Thank you! Appreciate the step-by-step instructions!

@PreMath

3 ай бұрын

Glad it was helpful! You are very welcome! Thanks ❤️🌹

I look forward to more exciting videos . ...And each day they have been progressively more exciting. I'm stoked! 🙂

@PreMath

3 ай бұрын

Glad to hear it! Excellent! Thanks ❤️

Very methodical manipulation. Thanks Professor!

@PreMath

3 ай бұрын

Glad you liked it! You are very welcome! Thanks ❤️

Well done!!

Excellent work,you made mathematics enjoyable

شكرا لكم على المجهودات . يمكن استعمال t=OF وAB=x وy ضلع الأزرق وzضلع الأصفر z=x-y x^2+(x-t)^2=26^2 x2+(y+t)^2=26^2 (x-y)^2+(x-y+t)^2=26^2 الفرق بين المعادلة الأولى والثانية نجد x-y=2t بالتعويض في الثالثة نجد t^2=52 والتعويض في الأولى نجد x^2=468 وفي الاخير y^2=208 z^2=52

Let a,b,c be the radii of green, yellow and blue squares, 26^2=a^2+(a+c)/2)^2=b^2+(b+(a-c)/2)^2, a=b+c, 3 equations 3 unknowns a,b,c。😅 It works, by carefully calculation, first get b=4sqrt(13), then c=2sqrt(13), a=b+c=6sqrt(13), their areas are 36*13, 16*13, 4*13.

@PreMath

3 ай бұрын

Thanks ❤️

@JLvatron

3 ай бұрын

That's what I thought! But the solution was almost as complex, anyway!

Let the sides of the blue square have length c. We know that c = a - b, so a = b + c and b = a - c. At 13:00, we find that a/b = 3/2, so a = 3b/2 and b = 2a/3. In a = 3b/2, replace b with a - c, so a = 3(a - c)/2, 2a = 3a - 3c and a = 3c. In b = 2a/3, replace a with b + c, so b = 2(b + c)/3, 3b = 2b + 2c and b = 2c. Apply the Pythagorean theorem to ΔABO: a² + b² = (26)², (3c)² + (2c)² = (26)² and 9c² + 4c² = (26)², 13c² = (26)(26), c² = (2)(26), c² = 52 and c = √(52). So, b = 2c = 2√(52) and a = 3c = 3√(52). We square each of these lengths a, b, c to get the areas of each of the squares. PreMath's k turns out to be c, the length of the side of the blue square.

Let's nome a the length of the green square and b the length of the blue square, then the length of the yellow square is a - b. We use an adapted orthonormal, center O and first axis (CB). EA = a + b and E and A are symetric by the second axis, so A((a+b)/2 ; a) Then D( (a + b)/2 -a - (a -b) ; a - b) or D((-3/2).(a - b) ; a - b) The equation of the circle is x^2 + y^2 = 26^2 = 676. D beeing on the circle we have: (9/4).(a - b)^2 + (a -b)^2 = 676, so (a - b)^2 = 676/ (13/4) = 208, the area of the yellow square is then 208. Now we have a - b = sqrt(208) = 4.sqrt(13) and a = b + 4.sqrt(13). A beeing on the circle we have: (a +b)^2 + a^2 = 676 or 5.a^2 + 2.a.b +b^2 = 676.4 = 2704, we replace a by b + 4.sqrt(13) and get: 5.(b^2 +208 +8.b.sqrt(13)) + 2.b.(b + 4.sqrt(13)) + b^2 = 2704 or 8.b^2 +48.b.sqrt(13) -1664 = 0 or b^2 +6.sqrt(13).b -208 = 0 Deltaprime = (-3.sqrt(13))^2 +208 = 325 = (5.sqrt(13)^2 Then b = -3.sqrt(13) + 5;sqrt(13) = 2.sqrt(13), the other possibility is rejected as beeing negative. So the area of the blue square is b^2 = 52. Finally a = b + 4.sqrt(13) = 6.sqrt(13) and the area of the green square is a^2 = 468.

@PreMath

3 ай бұрын

Excellent! Thanks ❤️

1) Let a= AB (side of the Green), b= CD (side of the Yellow), c=PF (side of the Blue), z= OF (offset) 2) a=b+c; => b= a-c; 3) Let us use Pythagorean theorem thrice: (a-z)^2+a^2=26^2 (1) (b+z)^2+b^2 =26^2 (2) (z+c)^2+a^2=26^2 (3) 4) from (1): (a-z)^2=676-a^2; from (3): (z+c)^2=676-a^2; (a-z)^2=(z+c)^2; a-z=z+c; a-c=2z; => b=2z; z = b/2; 5) let us put newfound z into (2): (b+b/2)^2+b^2=676; Area of the Yellow square: b^2 = 676/(1+9/4) = 208 sq units; b= sqrt(208) = 4*sqrt(13); 6) z= 2*sqrt(13) ; 7) Let us put z in (1): (a-2*sqrt(13))^2+a^2=676 => a= 6*sqrt(13) => Area of the Green square = a^2= 468 sq units 8) c= a-b= 6*sqrt(13)-4*sqrt(13)=2*sqrt(13) => Area of the Blue square = c^2 = 52 sq unts.

@PreMath

3 ай бұрын

Thanks ❤️

In first place one must note that Green Square Side (a) = Yellow Square Side (b) + Blue Square Side (c). It means that a^2 = (b + c)^2 = b^2 + 2*bc + c^2. The angle AOD = 90º, as they share the same hypotenuse (R) = OD = OA = 26. So, OB = DC = CE' (E' = upper right corner of Blue Square). OA // CE' Now, let's call the the Side Length of Blue Square "x" and Area of Blue Square "x^2". EA = 4x and AB = 3x By Pythagorean Theorem: (2x)^2 + (3x)^2 = 26^2 ; 4x^2 + 9x^2 = 676 ; 13x^2 = 676 ; x^2 = 676 / 13 ; x^2 = 52 ; x = sqrt(52) ; x ~ 7,2 lin un Now: 1) Blue Square Area = 52 sq un 2) Yellow Square Area = (4 * 52) sq un = 208 sq un 3) Green Square Area = (9 * 52) sq un = 468 sq un As the Area of the Semicircle is equal to 676Pi/2 = 338Pi ~ 1.062 sq un The Area occupied by the 3 Squares together is equal to 728 sq un.

This one made me laugh out loud to see how it all unfolded. Thanks for posting!

@PreMath

3 ай бұрын

Glad you enjoyed it! You are very welcome! Thanks ❤️

(52)^2=2704 (180°-2704)=√2524 √5^√5 √4^√6 √1^√1√2^√2 3^2 √1^√1 3^2 (x+2x-3)

Very good!!

@PreMath

3 ай бұрын

Thanks a lot!❤️

Magic!

Greate ... i like to clarify from the Area of a triangle we get (b * h) / 2 = bh/2, and for a aquere b * h=bh ..using a scaling factor k we get the areas of the triangle(At) or square(As) to be Triangle: (kb * kh) / 2 = k(b * h)/2 = k²bh/2 = At Square: (kb * kh) = k(b * h) = k²bh = As ..since k is a scling factor we can ignore if its a triangle or a squere...

@PreMath

3 ай бұрын

Thanks ❤️

2704/2=√1352 √13^1 4^√13 √1^√1 √4^√1 √2^2 1^2 (x+1x-2)

Sauber

getting challenging ;d

@PreMath

3 ай бұрын

Glad to hear that! Thanks ❤️

2704/3=√904 √300√^300 √2^√2 √5^√60 √5^√60 √1^√1 √1^√30^√2√ 1^√30^√2 √5^√6^√1 √5^√6^√1 √1^√3^√2 √1^3^2 √1^√1 3^2 (x+2x-3)

Strange puzzle with Strange solution 😢

easy

@PreMath

3 ай бұрын

Thanks ❤️

Too confusing

Taking a wild guess about proportions: blue square = 1², yellow square = 2² and green square = 3²... and it works!

@PreMath

3 ай бұрын

Thanks ❤️