Find the Area of the Red Circle | Using the Distance From the Square

Рет қаралды 13,169

Күн бұрын

Welcome to the latest video where we tackle an intriguing geometry problem: finding the area of a red circle nestled within a square. But here's the twist - we're not using traditional methods. Instead, we employ a clever strategy involving the distance from the square's edge to unravel the mystery of the circle's area.
Join me on this mathematical journey as we explore innovative techniques for solving geometric puzzles. Whether you're a student looking to sharpen your math skills or a curious mind eager to delve into mathematical concepts, this video offers a fresh perspective on problem-solving.
Visit our channel 👉 / @thephantomofthemath

Пікірлер: 34

@zahranf.a.9864 4 ай бұрын

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.

@MCPhssthpok 4 ай бұрын

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 4 ай 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.

@JobBouwman 4 ай бұрын

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 4 ай бұрын

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

@bkp_s Ай бұрын

Also got solution by applying the formula 4r^=a^+b^+c^+d^ As demonstrated by you in another lesson with same solution

@brickie9816 4 ай бұрын

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 4 ай бұрын

Pleasure

@STEAMerBear 4 ай бұрын

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 4 ай бұрын

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

@KipIngram 19 күн бұрын

Let R be the circle radius. Then the square side length is 2*R-2. Now consider the right triangle formed by the center of the circle, the top right corner of the square, and the middle of the right side of the square. The hypotenuse has length R, the vertical side has length R-1(half of the square side length) and the bottom side has length R-2 (obvious from the diagram). Use the Pythagorean theorem: R^2 = (R-1)^2 + (R-2)^2 R^2 = R^2 - 2*R + 1 + R^2 - 4*R + 4 0= R^2 -6*R + 5 0 = (R-5)*(R-1) R=1 is clearly an inappropriate solution; R must be greater than 2. So we take R=5, and therefore Red Circle Area = pi*5^2 = 25*pi Q.E.D.

@adampiechuta5774 4 ай бұрын

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

@michaelchampoux-n3y 4 ай бұрын

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 4 ай 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 4 ай бұрын

x²=4x |÷x (x≠0) => x=4

@lpi3 4 ай бұрын

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

@jacquespictet5363 Ай бұрын

Nice. As often in your videos, it might be more interesting to use an unknown lenght L instead of a given value, that can be used later as an example.

@ThePhantomoftheMath Ай бұрын

@@jacquespictet5363 Hm, that's an interesting idea. I will try that in some of the future videos.

@GirishManjunathMusic 4 ай бұрын

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 4 ай бұрын

That works just fine. Nice!

@himo3485 4 ай бұрын

Chord Theorem 2r-2 (2r-2)/2=2(r-1)/2=r-1 (r-1)(r-1)=2＊(2r-2) r^2-2r+1=4r-4 r^2-6r+5=0 (r-1)(r-5)=0 r=1 is rejected , thus r=5 area of Red Circle : 5＊5＊π=25π

@meedonexus 4 ай бұрын

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

@quigonkenny 4 ай бұрын

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 4 ай бұрын

Great job! That will work too!

@SumitVerma-lg3qh 10 күн бұрын

It can easily solved by using power point of circle => 2 * 2r-2 =( r-1)²

@MrPaulc222 4 ай бұрын

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 4 ай 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!