Calculate the Area of This Inscribed Square | 3 Easy Methods

In this video, we’ll tackle an interesting mathematical problem: how to calculate the area of a square inscribed in a quarter circle. I'll demonstrate 3 different methods to solve this problem, ensuring you grasp each approach with ease.
🔢 What You’ll Learn:
How to set up and solve the problem using algebraic methods.
A geometric approach to visualize and calculate the area.
An analytical method for a deeper understanding.
Each method is broken down step-by-step, making it easy to follow along and learn. Whether you're a student, teacher, or math enthusiast, you'll find valuable insights and techniques in this video.
📏 Chapters:
0:00 Presenting the Problem
0:30 Method 1: Using Perpendicular Bisector
4:19 Method 2: Using the Diagonal of the Square
7:45 Method 3: "Out-of-the-Box"
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Happy learning and see you in the video!

[@Esraa-pf5dg]

Very very very good،مبدع واللله

[@ThePhantomoftheMath]

شكرًا! :Hope I wrote this wright

[@Silver_crap]

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.

[@juanregidor540]

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.

[@ThePhantomoftheMath]

Thanks!

[@Esraa-pf5dg]

الطريقه الاخيره جميله جدا جدا

[@ThePhantomoftheMath]

شكرًا جزيلاً

[@kextrz]

Wow! The third method using Thales is by far the quickest and soundest method.

[@andrec.2935]

Lindíssimo problema e soluções!

[@ThePhantomoftheMath]

@@andrec.2935 Thanks 🤗

[@theupson]

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

[@ThePhantomoftheMath]

Analytical geometry method. Nice.

[@Aashu2631]

Nice one .......🎉

[@ThePhantomoftheMath]

Thank you! 🙂

[@dineshkumthekar3135]

10:23

[@paulwakeford8566]

But how do you know for sure that OCD is isocellous ?

[@ThePhantomoftheMath]

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.

[@jeremyjay380]

​@@ThePhantomoftheMath I don't think you get the joke.

[@ThePhantomoftheMath]

@@jeremyjay380 If it was a joke...you're right: I didn't get it 😖

[@dineshkumthekar3135]

😊😊😊

[@NotGleSki]

user the perpendicular bisector method

[@99jei63]

I got it by multiplying the perpendicular chords (i forgot what it's called 😂). Closer to the 3rd solution.

[@ianmunro5033]

The chord-chord power theorem. (10+3s/2)(10-3s/2)=(s/2)^2.

[@troymingming]

instead of a quarter circle, turn it into 60°

[@Grizzly01-vr4pn]

A = 200 - 100√3 ≈ 26.79 units²

[@troymingming]

how about 61°?

[@tetramur8969]

​@@troymingminghow about an arbitrary angle?

[@troymingming]

this is my own formula

[@Grizzly01-vr4pn]

@@troymingming ?

[@dineshkumthekar3135]

10:23