At a quick glance: Using the rule : two perpendicular chords of a circle, the products of the lengths of the intersecting sections of the two perpendicular lines are equal. Drawing a horizontal line through 'o' intersecting perpendicularly the right side of the square and dividing the side into two lengths of distance 'a' each .Hence 2 * a = square side length. 2 * a also equals the left part of the horizontal chord. Applying our rule of chords: 2 * a * 1 = a * a. Hence a = 2 and the square side length = 4. The area of the square is 4 * 4 = 16.
Solution using trigonometry: I'm considering a to be the point near the text a. Angle Bao = angle oaB since triangle Bao is isosceles triangle Angle AaB = Angle oBF since aBF is a right angle due to property of circle. Angle oBF = Angle BFo Hence, Angle AaB = Angle BFo Hence, tan of both will be equal. Assume x to be side of the square. Tans of both. x/(x/2) = (x/2)/1 x=4 Area of square is 16.
Let the circle have radius R and be centered at our origin. Then the top right of the square is at coordinates x = R-1, y = R - 0.5. By the Pythagoras's law, we have R^2 = (R-1)^2 + (R-0.5)^2 R^2 = R^2 - 2*R + 1 + R^2 - R + 0.25 R^2 = 2*R^2 - 3*R + 5/4 R^2 - 3*R + 5/4 = 0 This has roots of R = 2.5 and R= 0.5. The R=0.5 solution is clearly inappropriate, so we take R = 2.5. Now the square side length is 2*R - 1, so we plug in R = 2.5 and get side length 4, so Blue Square Area = 16 Q.E.D.
Y'all are making this way more complicated than it needs to be. Draw a triangle PBF, then let the line BE make two smaller triangles PEB and BEF. Per Thales' theorem of inscribed angles that lie on a diameter, we know that PBF is a right triangle. Therefore, our three triangles are similar, because the two smaller are also right triangles that also share one additional angle with PBF by virtue of it being the same corner. We also know that if triangles share the same angles, then the ratio of side length must also be the same, and if length BE is half of length PE, then length EF (1) must be half of BE. Therefore BE is 2, BC is 4, and area of the square is 16.
Just having some fun! "Jet Jackson" used that expression on his TV program in the 1950's. To some it was quite funny, being preposterous. You are probably too young (lucky). @@gelbkehlchen
Lado del cuadrado =2a Potencia del punto medio de la cuerda vertical respecto a la circunferencia =1*2a=a^2 >>> a=2 >>> Área del cuadrado =(2*2)^2=16 Un saludo cordial.
Let 2s×2s be the square, r be the radius of the circle, then 2s+1=2r, r=s+1/2, and (s+1/2)^2=r^2=s^2+(r-1)^2=s^2+(s-1/2)^2, s^2=2s, as s not zero, s=2, therefore the area of the square is 4×4=16.😊
Let AB meet the circle at G & draw a perpendicular from G to meet PO at H. Since AB is parallel to PF, ∠GOP=∠ BOF. So we get |AG| = |PH| = |EF| = 1. By the tangent-secant theorem from the point A, we get |AP|² = |AG|.|AB|. So (a/2)²=(1)(a). ∴ a²/4=a, so a(a-4)=0. Since a>0, a=4. ∴ area(ABCD)=16.
You must be kidding. We learned logs in 6th grade (age 12) and also how to use a slide rule. But we were smarter than most, especially in the sciences. @@84com83
Let's prove it. As BC is a chord we can draw radius of perpendicular to it. Also ad is tangent(given in diagram) so radius OT(say) is perpendicular to it. Now consider the parallel lines ad and BC. They have perpendicular OT and OF which pass through the same point O. Therefore, TF is a transversal and makes a line cutting the circle to form a chord which is the diameter as OT passes through O
Let 2s×2s be the square, r be the radius of the circle, then 2s+1=2r, r=s+1/2, and (s+1/2)^2=r^2=s^2+(r-1)^2=s^2+(s-1/2)^2, s^2=2s, as s not zero, s=2, therefore the area of the square is 4×4=16.😊
Let 2s×2s be the square, r be the radius of the circle, then 2s+1=2r, r=s+1/2, and (s+1/2)^2=r^2=s^2+(r-1)^2=s^2+(s-1/2)^2, s^2=2s, as s not zero, s=2, therefore the area of the square is 4×4=16.😊
At a quick glance: Using the rule : two perpendicular chords of a circle, the products of the lengths of the intersecting sections of the two perpendicular lines are equal. Drawing a horizontal line through 'o' intersecting perpendicularly the right side of the square and dividing the side into two lengths of distance 'a' each .Hence 2 * a = square side length. 2 * a also equals the left part of the horizontal chord. Applying our rule of chords: 2 * a * 1 = a * a. Hence a = 2 and the square side length = 4. The area of the square is 4 * 4 = 16.
That is how I solved it.
Its called the power chord theorem and they are not required to be perpendicular
Intersecting chords theorem is the way to go. Then it becomes 1 x a = a/2 x a/2, so a=4 and a^2 =16 thanks for a nice challenge!
Solution using trigonometry:
I'm considering a to be the point near the text a.
Angle Bao = angle oaB since triangle Bao is isosceles triangle
Angle AaB = Angle oBF since aBF is a right angle due to property of circle.
Angle oBF = Angle BFo
Hence, Angle AaB = Angle BFo
Hence, tan of both will be equal.
Assume x to be side of the square.
Tans of both.
x/(x/2) = (x/2)/1
x=4
Area of square is 16.
Per construction line(E,B) has length of the square root of a.
==> a/2 = sqrt(a)
sqrt(a) = 2
==> a^2 = 16
Let the circle have radius R and be centered at our origin. Then the top right of the square is at coordinates x = R-1, y = R - 0.5. By the Pythagoras's law, we have
R^2 = (R-1)^2 + (R-0.5)^2
R^2 = R^2 - 2*R + 1 + R^2 - R + 0.25
R^2 = 2*R^2 - 3*R + 5/4
R^2 - 3*R + 5/4 = 0
This has roots of R = 2.5 and R= 0.5. The R=0.5 solution is clearly inappropriate, so we take R = 2.5. Now the square side length is 2*R - 1, so we plug in R = 2.5 and get side length 4, so
Blue Square Area = 16
Q.E.D.
Nice video
Y'all are making this way more complicated than it needs to be. Draw a triangle PBF, then let the line BE make two smaller triangles PEB and BEF. Per Thales' theorem of inscribed angles that lie on a diameter, we know that PBF is a right triangle. Therefore, our three triangles are similar, because the two smaller are also right triangles that also share one additional angle with PBF by virtue of it being the same corner. We also know that if triangles share the same angles, then the ratio of side length must also be the same, and if length BE is half of length PE, then length EF (1) must be half of BE. Therefore BE is 2, BC is 4, and area of the square is 16.
This was the simplest explanation. Thanks
Solution:
x = side of the square.
Euclid's height theorem:
x*1 = (x/2)² |-x ⟹
(x/2)²-x = x²/4-x = x*(x/4-1) = 0 |either x1 = 0 [no importance for this task] or x/4-1 = 0 ⟹ x = 4 ⟹ x² = 16
Tooo complicated! Just use Logs.
@@eb1684 Please explain to me exactly!
Just having some fun!
"Jet Jackson" used that expression on his TV program in the 1950's.
To some it was quite funny, being preposterous.
You are probably too young (lucky).
@@gelbkehlchen
he is trying to look cool thinking we don't know bout logarithms😂@@gelbkehlchen
brainstorming , still too easy
I got the 1st equation on my own
Had to watch the viideo forr the 2nd one
4=PE=2*EB=4*EF (similar right triangles with tangent =1/2)
I wanted to say the Same Like Tom Bufford
Cool
Secant- tangent theorem is suitable to this problem: (a/2)^2=(1)(a) ====> a^2=16.
Better to use the intersecting chord theorem
Lado del cuadrado =2a
Potencia del punto medio de la cuerda vertical respecto a la circunferencia =1*2a=a^2 >>> a=2 >>> Área del cuadrado =(2*2)^2=16
Un saludo cordial.
?Que?
¿Qué quieres decir?@@eb1684
¿Qué quieres decir?@@eb1684
1:02 i didn't understand how BE = a/2, pls explain
PF is radius and its perpendicular to BC so its go through midpoint of BC which is E
Let 2s×2s be the square, r be the radius of the circle, then 2s+1=2r, r=s+1/2, and (s+1/2)^2=r^2=s^2+(r-1)^2=s^2+(s-1/2)^2, s^2=2s, as s not zero, s=2, therefore the area of the square is 4×4=16.😊
Let AB meet the circle at G & draw a perpendicular from G to meet PO at H.
Since AB is parallel to PF, ∠GOP=∠ BOF. So we get |AG| = |PH| = |EF| = 1.
By the tangent-secant theorem from the point A, we get |AP|² = |AG|.|AB|.
So (a/2)²=(1)(a). ∴ a²/4=a, so a(a-4)=0. Since a>0, a=4. ∴ area(ABCD)=16.
Its 16 the point 0 is from circle not on square
Good.
Why not just use logarithms?
How?
You must be kidding. We learned logs in 6th grade (age 12) and also how to use a slide rule. But we were smarter than most, especially in the sciences. @@84com83
Why it doesn't work lmao?
X=(x/2)²
X=4
X²=16🤔💚💙💜
Bro this can be solved using trigonometry
As
EF = 1 UNIT
ANGLE BEF = 90 DEGREES
THUS BF= 2 UNIT
AND BE= √3
THUS BC =2×√3
AREA ABCD = (2√3)^2=12 UNIT
its not a 30-60-90 triangle
profesor, practique caligrafía, sus letras y números son pesimos😅😅😅
it is not given that EF is on diameter. you just presumed it.
Let's prove it.
As BC is a chord we can draw radius of perpendicular to it.
Also ad is tangent(given in diagram) so radius OT(say) is perpendicular to it.
Now consider the parallel lines ad and BC. They have perpendicular OT and OF which pass through the same point O.
Therefore, TF is a transversal and makes a line cutting the circle to form a chord which is the diameter as OT passes through O
That isn't a square because only a square can the inscribed inside of a circle.
Let 2s×2s be the square, r be the radius of the circle, then 2s+1=2r, r=s+1/2, and (s+1/2)^2=r^2=s^2+(r-1)^2=s^2+(s-1/2)^2, s^2=2s, as s not zero, s=2, therefore the area of the square is 4×4=16.😊
Let 2s×2s be the square, r be the radius of the circle, then 2s+1=2r, r=s+1/2, and (s+1/2)^2=r^2=s^2+(r-1)^2=s^2+(s-1/2)^2, s^2=2s, as s not zero, s=2, therefore the area of the square is 4×4=16.😊
Isn’t this his solution where a is just substituted with 2s? It’s the same 2 equations. Do you find this easier to understand?