Lovely work, Professor!🥂❤

@PreMath

Жыл бұрын

Thank you! Cheers! Glad you think so! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

@bigm383

Жыл бұрын

@@PreMath 😀🥂🍺

@TDSONLINEMATHS

Жыл бұрын

Nice

There’s a lot of people making math videos on KZread 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.

Muito bom excelente aula parabéns

Very inspiring ! Would like to have more with brain teasing stuffs.

This one is easy. I solved! Thank you!

Boa Tarde Obrigado pelos Ensinamentos Deus Lhe Abençoe

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.

@PreMath

Жыл бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

@Mediterranean81

17 күн бұрын

same

Think you! It's the beauty and simple problem!

@PreMath

Жыл бұрын

Glad to hear that! Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

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.

@PreMath

Жыл бұрын

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 😀

@theoyanto

Жыл бұрын

Nice, that's one to remember, thanks👍🏻

Great explanation👍 Thanks for sharing😊😊

@PreMath

Жыл бұрын

My pleasure 😊 You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

This was a fun one.

Love and respect from Bangladesh

@PreMath

Жыл бұрын

So nice of you, dear Thanks for your continued love and support! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

Thanks for video.Good luck sir!!!!!!!!!!

@PreMath

Жыл бұрын

So nice of you Thanks for your continued love and support! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

Un problema apparentemente irrisolvibile brillantemente risolto. Complimenti!

@PreMath

Жыл бұрын

Sono contento che tu la pensi così! Grazie per il tuo feedback! Saluti! Sei fantastico. Continua così 👍 Amore e preghiere dagli Stati Uniti! 😀

Well explain sir

@PreMath

Жыл бұрын

Keep watching Thank you! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

Thankyou so much sir 🙏for your hardwork 🙏

@PreMath

Жыл бұрын

So nice of you. Thanks for your continued love and support! You are awesome, Rishu. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

Splendiferous, fantasmagorical, even though I make many errors, these problems are such great fun 😊👍🏻

@PreMath

Жыл бұрын

Glad you enjoyed it! Thanks for your feedback! Cheers! You are awesome, Ian. Keep it up 👍 Love and prayers from the USA! 😀

Very nice.

@PreMath

Жыл бұрын

Glad you think so! Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

Extremely beautiful ❤❤😊😊

@PreMath

Жыл бұрын

Glad you think so! Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

Good morning Sir

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

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.

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)

@PreMath

Жыл бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

I used a CAD program to solve this. Maybe it wasn't as precise but it was mighty close, within a rounding error.

Very easy question it's also in our maths module

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)

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

@PreMath

Жыл бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

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.

@PreMath

Жыл бұрын

Excellent!! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

Intersecting chords theorem can also apply here

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 ;-)

@triplem1812

Жыл бұрын

3 is impossible, just compare the length of the 3 with the length to the radius...

@-Seaheart-

Жыл бұрын

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

(r-3)^2+(r-6)^2=r^2, r^2-18r+45=0, (r-3)(r-15)=0, r=3,rejected, so r=15, the area is 225pi=706.5 approximately, done 😊

@PreMath

Жыл бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

circle of r=3 would have center 3 units left of A point

if u think this in graph paper it will be more easier to think

formula of a circle is Area=pie×radius 😊😊😊😊😊

707.14 here, but that's only because I did it on paper using 22/7 as pi.

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.

@b89john

Жыл бұрын

The area of course is 9*Pi and no cm^2🙃

@normanc918

Жыл бұрын

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.

@b89john

Жыл бұрын

@@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.

Math good.❤❤❤❤❤❤❤❤❤❤❤❤😂😂❤😂❤😂

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 ⭕ 👍

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...😂

r=15

@PreMath

Жыл бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

(2r-9)3=(r-6)^2; r= 15; A= πr^2= π15^2= 225π

@PreMath

Жыл бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

@alexniklas8777

Жыл бұрын

​@@PreMath Sir, look at my solution to the previous problem. You will have another way to solve this problem. Thanks sir! ❤

@PreMath

Жыл бұрын

@@alexniklas8777 Thanks, Alex. I'll do that as well. Keep rocking 👍

@alexniklas8777

Жыл бұрын

@@PreMath good 🤝

👉👈

Op

@PreMath

Жыл бұрын

Thank you! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀