6*4 = 2*x 24/2 = x = 12 We can put a parallel line on the right side that mirrors the left The internal horizontal segment is 10. We can another parallel line over the center that mirrors the bottom The internal vertical segment is 2. The diagonal is SQRT(100+4) =2 SQRT(26) z+ 2* SQRT(26) + z = diameter, radius is z + SQRT(26) and area is pi (z + SQRT(26))^2 = (z^2 + z* 2*SQRT(26) +26)*pi z * (z + 2*SQRT(26)) = 24 = z^2 + z * 2 *SQRT(26) Therefore area = (24+26) pi = 50pi units^2
I think there is another the method which is graphical: find the intersection of both lines perpendicular to the chords passing by their respective midpoints, giving so the geometric place of the center of the circle. By means of a compass (opened from center tu any of the ends of chords) draw the circumference then shade the area inscribed to find the solution. regards.
Thank you for your comment! You’re absolutely right about the graphical method. This method is a classic geometric approach and is very effective. Thanks for sharing!
Solution: r = radius of the circle. I place the intersection point of the two chords at the origin of a coordinate system. Then the left end point of the horizontal chord is at A = (-2;0), its right end point at C = (12;0), because the section of this chord results from the Chord theorem in mathematics, 4*6/2 = 12. The end points of the vertical chord are at B = (0;-4) and D = (0;6). The two perpendicular bisectors of these two chords intersect at the center of the circle. The horizontal perpendicular bisector then has the equation y = (6+(-4))/2 = 1 and the vertical perpendicular bisector then has the equation x = ((-2)+12)/2 = 5. The center is then at M = (5;1). According to the Pythagorean theorem, the radius of the circle is: MC² = r² = (12-5)²+(0-1)² = 7²+1 = 50. The area of the circle is then: 50*π ≈ 157.0796
This is trivially easy using the law of sines. For a triangle inscribed in a circle, with chords a, b, and c and opposite angles A, B, C, we have a/sin(A) = b/sin(B) = c/sin(C) = D, where D is the diameter of the circle. So it's easy to see that the angle formed at the red dot is arctan(6/2) + arctan(4/2) = 135 degrees. That chord is of length 10. So, by the above we have D = 10/sin(135) = 10*sqrt(2). So R = 5*sqrt(2), and area = 50*pi. Easy peasy - doesn't really need a 13 minute video.
@@KipIngram Hi! Thank you for your comment and for sharing your solution. You’re right-the Law of Sines can definitely be applied to this problem, and it’s a great method. 👍 Regarding the length of the video, I see your point, but I wanted to clarify that in those 13 minutes, I aimed to explain three different methods for solving the problem. I believe it’s important to make the content accessible to everyone, especially for those who might not be as proficient in math as you. I always try my best to ensure that the explanations are clear and understandable, and that takes time. Cheers!
@@ThePhantomoftheMath No need of taking additional bother for this.your vdo and replying any message are sufficient for me. I shall pick it up from there. Thanks once again for taking pain to response me. I am really grateful to you. 🙏
i found that points p1=(x,y), p2 = (x+2,y+6) and p3 = (x+2,y-4) all satisfy r = sq_root(x^2 + y^2) , and was able to solve it as a system of equations 0 = 4x + 4 +12y + 36 && 0 = 4x + 4 -8y + 16 -> x = -3y -10 && x = 2y -5 ---> y = -1 && x=-7 so p1 is at coords y = -1 x = -7 where 0,0 is the center of the circle. This gives a radius equal to sq((-1)+(-7)), and we can just square it and multiply with pi for 50pi I feel intuitively that the horizontal bit with unknown length should be 12, due to shape similarity with triangles on left hand side. I guess it holds, as one can calculate a vector (7,1) from p1 to center of circle given that the sublengths of the chords are known.
Points on circles are (0,6), (-2 0) ,(0 -4) Centre (-g, -f) Circle and equation x ^2+y^2+2gx +2fy +c=0 Putting the points in it 36+12f+c =0--(1) 4 - 4 g +c =0- (2) 16-8f +c =.0--(3) From 1 and 3 f = -1 from 3 c=-24 From 2 g = 5 (-g,-f ) =(-5,1) Distance between (radius) (0 ,6).& (-5 ,1) =√(25+25)=√50 Area of circle =50π sq unit. Comment please.
@@Programmingwithrohithcrossing chords theorem: 2x=6*4=12. Radius point is 1 above horizontal line and 7 left of the right end on the arc. R^2=1^2+7^2=50 therefore area equals 50Pi.
Please reduce the volume of your disgusting music down so we can hear you better. It is distracting, and in effect interferes with your presentation. Many videos on You Tube have nauseating and monotonous music. They think it enhances the video. It does NOT.
Thank you for the feedback-I understand that background music can sometimes be distracting. I’ve actually reduced the volume a lot in my newer videos, so hopefully, the explanations come through more clearly now. I appreciate you taking the time to share your thoughts, and I’ll continue to keep this in mind for future content!
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6*4 = 2*x
24/2 = x = 12
We can put a parallel line on the right side that mirrors the left
The internal horizontal segment is 10.
We can another parallel line over the center that mirrors the bottom
The internal vertical segment is 2. The diagonal is SQRT(100+4) =2 SQRT(26)
z+ 2* SQRT(26) + z = diameter, radius is z + SQRT(26) and area is pi (z + SQRT(26))^2 = (z^2 + z* 2*SQRT(26) +26)*pi
z * (z + 2*SQRT(26)) = 24 = z^2 + z * 2 *SQRT(26)
Therefore area = (24+26) pi = 50pi units^2
Thats my solution also.
who's gonna tell him he's finding the area of Super Smash Bros?
LOL 🤣🤣🤣
I think there is another the method which is graphical:
find the intersection of both lines perpendicular to the chords passing by their respective midpoints, giving so the geometric place of the center of the circle. By means of a compass (opened from center tu any of the ends of chords) draw the circumference then shade the area inscribed to find the solution. regards.
Thank you for your comment! You’re absolutely right about the graphical method. This method is a classic geometric approach and is very effective. Thanks for sharing!
I never knew about that 2nd method. Fascinating
Solution:
r = radius of the circle.
I place the intersection point of the two chords at the origin of a coordinate system. Then the left end point of the horizontal chord is at A = (-2;0), its right end point at C = (12;0), because the section of this chord results from the Chord theorem in mathematics, 4*6/2 = 12. The end points of the vertical chord are at B = (0;-4) and D = (0;6). The two perpendicular bisectors of these two chords intersect at the center of the circle. The horizontal perpendicular bisector then has the equation y = (6+(-4))/2 = 1 and the vertical perpendicular bisector then has the equation x = ((-2)+12)/2 = 5. The center is then at M = (5;1).
According to the Pythagorean theorem, the radius of the circle is:
MC² = r² = (12-5)²+(0-1)² = 7²+1 = 50.
The area of the circle is then: 50*π ≈ 157.0796
Vertical shift O is (6+4)/2 - 4 = 1. Then we have for blue chord and AC: (r-1)(r+1)=7*7, then r^2=50 and S=50*PI.
Applying the concepts of a circle passing via 3points,that's a nice job!
l learnt this in my school era❤❤
Love the ezplanation
Thank you. Glad you enjoyed it!
This is trivially easy using the law of sines. For a triangle inscribed in a circle, with chords a, b, and c and opposite angles A, B, C, we have a/sin(A) = b/sin(B) = c/sin(C) = D, where D is the diameter of the circle.
So it's easy to see that the angle formed at the red dot is arctan(6/2) + arctan(4/2) = 135 degrees. That chord is of length 10. So, by the above we have D = 10/sin(135) = 10*sqrt(2). So R = 5*sqrt(2), and area = 50*pi. Easy peasy - doesn't really need a 13 minute video.
@@KipIngram Hi! Thank you for your comment and for sharing your solution. You’re right-the Law of Sines can definitely be applied to this problem, and it’s a great method. 👍
Regarding the length of the video, I see your point, but I wanted to clarify that in those 13 minutes, I aimed to explain three different methods for solving the problem. I believe it’s important to make the content accessible to everyone, especially for those who might not be as proficient in math as you. I always try my best to ensure that the explanations are clear and understandable, and that takes time. Cheers!
Sir,
I was known to the formula applied in method two. Thanks a lot sir
@@bkp_s Really nice! You are one of the rare ones!
@@ThePhantomoftheMath I need to learn classic english too with you in this way
@@bkp_s English is not my native language (as you can probably tell 😹), but I'm happy to help if I can.
@@ThePhantomoftheMath No need of taking additional bother for this.your vdo and replying any message are sufficient for me. I shall pick it up from there. Thanks once again for taking pain to response me. I am really grateful to you. 🙏
i found that points p1=(x,y), p2 = (x+2,y+6) and p3 = (x+2,y-4) all satisfy r = sq_root(x^2 + y^2) , and was able to solve it as a system of equations
0 = 4x + 4 +12y + 36 && 0 = 4x + 4 -8y + 16 -> x = -3y -10 && x = 2y -5 ---> y = -1 && x=-7
so p1 is at coords y = -1 x = -7 where 0,0 is the center of the circle. This gives a radius equal to sq((-1)+(-7)), and we can just square it and multiply with pi for 50pi
I feel intuitively that the horizontal bit with unknown length should be 12, due to shape similarity with triangles on left hand side.
I guess it holds, as one can calculate a vector (7,1) from p1 to center of circle given that the sublengths of the chords are known.
Really nice! Well done!
Points on circles are
(0,6), (-2 0) ,(0 -4)
Centre (-g, -f)
Circle and equation
x ^2+y^2+2gx +2fy +c=0
Putting the points in it
36+12f+c =0--(1)
4 - 4 g +c =0- (2)
16-8f +c =.0--(3)
From 1 and 3
f = -1 from 3
c=-24
From 2
g = 5
(-g,-f ) =(-5,1)
Distance between (radius)
(0 ,6).& (-5 ,1)
=√(25+25)=√50
Area of circle =50π sq unit.
Comment please.
Yes! Everything checks out! The center is (5, 1), and the area of the circle is 50π square units. Great job!!! Nice take!
A line thru the center of circle and parallel to the "2" line would have equal distances along the "4", "6" lines; i.e. 5 and 5.
Yes...and it does just that at 3:49
I learned math form RUclips
Easy. 6×4=2X==>X=12, so 4R^2=6^2+4^2+2^2+12^2===>R=50 and area= 157
I did it in my head. Rsquared=7squared plus 1squared =50, area =50pi
What's the formula, what's the method?
@@Programmingwithrohithcrossing chords theorem: 2x=6*4=12. Radius point is 1 above horizontal line and 7 left of the right end on the arc. R^2=1^2+7^2=50 therefore area equals 50Pi.
I knew the formula (but also forgot it)
Yeah, it's not frequently used formula...
eat circumscribed polygons for lunch
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آسف! سأنتبه إلى هذه التفاصيل في المستقبل كلما سنحت لي الفرصة حتى تكون الرؤية أوضح.
Please reduce the volume of your disgusting music down so we can hear you better. It is distracting, and in effect interferes with your presentation. Many videos on You Tube have nauseating and monotonous music. They think it enhances the video. It does NOT.
Thank you for the feedback-I understand that background music can sometimes be distracting. I’ve actually reduced the volume a lot in my newer videos, so hopefully, the explanations come through more clearly now. I appreciate you taking the time to share your thoughts, and I’ll continue to keep this in mind for future content!
The original diagram the chords don't intersect at right angles so stop your nonsense
50pi