Horizontal chord from A and vertical diameter gives (by symmetry and intersecting chords theorem) 3·(2r−3) = (r−6)², which resolves to the same quadratic equation. Root r = 3 corresponds to degenerate case when red rectangle is a half of the square.
I always like to look at the alternative value given by the quadratic equation, rather than just discount it out of hand. The outer square has a side length of 2xR so if we look at that value of 3=R (that you discounted as not possible) we would have a square with side lengths of 6. So if you were to draw the square with side lengths of 6 you'd find that it fits over the brown rectangle perfectly, with the rectangle occupying exactly the top half of the square. If we were to draw an inscribed circle in that square, the bottom corner of the brown rectangle would, indeed, lie on the circle. So this is what that alternative value is telling us. So it's not that the value of 3 is wrong, it's just telling you an alternative solution. (albeit not fulfilling all of the parameters of the original problem)
It’s interesting how two different thought processes lead to the same equation. For this problem I plotted the circle in the x-y plane so the bottom left of the square is the origin. Then by symmetry and the given rectangle, we know one of the points of the circle is (6,3) We also see the circle is shifted to the right of the origin by r and also up by r. So the equation for the circle would be (x-r)^2 + (y-r)^2 = r^2 Using the point (6,3) leads to the same quadratic equation for r which gives r=3 and r=15. Rejecting 3 for the same reason and leading to 225pi for the area.
Many approaches are possible to find the solution to this problem! Excellent! Thanks for sharing! Cheers! You are awesome, Nate. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀
It's a bit easier with coordinategeometry: we are looking for a circle with one point at 6;3. So the circle is (x-r)^2+(y-r)^2=r^2, and x=6, y=3. We have to find r: (6-r)^2+(3-r)^2=r^2 => r^2-18r+45=0. We have two solutions: 3, 15; 3 rejected, the radius is 15 => area is 225pi.
I took a slightly different route to get the same result. Using the formula for a circle (x-a)^2+(y-b)^2=r^2. Then plugging in the three points: (6,-3), (r,0), (0,-r). Then realizing that |a|=|b|=|r|. Lets us solve for r=15.
Draw horizontal from BC to left parallel to top side. Draw a perpendicular AD from A to BC. Let R be the radius. AC sq = ADsq + CDSq. Rsq = (R - 3)sq + (R - 6)sq. Solve for R. R = 9 +/- 6. Reject 3. R = 15. Area = pi*225 = 706.85.
R= 3 is not impossible, it signifies radius of the little circle inscribed in the red box and which meets the requirements stated. You cannot fool mathematics - it insists on being consistent and it couldn't possibly know which of the two possible triangles you were thinking of ;-)
It’s impossible for the stated arrangement in which the circle must be inscribed in the square while being tangent to the point A and to me it also seems impossible mathematically too, since it would imply a triangle with only one of its sides equal to 0
I think the result r=3 is too easily rejected. When r=3, the rectangle is exactly half the area of the square: length 6cm (diameter) and height 3cm (radius). The area of the circle is 9cm^2. While the solution does not accord with your diagram, it is nonetheless valid.
@@normanc918 the numbers 3 and 6 refer to the proportions of the rectangle not to a coordinate. Using coordinates with the bottom left corner of the square being (0,0) is a good way of approaching the problem leading to an equation for the circle radius r of (x-r)² + (y-r)² = r² which means the corner of the rectangle touching the circle in the diagram is (2r -6, 2r-3) or when r=15 (24,27). When r=3 the corner of the rectangle touching the circle is at (6,3) which for it to be a point on the circle will satisfy the equation (x-r)² + (y-r)² = r², or (x-3)² + (y-3)² =9. As you can see this point is on the circle and it is also a corner of the rectangle. Just not the same corner. The problem does not specify a corner in contact with the circle so there are two valid answers hence my observation that the solution r=3 was dismissed without a clear explanation as to why.
This one is obviously simple, just Pythagoras with (r-6)² + (r-3)² = r² I'll give that to my students next time and see if they write down *both* solutions or if they check that r = 3 isn't possible...😂
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Lovely work, Professor!🥂❤
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@@PreMath 😀🥂🍺
Nice
There’s a lot of people making math videos on RUclips and after some time of watching different people, I think you might be the best.
Horizontal chord from A and vertical diameter gives (by symmetry and intersecting chords theorem) 3·(2r−3) = (r−6)², which resolves to the same quadratic equation. Root r = 3 corresponds to degenerate case when red rectangle is a half of the square.
I always like to look at the alternative value given by the quadratic equation, rather than just discount it out of hand. The outer square has a side length of 2xR so if we look at that value of 3=R (that you discounted as not possible) we would have a square with side lengths of 6. So if you were to draw the square with side lengths of 6 you'd find that it fits over the brown rectangle perfectly, with the rectangle occupying exactly the top half of the square. If we were to draw an inscribed circle in that square, the bottom corner of the brown rectangle would, indeed, lie on the circle. So this is what that alternative value is telling us.
So it's not that the value of 3 is wrong, it's just telling you an alternative solution. (albeit not fulfilling all of the parameters of the original problem)
This one is easy. I solved! Thank you!
Un problema apparentemente irrisolvibile brillantemente risolto. Complimenti!
Sono contento che tu la pensi così!
Grazie per il tuo feedback! Saluti!
Sei fantastico. Continua così 👍
Amore e preghiere dagli Stati Uniti! 😀
Muito bom excelente aula parabéns
Very inspiring ! Would like to have more with brain teasing stuffs.
Think you! It's the beauty and simple problem!
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Thanks for video.Good luck sir!!!!!!!!!!
So nice of you
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Boa Tarde
Obrigado pelos Ensinamentos
Deus Lhe Abençoe
It’s interesting how two different thought processes lead to the same equation.
For this problem I plotted the circle in the x-y plane so the bottom left of the square is the origin.
Then by symmetry and the given rectangle, we know one of the points of the circle is (6,3)
We also see the circle is shifted to the right of the origin by r and also up by r.
So the equation for the circle would be (x-r)^2 + (y-r)^2 = r^2
Using the point (6,3) leads to the same quadratic equation for r which gives r=3 and r=15. Rejecting 3 for the same reason and leading to 225pi for the area.
Many approaches are possible to find the solution to this problem!
Excellent!
Thanks for sharing! Cheers!
You are awesome, Nate. Keep it up 👍
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Stay blessed 😀
Nice, that's one to remember, thanks👍🏻
Nice and awesome, many thanks, Sir!
sin(φ) = (a - 6)/a
cos(φ) = (a - 3)/a
sin^2(φ) + cos^2(φ) = 1 → (a - 6)^2 + (a - 3)^2 = a^2 → a = 15 → πa^2 = 225π
btw: sin(φ) = 3/5 → ∆ = pyth. triple (9-12-15)
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Thankyou so much sir 🙏for your hardwork 🙏
So nice of you.
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Well explain sir
Keep watching
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It's a bit easier with coordinategeometry: we are looking for a circle with one point at 6;3. So the circle is (x-r)^2+(y-r)^2=r^2, and x=6, y=3. We have to find r: (6-r)^2+(3-r)^2=r^2 => r^2-18r+45=0. We have two solutions: 3, 15; 3 rejected, the radius is 15 => area is 225pi.
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same
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So nice of you, dear
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Great explanation👍
Thanks for sharing😊😊
My pleasure 😊
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This was a fun one.
I used a CAD program to solve this. Maybe it wasn't as precise but it was mighty close, within a rounding error.
Splendiferous, fantasmagorical, even though I make many errors, these problems are such great fun 😊👍🏻
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Good morning Sir
I took out my dividers and found that 2 units of 6 plus 1 unit of 3 made the radius of 15. The rest was easier.
Intersecting chords theorem can also apply here
Very easy question it's also in our maths module
After seeing the Pythagorean Theorem for solving the radius, I knew where this is going so I knew the answer before I could finish the video
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Applying intersecting chords rule we get
3*(2r-3) = (r-6)*(r-6)
Simplifying this
r^2-15r+45=0
r=3 or 15
Area 225pi
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I took a slightly different route to get the same result. Using the formula for a circle (x-a)^2+(y-b)^2=r^2. Then plugging in the three points: (6,-3), (r,0), (0,-r). Then realizing that |a|=|b|=|r|. Lets us solve for r=15.
Draw horizontal from BC to left parallel to top side. Draw a perpendicular AD from A to BC.
Let R be the radius. AC sq = ADsq + CDSq. Rsq = (R - 3)sq + (R - 6)sq.
Solve for R. R = 9 +/- 6. Reject 3. R = 15. Area = pi*225 = 706.85.
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Extremely beautiful ❤❤😊😊
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R= 3 is not impossible, it signifies radius of the little circle inscribed in the red box and which meets the requirements stated. You cannot fool mathematics - it insists on being consistent and it couldn't possibly know which of the two possible triangles you were thinking of ;-)
3 is impossible, just compare the length of the 3 with the length to the radius...
It’s impossible for the stated arrangement in which the circle must be inscribed in the square while being tangent to the point A and to me it also seems impossible mathematically too, since it would imply a triangle with only one of its sides equal to 0
I think the result r=3 is too easily rejected. When r=3, the rectangle is exactly half the area of the square: length 6cm (diameter) and height 3cm (radius). The area of the circle is 9cm^2. While the solution does not accord with your diagram, it is nonetheless valid.
The area of course is 9*Pi and no cm^2🙃
If r =3, then point A (3,6) cannot exist. For point A to be there, r must be more than 6. So r=15 must be true.
@@normanc918 the numbers 3 and 6 refer to the proportions of the rectangle not to a coordinate. Using coordinates with the bottom left corner of the square being (0,0) is a good way of approaching the problem leading to an equation for the circle radius r of (x-r)² + (y-r)² = r² which means the corner of the rectangle touching the circle in the diagram is (2r -6, 2r-3) or when r=15 (24,27). When r=3 the corner of the rectangle touching the circle is at (6,3) which for it to be a point on the circle will satisfy the equation (x-r)² + (y-r)² = r², or (x-3)² + (y-3)² =9. As you can see this point is on the circle and it is also a corner of the rectangle. Just not the same corner. The problem does not specify a corner in contact with the circle so there are two valid answers hence my observation that the solution r=3 was dismissed without a clear explanation as to why.
formula
of a circle is Area=pie×radius
😊😊😊😊😊
circle of r=3 would have center 3 units left of A point
(2r-9)3=(r-6)^2; r= 15; A= πr^2= π15^2= 225π
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@@PreMath Sir, look at my solution to the previous problem. You will have another way to solve this problem.
Thanks sir! ❤
@@alexniklas8777 Thanks, Alex. I'll do that as well. Keep rocking 👍
@@PreMath good 🤝
if u think this in graph paper it will be more easier to think
I Paused The Video.
Counting by hand, pen & paper.
(R-6)" + (R-3)" = R"
R" - 18R + 45 = 0 ➡ skip, skip
(R-15)•(R - 3) = 0 ➡ R = 15 ✅
A = π•R" = 225•π = 706,86 sq units
My phi is *3,1416*
--- --- ---
PS: duration could be shorter -2'
NB: Nice Maths Problem ⭕ 👍
707.14 here, but that's only because I did it on paper using 22/7 as pi.
r=15
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Math good.❤❤❤❤❤❤❤❤❤❤❤❤😂😂❤😂❤😂
This one is obviously simple, just Pythagoras with (r-6)² + (r-3)² = r²
I'll give that to my students next time and see if they write down *both* solutions or if they check that r = 3 isn't possible...😂
Op
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