Consider the right triangle formed by the center of the circle, the top left corner of the square, and the bottom right corner of the square. The hypotenuse is a radius, so length 10. If square side length is S, then the diagonal of the square is sqrt(2)*S. And the side out from the center of the circle to the top left of the square is length S/sqrt(2). Use the Pythagorean theorem: (S*sqrt(2))^2 + (S/sqrt(2))^2 = 10^2 2*S^2 + S^2/2 = 100 2.5*S^2 = 100 S^2 = Area = 100/2.5 = 40 Q.E.D.
I solved it using co ordinate geometry,consider the right angle of circle 0,0 by inspection it can be easily seen all the vertices of square are (0,a),(a,0),(2a,a),(a,2a).last two lies on the circle, gives 4a^2+a^2=100 5a^2=100 a^2=20 2a^2=40 a=distance from origin to first vertix on axises.
defining the center as (0,0) and y as the distance below the center, if the edge length of the square is X, the coordinates of two right side vertices are (X/2, X/2) and (X/2, Y). since the distance between those points is also X, and the second point lies on the circle you get the system: Y-X/2 = X (X/2)^2+Y^2 = 100, very easily solved for X^2
Hi. The line segment DO is equal to the line segment OC. This equality arises from the symmetry of the construction and the perpendicular height from O to CD. Line AP is the perpendicular bisector of chord AB, ensuring symmetry. Additionally, because ABCD is a square inscribed in a quarter circle, there is only one way to inscribe it, reinforcing the symmetrical properties. Thus, OD and OC are equal in length, making triangle OCD isosceles.
Consider the right triangle formed by the center of the circle, the top left corner of the square, and the bottom right corner of the square. The hypotenuse is a radius, so length 10. If square side length is S, then the diagonal of the square is sqrt(2)*S. And the side out from the center of the circle to the top left of the square is length S/sqrt(2). Use the Pythagorean theorem:
(S*sqrt(2))^2 + (S/sqrt(2))^2 = 10^2
2*S^2 + S^2/2 = 100
2.5*S^2 = 100
S^2 = Area = 100/2.5 = 40
Q.E.D.
Bunch of options are illuminating class of teaching. 🙏great sir!!!
I solved it using co ordinate geometry,consider the right angle of circle 0,0 by inspection it can be easily seen all the vertices of square are (0,a),(a,0),(2a,a),(a,2a).last two lies on the circle, gives 4a^2+a^2=100
5a^2=100
a^2=20
2a^2=40
a=distance from origin to first vertix on axises.
@@Silver_crap Nice method! I like it! Great job!
Very very very good،مبدع واللله
شكرًا! :Hope I wrote this wright
Incredible problem! I thought there was a way to get the value of a^2 by extending to a circle and you showed me. Keep it up.
Thanks!
defining the center as (0,0) and y as the distance below the center, if the edge length of the square is X, the coordinates of two right side vertices are (X/2, X/2) and (X/2, Y). since the distance between those points is also X, and the second point lies on the circle you get the system:
Y-X/2 = X
(X/2)^2+Y^2 = 100, very easily solved for X^2
Analytical geometry method. Nice.
Wow! The third method using Thales is by far the quickest and soundest method.
الطريقه الاخيره جميله جدا جدا
شكرًا جزيلاً
Lindíssimo problema e soluções!
@@andrec.2935 Thanks 🤗
But how do you know for sure that OCD is isocellous ?
Hi. The line segment DO is equal to the line segment OC. This equality arises from the symmetry of the construction and the perpendicular height from O to CD. Line AP is the perpendicular bisector of chord AB, ensuring symmetry. Additionally, because ABCD is a square inscribed in a quarter circle, there is only one way to inscribe it, reinforcing the symmetrical properties. Thus, OD and OC are equal in length, making triangle OCD isosceles.
@@ThePhantomoftheMath I don't think you get the joke.
@@jeremyjay380 If it was a joke...you're right: I didn't get it 😖
Nice one .......🎉
Thank you! 🙂
user the perpendicular bisector method
10:23
Three ways to solve this question and I couldn't find even one😢 doesn't support my self esteem... Great video, though.
😊😊😊
I got it by multiplying the perpendicular chords (i forgot what it's called 😂). Closer to the 3rd solution.
The chord-chord power theorem. (10+3s/2)(10-3s/2)=(s/2)^2.
instead of a quarter circle, turn it into 60°
A = 200 - 100√3 ≈ 26.79 units²
how about 61°?
@@troymingminghow about an arbitrary angle?
this is my own formula
@@troymingming ?
10:23