Find the Area of the Red Circle | Using the Distance From the Square
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- čas přidán 28. 04. 2024
- 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
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.
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.
Absolutely, Thales is another way of thinking! Realy nice!
Something all you need to solve a geometry problem is to add a few more lines. Thank you for your insight on this problem
Pleasure
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.
Ty friend! Feel free to borrow what ever you like.
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
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.
That works just fine. Nice!
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.
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!
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
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.
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!
x²=4x |÷x (x≠0) => x=4
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
I guess pie×5×5?
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
My brain hurts...
🤣🤣🤣 Sorry for that!
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.👍
Lol 😹