ∆GEC are similar triangles. Let OD = x.
CD/EC = EC/CG
(r+x)/12√3 = 12√3/2r
2r(r+x) = 144(3) = 432
r(r+x) = 216
As ABCD is a rectangle and OF, as a radius intersecting a tangent, is perpendicular to AB, the height of ABCD is DA = BC = OF = r, and the width of ABCD is OC+OD = AB = r+x.
So the area of ABCD is r(r+x) = 216 units.
Congratulations for your videos. But this one I'd solve using another way. The rectangle width (w) is equal to the radius (r). The rectangle length (l) is a cathetus from CDE triangle. So l^2 + (DE)^2 = (CE)^2=432. DE is the geometric mean of l and (2r-l). So (DE)^2 = l(2r-l). If we use the first equation, we have l^2 + l(2r-l) = 432, so l^2 + 2lr - l^2 = 432. So, 2lr=432. But r=w, so wl = 216. And that's the answer.
Thanks for sharing! Cheers! 😀
Интересно, что к этому времени только три пользователя догадались, что задачу можно решить простым и быстрым способом, основанном на том, что результат не зависит от положения точки E на стороне AD (угол DCE может изменятся от 0 до 45°, соотношение сторон ABCD - от 1:1 до 2:1).
Все они рассмотрели один крайний случай: E совпадает с A (угол DCE=45°, ABCD - квадрат).
Мне остаётся только рассмотреть другой крайний случай: E совпадает с D (угол DCE=0, ABCD - прямоугольник 2:1).
[ABCD]=12sqrt3×6sqrt3=216
Что и требовалось доказать!
Amazing. You don't need to calculate the radius of the semi-circle
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Let the point where Diameter CD ends be called P. Let length and breadth of rectangle be l and r respectively.
Now ED is perpendicular to CP
and Triangle PEC is right angled at P. (Angle inside semicircle)
Hence DE^2 = PD * DC (Property of similar triangles)
=> EC^2 - DC^2 = PD * DC
=> 432-l^2 = l* (2r-l)
Hence rl = 216. (after canceling out l^2)
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Another way:
Set the point G to have OG symmetric OC. CG is the diameter of the circle.
If DG =x , we have ED*2=DG×DC=x(2r_x)
ED*2+DC*2=EC*2=(12×3^2)*2
x(2r_x)+(2r_x)*2=144×3
2rx_x*2+4r*2_4rx+x*2=432
4r*2_2rx=432
2r*2_rx =216
r(2r_x) =216
AD×CD=216
Blue area=216
nice
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Fantastic ❤❤
Keep going ❤
If x+1/x=5
Then 2/(12x-7)=?
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[ rectangle ABCD]
= BC x CD = CO x CD = CF x CD /2
Herein F is other end of diameter joining C and O.
= CE^2 , ∆CEF is right angular, & CD perpendicular to hypotenuse CF
Hereby [ rectangle ABCD] = CE^2/2
216m hk bus
ThanksGoogle
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με ομοιοτητες τριγωνων:EC/2r=DC/EC,αρα 2xrxDC=ECXEC,αρα 2xrxDC=144x3=432,αρα 2XA=432,αρα Α=216
👏
c = 12 √3 = 20,785 cm
Similarly of triangles:
(c/2) / R = (R+x) / c
(R+x). R = (c/2). c = Area
(R+x). R = c² / 2
(R+x). R = 20,785² / 2
Area = 216 cm² ( Solved √ )
The area of this rectangle with base R+x and height R
is equal to:
The area of a square with diagonal 12√3 !!!!!!
A=72*3
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يمكن استعمال cosa و cos2a نجد s=216
We don't get other dimensions of this figure
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Ho ottenuto lo stesso risultato, molto più velocemente, con le proporzioni tra il rettangolo DCE e EC/2 OC . DC x OC = 6√3 x 12√3 = 216
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216
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