Once you know r^2 = 85/4, you can just leave it at that and not bother taking the square root of both sides. You're just going to use r^2 when you calculate the area of the circle.
It is amazing how the process is fluid once you have established a line of attack. Worked it out only after seeing the two triangles. This is a great brainteaser.
Very clever and a simpler method than I attempted. Could have been even simpler, once you had found the value of r square, you just needed to multiply by pi to get the area, no need to get the square root and then square it again.
May I suggest another way (I think: simpler): Sides of squares are 2. Connect A to C, it will touch the bottom-left corner of the square with the center - Mark as F. CF squared is 2²+4²=20. Construct another square to the right of the one with the center. Mark it's upper-right corner as G. Connect G and F. GF is perpendicular to CG, and will go through the center. GF squared is also 20. OF = 1/4 of GF. OF squared is (1/4)² = 1/16 of GF = 20/16 = 5/4. r² = OF² + CF² = 5/4 + 20 = 21.25 (no need to take the square root, see next). Area = πr² = 21.25π ~ 66.759 (by calculator. using 3.14 on paper: 66.725).
Thank you for your amazing video. This also trigger me to find alternative solution. Draw a horizontal line at A, a vertical line at B and intersect circle at D Use Chrod theorem to determine shortest chrod Then find the diameter using pythagorean theorem of triangle DBC And hence to get the area of the circle
A faster way would be: Consider the triangle B = (0,0), C = (2,0), A = (-2,8). Its area is Area = 8, and its sides can be calculated as a = 2, b = sqrt(80), c = sqrt(68) using the Pythagorean theorem. Then the radius of the circumscribed circle is given by a*b*c / 4*Area = sqrt(85) / 2
Nice challenge, awesome way to solve it. Sir, you are master of circle and rectangled triangle! 🙂 Another way: draw x-y-coordinate-system with x axis = BC, y axis = (C-B)/2 to O → A = A(-3;8), B = B(-1;0), C = C(1;0) → tan(φ) = 4 → f(x) = (1/4)x + 9/2 → x = 0 → f(x) = 9/2 → 8 - (9/2) = 7/2 →(7/2)^2 + 9 = 85/4 = r^2 → r = OA = √85/2 → πr^2 = π(85/4) ≈ 66,7588 sq units 🙂 fast lane: Because the vertical line of the circle is between B and C → horizontal line AA' (secant) = 6 → chords theorem: 2(4) = 8(1) → (9/2)^2 + 1 = 85/4 = r^2 → area circle = (85/4)π
This guy is not only a fantastic teacher of mathematics , but also great at giving video presentations. .. Teacher i do apologize if i asked an awkward question a few days ago. pleased forgive me. and continue as you are.
So nice of you Paul! You are awesome!!! I really appreciate that. Making quality videos is a very time consuming and meticulous task! Your devout supportive and cheerful comments keep me going... Take care dear and stay blessed😃 Keep smiling😊
I use circle equation, sir. x²+y²+Ax+By+C=0. The use the three point given to solve the equation. Let B(0,0) so C(2,0) and A (-2,8). Cirlce eq: x² + y² - 2x - 9y = 0. Then (x-1)² + (y-4½)² = 85/4.
I found two easier solutions. Both used the perpendicular radii to a chord theorem and the intersecting chords theorem. The horizontal chord from point A is 6 and the vertical chord from point C is 9. A radius through the horizontal chord bisects into two equal segments of 3. A horizontal radius to the vertical chord bisects it into two segments of 4 1/2. However, one could use the same method to find the diameter using the chord of 9 as the longer side of a right triangle.
The horizontal chord measurement for point A is correct and can be explained by symmetry, but how do you get the vertical chord measurement for C? How do you say it equals 9?
I love Pythagoras but coordinate points and the circle formula is easier: T1(1,0), T2(-1,0), T3(3,8) , O(0,k) -> h=0 (because h is between -1 and 1) 1² + k² = r² 3² + (8-k)² = r² 1 + k² = 9 + 64 - 16k + k² 16k = 72, k = 9/2 r² = 81/4 +1, r² = 85/4 r = √(85/4), A= 85/4 π
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Make a line with connecting A and C. This line passes through one corner of the square which contains O. Then call that corner D. And also call a right corner the square E. Make a vertical line from O to line DE and call the crossing point F. Now preparation has been completed. We can find a right triangle ADO. AD=2 x square root 5. We can find DF : OF=2:1. So, DO x DO = 1x1 + (1/2)x(1/2)=(5/4). AS AO x AO=AD x AD + DO x DO=20 + (5/4) = (85/4). Area of the circle is (85/4) x pi. Anyway sorry for my poor English.
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I follow the explanation you have given but I am curious about the point you have marked as O. I think you may have assumed it to be the centre of the circle but it doesn't say that in the question. If the point O is not at the centre, you can't draw a line to bisect the chord. Am I missing something?
"O" is a completely arbitrary point if you don't assume it to be the centre, with no restrictions on its position whatsoever. If you prefer, you could simply create another point defined as the centre and call it "M", then follow the method shown. Note that the positioning of the centre on the diagram is unimportant. If the centre is actually below the top of the second box, then x will result in a negative value. Personally, I found it easier to use a different method that doesn't require the centre. Defining point B as your datum (0,0) allows you to know the coordinates of three points on the circle - A (-2,8), B (0,0) and C (2,0). You can then solve simultaneous equations by plugging these x and y values into the formula of a circle - (x-a)^2 + (y-b)^2 = r^2
@@adamrobinson6951 Thank you for your reply. My point was that O is ASSUMED to be the centre of the circle but this assumption wasn't stated. If it was not the centre, you wouldn't get a right-angled triangle at the intersection with the line so you wouldn't be able to use pythagoras as shown in the demonstration. I agree that other methods would work just as easily, and possibly even easier, but my question is about the assumption that O is at the centre of the circle. Am I missing the proof that O is indeed the centre?
@@ApiaryManager O isn't specified to be in the centre. Therefore, there are no constraints whatsoever on O, and it's positioning will neither affect nor be affected by the geometry of the rest of the question. It is, however, helpful to draw lines to the centre of the circle when solving this problem. If you prefer, you could label the centre using some other term and ignore the existence of O entirely. In this case, O was defined as the centre by the problem solver because doing so didn't affect anything else in the question. Note: this does mean that the position of O in the diagram may potentially be misleading, as the centre may fall in a different square or outside the squares entirely. However, in that case the maths would correct for that issue by returning negative values for our geometric variables.
Why not using the rule in any rectangle Ab/sin(c) =2R? Sin(c) =8/AC Ac calculates by phithagurth (4 and 8) AB is calculates by phithagurth (2 and 8) And it is done without any extra complication
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My solution: Put the figure into (x, y) coordinates. A:: x=0 B and C :: y=0 A = (0, 8) ; B = (2, 0 ) ; C = (4, 0) Middle on BC :: E = (3, 0) 90 degrees on BC at E, to center of circle O O :: x = 3 Middle on AB :: M = (1, 4) x-difference M to O = +2 AB-> = [ -8 , 2 ] Turn 90 degree = [ 8, 2 ] Scale to x=2 ==> [ 2 , 0.5 ] O = M + [ 2 , 0.5 ] = ( 1+2 , 4+0.5) = (3 , 4.5) BO-> = [1 , 4.5] r^2 = | BO-> |^2 = (1 + 4.5^2) = 21.25 Area = PI × r^2 = PI × 21.25
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O is centered left-to-right in the square because the left and right sides of the square it's in are colinear with the left and right sides of the bottom square and B and C are points on the circle. O is equidistant from B and C because... circle. While that's far from a water-tight mathematical proof, it should be enough to convince you that O is centered left-to-right in its square.
I see your point, I think it's not conventional to assume positions of points in a graphic if it's not directly expressed in measurements. Perhaps the drawing should've demonstrated 'O' was 1 unit away from both the left and right side of that square.
@@bjorn6084 If you draw the perpendicular bisector of the chord BC, you know it must pass through the centre of the circle. Extending downwards the two vertical sides of the square containing the centre will produce two parallel lines one of which must pass through B, and the other must pass through C because the parallel lines must remain 2 units apart, by definition. Since the perpendicular bisector of the chord BC is 1 unit away from B and from C (because it's a bisector), it must remain equidistant from the two outer lines we constructed because it is parallel to both (because the angle each of them make with BC is a right angle). That shows that the centre of the square is equidistant from the vertical sides of the square containing it.
There is something.. How to check that OEC is right triangle? It looks proven with the Chord theorem. Even the theory here states the radius divides the chord into two equal parts rather than that radius to chord is always perpendicular to a chord (even it is).
It's actually completely equivalent in the calculations involved. Set the coordinate origin at the midpoint of BC, and let the centre then be at (0, Y) - we know the y-axis goes through the centre because it's the perpendicular bisector of the chord BC (the length I call Y is equal to what is x+4 in the video). A is (-3, 8); B is (-1, 0); C is (1, 0). The equation of the circle is then x^2 + (y-Y)^2 = r^2. Using point C we get: 1^2 + (0 - Y)^2 = r^2 so r^2 = Y^2 + 1 Using point A we get: 3^2 + (8-Y)^2 = r^2 so r^2 = 9 + 64 - 16Y + Y^2 = Y^2 -16Y + 73 That gives two expressions in Y for r^2, so we can equate them: Y^2 + 1 = Y^2 -16Y + 73 16Y = 72 so Y=9/2 and because r^2 = Y^2 +1, we now know that r^2 = 81/4 + 1 = 85/4 and the area is pi.r^2 = 81.pi/4 ~= 66.76 to 4 sig.fig.
I drew AC (slope -2) and line bisecting it (slope +1/2) which passes thru O then R^2 = 3^2 +3.5^2 = 21.25 thus circle area PI*R^2 = 21.25 *PI which is actually 66.75884.... not 66.73
Exactly what I did except I used AB. Got the same result as you. If we solve for r =(root85)/4 we get the same answer we obtained. Sloppy finish. Our solution is more elegant.
I always feel these sorts of lessons would be easier to remember if the example given were a practical real-world example rather than an arbitrary hypothetical.
Since for a square with sides of x the diagonal is x*2^.5 AB = ((2*2^.5 + 2*2^.5 + 2^.5)^2 + (2*2^.5 + 2^.5)^2)^.5 AB = (25*2 + 9*2)^.5 = 2*17^.5 R^2 = (AB/2)^2 + x^2 = 17 + x^2 And since... R^2 = 1^2 + y^2 = 1 + y^2 We get... y^2 = 4^2 + x^2 Therefore... x=3, y=5, and R=26^.5
I see my mistake now... My solution doesn't factor in that the center of circle is centered horizontally in the square and my choices for x and y are not the only values that work. My bad. Great video!
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So basically, math is nothing more than realising there is much more information than initially given in the question. You just have to figure out which information is hidden, what is useful, and then applying the rules ...
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@@PreMath thank u yours explanation on typical problems was very beautiful and also keeps us curious.. So i always learn from you.. I teach these problems in school and became your student. Though iam a teacher
OTHER WAY: In the vertical direction, you will have : 4*2= AD+OE= 8 AD=sqrt(r²-3²) with Pythagore in ADO OE=sqrt(r²-1²) with Pythagore in OEC So: sqrt(r²-3²)+sqrt(r²-1²)= 8 Solve it... (I elevated with ² two times), you find r²=85/4 (By the way, I also like very much @Fikri Sabit 's method in the comments !)
Set a coordinate system to be B(0,0), C(2,0) and A(-2,8) (x-a)^2+(y-b)^2=r^2 defines a circle in a rectangular coordinate system. Substitute coordinates of those 3 points in the formula of the circle. This gives you 3 equations with unknowns a, b and r. The equations are very easy to solve in a minute, so you get r^2 and accordingly area of the circle. Sorry for my poor English.
Dear Fikri, great tip for the Coordinate Geometry. I'll make a video on this problem again using this method pretty soon. By the way, your English is very good. Don't underestimate your potentials. You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃 Please keep giving feedback as often as you can...
@@PreMath I took A as origin (0,0) Then co-ordinates of B&C will be respectively (8,2) and (8,4). A, B and C are on the circumference of the circle x^2+y^2+2gx+2fy+c=0.Substituting the co-ordinates of A,B and C, and solving the equations, we get c=0, f=-3 and g=-3.5 from which, radius works out to (g^2+f^2)^1/2=1/2(85)^1/2. Area of the circle works out to 85pi/4 square units. I feel that this is a much easier solution.
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😍Thanks a lot sir for giving us such an amazing , informative video
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Once you know r^2 = 85/4, you can just leave it at that and not bother taking the square root of both sides. You're just going to use r^2 when you calculate the area of the circle.
@yo yo He proves that at 2:20 my dude.
Yes - that's a lot of unnecessary math after you have found r^2. Would have been more educational as well.
I tried it from the thumbnail using coordinate values and the formula for a circle, and got the same answer, which made my day a little happier. :)
Me too. This is the best way of solving this
How do we know that small section of side DO is equal to 1?
It is amazing how the process is fluid once you have established a line of attack. Worked it out only after seeing the two triangles. This is a great brainteaser.
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Very clever and a simpler method than I attempted. Could have been even simpler, once you had found the value of r square, you just needed to multiply by pi to get the area, no need to get the square root and then square it again.
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تمرين جميل جيد . شرح واضح مرتب .شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم. تحياتنا لكم من غزة فلسطين .
May I suggest another way (I think: simpler):
Sides of squares are 2.
Connect A to C, it will touch the bottom-left corner of the square with the center - Mark as F.
CF squared is 2²+4²=20.
Construct another square to the right of the one with the center.
Mark it's upper-right corner as G.
Connect G and F.
GF is perpendicular to CG, and will go through the center.
GF squared is also 20.
OF = 1/4 of GF.
OF squared is (1/4)² = 1/16 of GF = 20/16 = 5/4.
r² = OF² + CF² = 5/4 + 20 = 21.25 (no need to take the square root, see next).
Area = πr² = 21.25π ~ 66.759 (by calculator. using 3.14 on paper: 66.725).
I did practically the same, but between AB, I have a different comment on the details.
Thank you for your amazing video. This also trigger me to find alternative solution.
Draw a horizontal line at A, a vertical line at B and intersect circle at D
Use Chrod theorem to determine shortest chrod
Then find the diameter using pythagorean theorem of triangle DBC
And hence to get the area of the circle
Great tip! Thanks Alex for the feedback. You are awesome 👍 Take care dear and stay blessed
Excellent explanation. Marginally different here but explained by my using 22/7 as pi and calculating by hand and calling 11/14 as 0.77.
Glad it helped!
There should be additional information to be able to stablish OE as perpendicular to BC, no?
Once again, brilliant example from our Alchemist of Numbers 👍🏻
They recommend your videos and I just can’t ignore it!
A faster way would be: Consider the triangle B = (0,0), C = (2,0), A = (-2,8). Its area is Area = 8, and its sides can be calculated as a = 2, b = sqrt(80), c = sqrt(68) using the Pythagorean theorem. Then the radius of the circumscribed circle is given by a*b*c / 4*Area = sqrt(85) / 2
interesting!
Extremely beautiful sir ..so interesting 🤔🤔🤔🙏🙏🤔🙏🙏🙏🙏
Nice challenge, awesome way to solve it.
Sir, you are master of circle and rectangled triangle! 🙂
Another way:
draw x-y-coordinate-system with x axis = BC, y axis = (C-B)/2 to O →
A = A(-3;8), B = B(-1;0), C = C(1;0) → tan(φ) = 4 → f(x) = (1/4)x + 9/2 →
x = 0 → f(x) = 9/2 → 8 - (9/2) = 7/2 →(7/2)^2 + 9 = 85/4 = r^2 →
r = OA = √85/2 → πr^2 = π(85/4) ≈ 66,7588 sq units 🙂
fast lane:
Because the vertical line of the circle is between B and C →
horizontal line AA' (secant) = 6 → chords theorem: 2(4) = 8(1) →
(9/2)^2 + 1 = 85/4 = r^2 → area circle = (85/4)π
Relaxing, better than massage, thanks
This guy is not only a fantastic teacher of mathematics , but also great at giving video presentations. .. Teacher i do apologize if i asked an awkward question a few days ago. pleased forgive me. and continue as you are.
So nice of you Paul! You are awesome!!! I really appreciate that. Making quality videos is a very time consuming and meticulous task! Your devout supportive and cheerful comments keep me going... Take care dear and stay blessed😃 Keep smiling😊
I use circle equation, sir.
x²+y²+Ax+By+C=0.
The use the three point given to solve the equation. Let B(0,0) so C(2,0) and A (-2,8).
Cirlce eq: x² + y² - 2x - 9y = 0.
Then (x-1)² + (y-4½)² = 85/4.
I found two easier solutions. Both used the perpendicular radii to a chord theorem and the intersecting chords theorem. The horizontal chord from point A is 6 and the vertical chord from point C is 9. A radius through the horizontal chord bisects into two equal segments of 3. A horizontal radius to the vertical chord bisects it into two segments of 4 1/2. However, one could use the same method to find the diameter using the chord of 9 as the longer side of a right triangle.
The horizontal chord measurement for point A is correct and can be explained by symmetry, but how do you get the vertical chord measurement for C? How do you say it equals 9?
you can't assume DO = 3 if you're not told that O is in the middle (horizontally) of that square
I love Pythagoras but coordinate points and the circle formula is easier:
T1(1,0), T2(-1,0), T3(3,8) , O(0,k) -> h=0 (because h is between -1 and 1)
1² + k² = r²
3² + (8-k)² = r²
1 + k² = 9 + 64 - 16k + k²
16k = 72, k = 9/2
r² = 81/4 +1, r² = 85/4
r = √(85/4), A= 85/4 π
Very useful problems. Thank you very much.
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Good explanation 👍
Glad you liked it!
Make a line with connecting A and C. This line passes through one corner of the square which contains O. Then call that corner D. And also call a right corner the square E. Make a vertical line from O to line DE and call the crossing point F. Now preparation has been completed. We can find a right triangle ADO. AD=2 x square root 5. We can find DF : OF=2:1. So, DO x DO = 1x1 + (1/2)x(1/2)=(5/4). AS AO x AO=AD x AD + DO x DO=20 + (5/4) = (85/4). Area of the circle is (85/4) x pi. Anyway sorry for my poor English.
Delicate solution
Thank you 👍
Very good explanation sir. I like very much the way you teach the math problems
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@@PreMath 👍👍👍
at 7:24 how do you know that short side of segment DO is one??? It seems you are imparting "information" from eyeballing the diagram.
I follow the explanation you have given but I am curious about the point you have marked as O. I think you may have assumed it to be the centre of the circle but it doesn't say that in the question. If the point O is not at the centre, you can't draw a line to bisect the chord. Am I missing something?
"O" is a completely arbitrary point if you don't assume it to be the centre, with no restrictions on its position whatsoever. If you prefer, you could simply create another point defined as the centre and call it "M", then follow the method shown.
Note that the positioning of the centre on the diagram is unimportant. If the centre is actually below the top of the second box, then x will result in a negative value.
Personally, I found it easier to use a different method that doesn't require the centre. Defining point B as your datum (0,0) allows you to know the coordinates of three points on the circle - A (-2,8), B (0,0) and C (2,0). You can then solve simultaneous equations by plugging these x and y values into the formula of a circle - (x-a)^2 + (y-b)^2 = r^2
@@adamrobinson6951 Thank you for your reply. My point was that O is ASSUMED to be the centre of the circle but this assumption wasn't stated. If it was not the centre, you wouldn't get a right-angled triangle at the intersection with the line so you wouldn't be able to use pythagoras as shown in the demonstration.
I agree that other methods would work just as easily, and possibly even easier, but my question is about the assumption that O is at the centre of the circle. Am I missing the proof that O is indeed the centre?
@@ApiaryManager O isn't specified to be in the centre. Therefore, there are no constraints whatsoever on O, and it's positioning will neither affect nor be affected by the geometry of the rest of the question.
It is, however, helpful to draw lines to the centre of the circle when solving this problem. If you prefer, you could label the centre using some other term and ignore the existence of O entirely. In this case, O was defined as the centre by the problem solver because doing so didn't affect anything else in the question.
Note: this does mean that the position of O in the diagram may potentially be misleading, as the centre may fall in a different square or outside the squares entirely. However, in that case the maths would correct for that issue by returning negative values for our geometric variables.
Why not using the rule in any rectangle
Ab/sin(c) =2R?
Sin(c) =8/AC
Ac calculates by phithagurth
(4 and 8)
AB is calculates by phithagurth
(2 and 8)
And it is done without any extra complication
beautiful method!
You are good tetcher
Let A = (0,8), B = (2,0) C = (4,0) and O = (3,k)
OA = OB can find k. Then radius can be found.
Good one!
Means a lot ❤
What do you mean when you say A, B and C are points of Tangency ?! You could have just said 'they are points on the circle'
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How do you presume that O is the center?
That needs to be declared In a given.
Amazing explanation👍
Thank you so much for sharing this great video😀
Thank you so much for your continued love and support. Take care dear and stay blessed😃 You are awesome. Keep smiling😊 Enjoy every moment of your life 🌻
If you already had r^2, why take the square root to just turn around and square it again in the final step?
watched the video and liked it
My solution:
Put the figure into (x, y) coordinates.
A:: x=0
B and C :: y=0
A = (0, 8) ; B = (2, 0 ) ; C = (4, 0)
Middle on BC :: E = (3, 0)
90 degrees on BC at E, to center of circle O
O :: x = 3
Middle on AB :: M = (1, 4)
x-difference M to O = +2
AB-> = [ -8 , 2 ]
Turn 90 degree = [ 8, 2 ]
Scale to x=2 ==> [ 2 , 0.5 ]
O = M + [ 2 , 0.5 ] = ( 1+2 , 4+0.5) = (3 , 4.5)
BO-> = [1 , 4.5]
r^2 = | BO-> |^2 = (1 + 4.5^2) = 21.25
Area = PI × r^2 = PI × 21.25
😍amazing
Love it!
Thank you sir
You are very welcome Gowri!
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There is a presumption to your solution i think. Do we have proof that the 3 bottom squares are colinear
Excellent prooof
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Brilliant!
Once r sq = 85/4 then striaght away area = pixrsq so pix85/4
The solution is correct under the assumption that point O is located horizontally in the middle of the square. And this is not necessary!
how you know that BE=1 ??????????? and the point C is exactly in the center of B and C ?????
What does a wrong triangle look like? Off centre maybe?
So. We are assuming that the centre of the circle O is the middle of the square???
O is centered left-to-right in the square because the left and right sides of the square it's in are colinear with the left and right sides of the bottom square and B and C are points on the circle. O is equidistant from B and C because... circle. While that's far from a water-tight mathematical proof, it should be enough to convince you that O is centered left-to-right in its square.
I see your point, I think it's not conventional to assume positions of points in a graphic if it's not directly expressed in measurements. Perhaps the drawing should've demonstrated 'O' was 1 unit away from both the left and right side of that square.
@@bjorn6084 If you draw the perpendicular bisector of the chord BC, you know it must pass through the centre of the circle.
Extending downwards the two vertical sides of the square containing the centre will produce two parallel lines one of which must pass through B, and the other must pass through C because the parallel lines must remain 2 units apart, by definition.
Since the perpendicular bisector of the chord BC is 1 unit away from B and from C (because it's a bisector), it must remain equidistant from the two outer lines we constructed because it is parallel to both (because the angle each of them make with BC is a right angle).
That shows that the centre of the square is equidistant from the vertical sides of the square containing it.
This problem can be solved by the intersecting chord theorem, the procedure is simpler
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@@italixgaming915 This solution is easier than the one shown in the video?
Amazing solution. Thank you, Sir.
Ttank for premath
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Mind blown.
OE=8 so,r=8 and the area of the circle is r*r*3,14=200,96
There is something..
How to check that OEC is right triangle?
It looks proven with the Chord theorem.
Even the theory here states the radius divides the chord into two equal parts rather than that radius to chord is always perpendicular to a chord (even it is).
3^2 + 4^2 = c^2
9 + 16 = 25
c^2 = 25
c = Square root of 25 = 5
Using coordinate geometry seems a simpler way to solve this problem.
Good tip
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It's actually completely equivalent in the calculations involved. Set the coordinate origin at the midpoint of BC, and let the centre then be at (0, Y) - we know the y-axis goes through the centre because it's the perpendicular bisector of the chord BC (the length I call Y is equal to what is x+4 in the video).
A is (-3, 8); B is (-1, 0); C is (1, 0). The equation of the circle is then x^2 + (y-Y)^2 = r^2.
Using point C we get:
1^2 + (0 - Y)^2 = r^2
so r^2 = Y^2 + 1
Using point A we get:
3^2 + (8-Y)^2 = r^2
so r^2 = 9 + 64 - 16Y + Y^2 = Y^2 -16Y + 73
That gives two expressions in Y for r^2, so we can equate them:
Y^2 + 1 = Y^2 -16Y + 73
16Y = 72 so Y=9/2 and because r^2 = Y^2 +1, we now know that r^2 = 81/4 + 1 = 85/4 and the area is pi.r^2 = 81.pi/4 ~= 66.76 to 4 sig.fig.
Draw another square to the right corner of square 'A'. The solution will be two lines, without any 'x'
I drew AC (slope -2) and line bisecting it (slope +1/2) which passes thru O then R^2 = 3^2 +3.5^2 = 21.25 thus circle area PI*R^2 = 21.25 *PI which is actually 66.75884.... not 66.73
Exactly what I did except I used AB. Got the same result as you.
If we solve for r =(root85)/4 we get the same answer we obtained. Sloppy finish.
Our solution is more elegant.
I always feel these sorts of lessons would be easier to remember if the example given were a practical real-world example rather than an arbitrary hypothetical.
For some people, that would be true.
@@MarieAnne. Myself included.
pi*r^2=Circle’s area
Can you correct my approach... I get 26 Pi.
Since for a square with sides of x the diagonal is x*2^.5
AB = ((2*2^.5 + 2*2^.5 + 2^.5)^2 + (2*2^.5 + 2^.5)^2)^.5
AB = (25*2 + 9*2)^.5 = 2*17^.5
R^2 = (AB/2)^2 + x^2 = 17 + x^2
And since...
R^2 = 1^2 + y^2 = 1 + y^2
We get...
y^2 = 4^2 + x^2
Therefore...
x=3, y=5, and R=26^.5
I see my mistake now... My solution doesn't factor in that the center of circle is centered horizontally in the square and my choices for x and y are not the only values that work. My bad. Great video!
Right triangle
not enough information
👍
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So basically, math is nothing more than realising there is much more information than initially given in the question. You just have to figure out which information is hidden, what is useful, and then applying the rules ...
20,675
neat
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You’ve made the assumption that all the squares are perpendicular to each other which is not one of the givens.
Parecia difícil.
5
No. I can not find the area of the circle.
32π
Super problem
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@@PreMath thank u yours explanation on typical problems was very beautiful and also keeps us curious.. So i always learn from you.. I teach these problems in school and became your student. Though iam a teacher
Pythagorean Theorem
Nah, just guess it, I'm not a rocket scientist
c = 5
OTHER WAY:
In the vertical direction, you will have : 4*2= AD+OE= 8
AD=sqrt(r²-3²) with Pythagore in ADO
OE=sqrt(r²-1²) with Pythagore in OEC
So: sqrt(r²-3²)+sqrt(r²-1²)= 8
Solve it... (I elevated with ² two times), you find r²=85/4
(By the way, I also like very much @Fikri Sabit 's method in the comments !)
Pythagoras
Calculating r is useless, as such.
YeaH ---- diaMetre squarE --- buT main denoTioN oF toughesT ParT is finding radiuS --- CaN You! Everywhere whY comParison isnT on maiN parT !
Set a coordinate system to be B(0,0), C(2,0) and A(-2,8)
(x-a)^2+(y-b)^2=r^2 defines a circle in a rectangular coordinate system.
Substitute coordinates of those 3 points in the formula of the circle. This gives you 3 equations with unknowns a, b and r.
The equations are very easy to solve in a minute, so you get r^2 and accordingly area of the circle.
Sorry for my poor English.
Dear Fikri, great tip for the Coordinate Geometry. I'll make a video on this problem again using this method pretty soon.
By the way, your English is very good. Don't underestimate your potentials. You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
Please keep giving feedback as often as you can...
@@PreMath I took A as origin (0,0) Then co-ordinates of B&C will be respectively (8,2) and (8,4). A, B and C are on the circumference of the circle x^2+y^2+2gx+2fy+c=0.Substituting the co-ordinates of A,B and C, and solving the equations, we get c=0, f=-3 and g=-3.5 from which, radius works out to (g^2+f^2)^1/2=1/2(85)^1/2. Area of the circle works out to
85pi/4 square units. I feel that this is a much easier solution.
shouldnt get square root of r^2, useless operation
h=√(R²-1)-4 √130/3=R
2√24+√(7²-5²)=6√6
7x+80,5=210 x=129,5/7=18,5
(3+√2+√3+√5(/(3+√2+√3+√5+√2(3+√2+√ 3+√5)(=1/(1+√2)=√2-1
daddy
Красивая задачка.
Ttank for premath
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