Keep watching... I call these math puzzles "gymnastics for the mind!" They make us think and improve mental agility! We are all lifelong learners! Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍
Very clever of you to construct the 2 triangles ΔDPH and EQF congruent to right triangle ΔHBE! That construction provides the key to solving the problem. However, I noticed a couple shortcuts that were missed. At 5:00, we focus on ΔDPB. We could, perhaps should, recognize that this is the familiar 3-4-5 right triangle, with side lengths multiplied by 3, and quickly determine that length a = 12. At 8:45, we could observe that ΔAQF is an isosceles right triangle, so its hypotenuse, AF, is side times √2, or 3√2. Another observation: At 10:00, the green square is back but ΔAQF does not show. Actually, point Q is on line segment AB. We could divide the green square into 4 congruent triangles, one of them ΔAQF. The area of ΔAQF is (1/2)(3)(3) = 4.5 and 4 of them have total area 18. The caution that "This diagram may NOT be 100% be 100% true to the scale!" is well founded. If it were true to the scale, point Q would be on AB, but that would provide additional information to help solve the problem!
@@ybodoN Proportion of areas of squares solution. That's a fascinating problem and here is my solution. Let me use the term "ratio" instead of "proportion". As a preface, ratio of areas of PreMath's squares, 90 and 18, is 90/18 = 5 However, let's solve the general case. Let the length of equal line segments AQ, FQ, BE and PH be designated b and the length of equal line segments EQ, BH and PD be designated a. Let the quarter circle have radius r. From line segment AB, r = a + 2b. From ΔBPD, r² = (a + b)² + a² = a² + 2ab + b² + a² = 2a² + 2ab + b². From r = a + 2b, squaring both sides, r² = a² + 4ab + 4b², which we substitute into the second equation: a² + 4ab + 4b² = 2a² + 2ab + b² Consolidating terms to the left side: -a² + 2ab + 3b² = 0 Divide both sides by -b²: (a/b)² - 2(a/b) - 3 = 0 Let a/b = x and solve using the quadratic formula: x =( 2 +/- √(4 -(4)(1)(-3))/2 = (2 +/- √(4 + 12))/2 = (2 +/- √(16))/2 = (2 +/- 4)/2 producing two solutions, x = -1 and x = 3. Distances must be positive, so we discard x = -1 So x = a/b =3 and a = 3b. Now let the length of the side of the square be designated s. From ΔFQE, s² = a² + b². Replace a by 3b and s² = (3b)² + b² = 10b². Now examine the small square. It can be divided into 4 isosceles right triangles with sides of length b, so its area is 4(1/2)(b)(b) = 2b². So, s², the area of the large square, is 5(2b²), or 5 times the area of the small square.
1/ Draw the full circle and extend DP to the right cutting the circle at point D'. P is the midpoint of PP' and just label CP as x we have x(2R-x)=sq DP (chord theorem)------> x(30-x)=81------>sqx-30x+81=0 -----> x=3----->PH=3 The area of the yellow square= sq9+sq3=81+9=90 sq units 2/ AQ=FQ=3 so AF=3sqrt of 2------> the area of of the green square= 9x2=18 sq units
Thank you, sir, for this analysis. There is a note about the position of the green triangle in the figure. In line with the detected results, the straight line (AB) must pass through the diameter of the green square. Therefore, the position of the green triangle in the figure is wrong. I am a fan of this channel, and thank you, sir, for this wonderful channel.
@@PreMathTo be honest, this diagram threw me off for the longest time. If it were aligned with a diagonal, I would have got the without having wasted three days on this easy problem. Please correct mistakes like this as soon as you find it. I don't want to waste problems like this where you don't diagram properly! If I find that this happens again, I am unsubscribing from your channel.
Draw a line BD, which is the radius 15. DP is 9 by symmetry. By Pythagoras BP is 12 (144+81=225). Thus PH is 3 and the sidelength of the yellow square is sqrt(9+81), which gives us the area of the yellow square at 90. The sidelength of the green square is 3*sqrt(2) (as AQ and FQ are both 3), the Area is thus 9*2=18. Piece of cake!
@ 0:26 ... Of course I am just kidding, ... but first observation , how many Yellow and Green squares are there? I only see one of each because back in August when a Dallas woman (Tiffany Gomaz) claimed that a passenger at the back of an American Airlines flight was not real , I would've been off the plane right behind her! 🙂
Heart blowing concept sir ❤❤❤
Please teach me how to learn like this while solving such twisted questions to solve it.
Keep watching...
I call these math puzzles "gymnastics for the mind!" They make us think and improve mental agility! We are all lifelong learners!
Thanks for your feedback! Cheers! 😀
You are awesome. Keep it up 👍
@ 1:40 , everything falls into place once complementary angles Alpha and Beta are determined. ...Soooooooo cool! 🙂
Excellent!
Thanks for your feedback! Cheers! 😀
You are awesome. Keep it up 👍
Fun problem!
Thanks for your feedback! Cheers! 😀
You are awesome. Keep it up 👍
Very clever of you to construct the 2 triangles ΔDPH and EQF congruent to right triangle ΔHBE! That construction provides the key to solving the problem. However, I noticed a couple shortcuts that were missed. At 5:00, we focus on ΔDPB. We could, perhaps should, recognize that this is the familiar 3-4-5 right triangle, with side lengths multiplied by 3, and quickly determine that length a = 12. At 8:45, we could observe that ΔAQF is an isosceles right triangle, so its hypotenuse, AF, is side times √2, or 3√2.
Another observation: At 10:00, the green square is back but ΔAQF does not show. Actually, point Q is on line segment AB. We could divide the green square into 4 congruent triangles, one of them ΔAQF. The area of ΔAQF is (1/2)(3)(3) = 4.5 and 4 of them have total area 18.
The caution that "This diagram may NOT be 100% be 100% true to the scale!" is well founded. If it were true to the scale, point Q would be on AB, but that would provide additional information to help solve the problem!
Very nice solution, maybe the fastest one!
With point Q on AB, the question could be "in what proportion are the areas of the two squares", without giving the measure of any length. 😉
@@ybodoN Proportion of areas of squares solution. That's a fascinating problem and here is my solution. Let me use the term "ratio" instead of "proportion". As a preface, ratio of areas of PreMath's squares, 90 and 18, is 90/18 = 5 However, let's solve the general case. Let the length of equal line segments AQ, FQ, BE and PH be designated b and the length of equal line segments EQ, BH and PD be designated a. Let the quarter circle have radius r. From line segment AB, r = a + 2b. From ΔBPD, r² = (a + b)² + a² = a² + 2ab + b² + a² = 2a² + 2ab + b². From r = a + 2b, squaring both sides, r² = a² + 4ab + 4b², which we substitute into the second equation:
a² + 4ab + 4b² = 2a² + 2ab + b²
Consolidating terms to the left side:
-a² + 2ab + 3b² = 0
Divide both sides by -b²:
(a/b)² - 2(a/b) - 3 = 0
Let a/b = x and solve using the quadratic formula:
x =( 2 +/- √(4 -(4)(1)(-3))/2 = (2 +/- √(4 + 12))/2 = (2 +/- √(16))/2 = (2 +/- 4)/2 producing two solutions, x = -1 and x = 3. Distances must be positive, so we discard x = -1
So x = a/b =3 and a = 3b. Now let the length of the side of the square be designated s. From ΔFQE, s² = a² + b². Replace a by 3b and s² = (3b)² + b² = 10b². Now examine the small square. It can be divided into 4 isosceles right triangles with sides of length b, so its area is 4(1/2)(b)(b) = 2b². So, s², the area of the large square, is 5(2b²), or 5 times the area of the small square.
Thanks for your feedback! Cheers! 😀
You are awesome. Keep it up 👍
1/ Draw the full circle and extend DP to the right cutting the circle at point D'. P is the midpoint of PP'
and just label CP as x
we have x(2R-x)=sq DP (chord theorem)------> x(30-x)=81------>sqx-30x+81=0 -----> x=3----->PH=3
The area of the yellow square= sq9+sq3=81+9=90 sq units
2/ AQ=FQ=3 so AF=3sqrt of 2------> the area of of the green square= 9x2=18 sq units
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Thank you, sir, for this analysis. There is a note about the position of the green triangle in the figure. In line with the detected results, the straight line (AB) must pass through the diameter of the green square. Therefore, the position of the green triangle in the figure is wrong. I am a fan of this channel, and thank you, sir, for this wonderful channel.
Thanks for your feedback! Cheers! 😀
You are awesome. Keep it up 👍
@@PreMathTo be honest, this diagram threw me off for the longest time. If it were aligned with a diagonal, I would have got the without having wasted three days on this easy problem.
Please correct mistakes like this as soon as you find it. I don't want to waste problems like this where you don't diagram properly!
If I find that this happens again, I am unsubscribing from your channel.
I figured it out.
Thanks!
Great!
Thanks for your feedback! Cheers! 😀
You are awesome. Keep it up 👍
Draw a line BD, which is the radius 15. DP is 9 by symmetry. By Pythagoras BP is 12 (144+81=225). Thus PH is 3 and the sidelength of the yellow square is sqrt(9+81), which gives us the area of the yellow square at 90. The sidelength of the green square is 3*sqrt(2) (as AQ and FQ are both 3), the Area is thus 9*2=18. Piece of cake!
Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Sir can you please tell me what are the tags you use in your videos please 😢
Hello dear, please look at right below description of the video...
@@PreMath thank you sir
I was on the right path but ended up with one equation 2 unknowns. I plugged in your answers and it worked out.
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Καλησπέρα σας. Με το GEOGEBRA, η 3η κορυφή του πράσινου τετραγώνου είναι σημείο του ΑΒ. Νομίζω ότι πρέπει να διορθωθεί. Ευχαριστώ.
@ 0:26 ... Of course I am just kidding, ... but first observation , how many Yellow and Green squares are there? I only see one of each because back in August when a Dallas woman (Tiffany Gomaz) claimed that a passenger at the back of an American Airlines flight was not real , I would've been off the plane right behind her! 🙂
Thanks for your feedback! Cheers! 😀
You are awesome. Keep it up 👍
90...18
Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍