Solving the quadratic becomes a lot easier when you realise you have a right triangle with consecutive side lengths so one of the solutions must be from a 3-4-5 right triangle with r=5.

@ThePhantomoftheMath

17 күн бұрын

Yes, absolutely. In this particular case, you can definitely do that. Just be careful: it's not always going to be a 3-4-5 scenario. Nice thinking btw.

Thales tells that for the side of the square we have: 2 : s/2 = s/2 : s , hence s = 8. This means that the diameter = 10.

@ThePhantomoftheMath

17 күн бұрын

Absolutely, Thales is another way of thinking! Realy nice!

you could calculate this faster, using a chord equation. let a= side of square we have a*2=(a/2)^2 so a is 8. diameter of circle is 8+2=10 so radius is 5

Well done! I may borrow this to do with my students. I just subscribed to your channel. It reminds me somewhat of “Math and Engineering,” which I definitely recommend.

@ThePhantomoftheMath

19 күн бұрын

Ty friend! Feel free to borrow what ever you like.

Something all you need to solve a geometry problem is to add a few more lines. Thank you for your insight on this problem

@ThePhantomoftheMath

7 күн бұрын

Pleasure

Case R=1 is basically square is dot on the left of circle with radius 1.

use intersecting chord theorem with a = ½ (side of square) and we will get : a.a=2.2a a²=4a a=0, a=4 Since a = ½ (side of square), the side of square is 8 (it couldn't be 0), and diameter of the circle is 8+2 which is 10. So the area of the circle is 25π.

I chose simply to complete the square -- 5 => -5, add 9 to both sides and r-3 = +/- 2 =. r = 3 +/- 2

Move the square right by 1 and you have a gap of 1 on each side and at top and bottom. The centre of both square and circle are now at the same point. Half the square's side can be called r-1. Now put the square back where you found it. Circle's centre to top right corner is r.. Go half way down the vertical line for a side of r-1. The remaining side is r-2. (r-1)^2 + (r-2)^2 = r^2 r^2 - 2r + 1 + r^2 - 4r + 4 = r^2 2r^2 - 6r + 5 = r^2 r^2 - 6r + 5 = 0 As the coefficients add up to 0, (r-1) is a factor. (r-1)(r-5)=0, so r=1 or r=5 r must be greater than 1 due to the given line of 2, so r=5. Circle's area is 25pi sq un. Approx 78.54 sq un I have now checked the video and appreciate what you did, but I think my way was cleaner and simpler, i.e. moving the square to start to give simple proof that half the square's side = r-1.

@ThePhantomoftheMath

6 күн бұрын

This is why I love math so much: there are so many ways to solve a single problem. For me, that's the beauty of math. Your method is certainly awesome, and it works perfectly. Thank you so much for sharing it!

so taking side-length of the square as a: From construction the diameter of the circle is d = a + 2 Then by intersecting chords theorem, 2·a = (½a)·(½a) 2a = ¼a² 8a = a² a² - 8a = 0 a = 0, or 8 area of the circle = ¼πd² = ¼π(8 + 2)² = ¼π(10)² area = 25π sq units.

@ThePhantomoftheMath

2 күн бұрын

That works just fine. Nice!

Assume x= square side length two intersecting lines inside circle law (x/2)*(x/2)=2x (x*x)/4=2x X=8 r=8+2 Area=25pi

Let the center of the circle be O, the point of tangency between the circle and the left side of the square be M, the point opposite M on the square be N, and the top right and bottom right vertices of the square (where they contact the circle) be A and B respectively. Draw MN, NA, and AO. Let NA = x, so the side length of the square is 2x. Note that MN = 2x, as it is parallel to the top and bottom sides of the square and perpendicular to and intersects the right and left sides. MN is also 2 units less than the radius of the circle, as M is on the circumference, N is 2 units from the opposite circumference, and MN is colinear with O. Thus 2r = 2x+2 ==> r = x+1 ==> x = r-1. Troangle ∆ONA: ON² + NA² = OA² (r-2)² + (r-1)² = r² r² - 4r + 4 + r² - 2r + 1 = r² r² - 6r + 5 = 0 (r-1)(r-5) = 0 r = 1 ❌ | r = 5 2 A = πr² = π5² = 25π

@ThePhantomoftheMath

14 күн бұрын

Great job! That will work too!

No need to be solving a quadratic equation. The short line segment being to the furthest point from the edge of the circle is bisecting the side of the square. Then it's a simple intersecting cords problem. Let the side of the square = 2x. x^2=2*2x x^2=4x x^2-4x=0 x(x-4)=0 Xcannot be 0 x=4 is our choice. D=2x+2=2*4+2=10 R=D/2=10/2=5 A=5^2Pi=25Pi.

@ThePhantomoftheMath

17 күн бұрын

That's also a realy nice approach if you want to avoid quadratic equations. (I like quadratic equations... it's... you know... my thing 😂). But thank you for sharing another nice method for those who don't like them! Good job!

@rabotaakk-nw9nm

17 күн бұрын

x²=4x |÷x (x≠0) => x=4

I guess pie×5×5?

ㅇ

It all looks Very complicated. And I don’t Know what any of it means . But I think you left out the answer…78.75.👍

@ThePhantomoftheMath

22 күн бұрын

Lol 😹