At about 9:45, the value of b has been determined to be 100. Construct a line segment from the upper left corner of the rectangle to the center of the small circle. Two congruent right triangles have been formed with short side length b, long side length 3b and common hypotenuse. Let the angle between the hypotenuse and long side of each triangle be designated Θ. Then, tan(Θ) = b/3b = 1/3. From the tangent double angle formula, tan(2Θ) = 2tan(Θ)/(1 - tan²(Θ)), tan(2Θ) = 2(1/3)/(1 - (1/3)²) = (2/3)/(1 - 1/9) = (2/3)/(8/9) = 18/24 = 3/4. Consider the larger right triangle with sides (b + 100 + c), (3b + b) and hypotenuse (3b + c + 100) and angle 2Θ at the upper left corner of the rectangle. The short side is tan(2Θ) times the long side, or (3/4)(3b + b) = (3/4)(4b) = 3b. So, we replace its length (b + 100 + c) with 3b and the bottom side of the rectangle, its base, has length 3b + 100 + 2b = 5b + 100. Since b = 100, 5b + 100 = (5)(100) + 100 = 600. The left side, its height, has length 4b = 400. The rectangle's area is (base)(height) = (600)(400) = 240,000, as Math and Engineering also found at about 13:45.
Oh wow, when I first started reading your methodology, you said construct a line segment from the upper left corner to the center of the circle, I don't know why I mistook that for the upper right corner, I was trying to process the rest in my head, but nothing was adding up for me, then something else took my attention, I just came back and checked now then I realized made a mistake from the beginning, The line segment was from the upper left corner and when I used that and followed the rest of the solution, it was perfect and Interesting to use trigonometry, thank you so much
Very cleverly found, but personally I don't recall the double angle formula and I favour the pythagorean theorem here because it very elegantly eliminates most of the squares and leaves you with a clear equation for c.
Area 2nd circle = 4*Area 1st Circle --> 1st circle has radius r -> 2nd circle has radius 2r. Then vertical axis has length 2r*2 = 4r. Then 1st circle has top tangent 4r-r = 3r. Let bottom right tangent 1st circle =a, then (4r) ^2 + (r+a) ^2 = (3r+a) ^2 -> 16r^2 + (r^2 + 2ar + a^2) = 9r^2 + 6ar + a^2 -> (16+1-9) r^2 + (2-6) a*r + (1-1) a^2 = 0 --> 8r^2 = 4ar -> 8r=4a -> a=2r --> left triangle has dimensions 4r, 3r, and 5r. Now let left tangent of big circle = u, then u + 2r = 3r + 100 + 2r = 100 + 5r --> u = 5r + 100 - 2r = 3r + 100. Analyzing the diagonal, u+100 = 5r --> u = 5r-100 = 3r+100 -> 2r=200 --> r=100 -> u=3*100+100 = 400. Then Vertical side = 4*100 = 400; Horiz. side = u+2r = 400+2*100 = 600 ==> Area = 400*600 = 240000 u^2 (sq. units)
At about 9:45, the value of b has been determined to be 100. Construct a line segment from the upper left corner of the rectangle to the center of the small circle. Two congruent right triangles have been formed with short side length b, long side length 3b and common hypotenuse. Let the angle between the hypotenuse and long side of each triangle be designated Θ. Then, tan(Θ) = b/3b = 1/3. From the tangent double angle formula, tan(2Θ) = 2tan(Θ)/(1 - tan²(Θ)), tan(2Θ) = 2(1/3)/(1 - (1/3)²) = (2/3)/(1 - 1/9) = (2/3)/(8/9) = 18/24 = 3/4. Consider the larger right triangle with sides (b + 100 + c), (3b + b) and hypotenuse (3b + c + 100) and angle 2Θ at the upper left corner of the rectangle. The short side is tan(2Θ) times the long side, or (3/4)(3b + b) = (3/4)(4b) = 3b. So, we replace its length (b + 100 + c) with 3b and the bottom side of the rectangle, its base, has length 3b + 100 + 2b = 5b + 100. Since b = 100, 5b + 100 = (5)(100) + 100 = 600. The left side, its height, has length 4b = 400. The rectangle's area is (base)(height) = (600)(400) = 240,000, as Math and Engineering also found at about 13:45.
Oh wow, when I first started reading your methodology, you said construct a line segment from the upper left corner to the center of the circle, I don't know why I mistook that for the upper right corner, I was trying to process the rest in my head, but nothing was adding up for me, then something else took my attention, I just came back and checked now then I realized made a mistake from the beginning,
The line segment was from the upper left corner and when I used that and followed the rest of the solution, it was perfect and Interesting to use trigonometry, thank you so much
Very cleverly found, but personally I don't recall the double angle formula and I favour the pythagorean theorem here because it very elegantly eliminates most of the squares and leaves you with a clear equation for c.
Area 2nd circle = 4*Area 1st Circle --> 1st circle has radius r -> 2nd circle has radius 2r. Then vertical axis has length 2r*2 = 4r. Then 1st circle has top tangent 4r-r = 3r. Let bottom right tangent 1st circle =a, then (4r) ^2 + (r+a) ^2 = (3r+a) ^2 -> 16r^2 + (r^2 + 2ar + a^2) = 9r^2 + 6ar + a^2 -> (16+1-9) r^2 + (2-6) a*r + (1-1) a^2 = 0 --> 8r^2 = 4ar -> 8r=4a -> a=2r --> left triangle has dimensions 4r, 3r, and 5r. Now let left tangent of big circle = u, then u + 2r = 3r + 100 + 2r = 100 + 5r --> u = 5r + 100 - 2r = 3r + 100. Analyzing the diagonal, u+100 = 5r --> u = 5r-100 = 3r+100 -> 2r=200 --> r=100 -> u=3*100+100 = 400. Then Vertical side = 4*100 = 400; Horiz. side = u+2r = 400+2*100 = 600 ==> Area = 400*600 = 240000 u^2 (sq. units)
Ok, I was busy and didn't get to look at this, unill now, the procedure is perfect, thanks for sharing.