Can You Solve This Shaded Area With Limited Data

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Can You Solve This Shaded Area With Limited Data

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The Phantom of the Math

Күн бұрын

Пікірлер: 31

@bkp_s 29 күн бұрын

Sir, paused the vdo to do it on my own. And I am glad to inform you that the process adopted by me was in true sense parallel to the way you described. Thanks sir

@tomtke7351 Ай бұрын

Actually is an easier problem... First define shapes: C.L : left circle C.R : right circle R : rectangle Observed dimensions: (-1-) C.L diameter = C.R radius (-2-) R area = 8 × C.L.diameter = 16 × C.L.r (-3-) Shaded area = R.area - (C.L.area+C.R.area) C.L.area = (1/2)(pi)(C.L.r^2) C.R.area = (1/4)(pi)(C.R.r^2) (-4-) The two circles touch where a line intersects this touch point. This line (from both circle centers) starts lower right corner of R and ends at center of R's left side. (-5-) This line has length = 2 radius lengths = C.L.r + C.R.r (-6-) Line, above, and R's bottom line = 8 form a triangle (-7-) This triangle has a left side of length: C.L.r (-8-) Triangle summary Left: C.L.r Bottom: 8 Hypotenuse: C.L.r + C.R.r Pythagoreans theorem: 8^2 + C.L.r^2 = (C.L.r+C.R.r)^2 64+C.L.r^2 =C.L.r^2+2(C.L.r)(C.R.r)+C.R.r^2 64= 2(C.L.r)(C.R.r)+C.R.r^2 eq.1:C.R.r^2+2(C.L.r)(C.R.r)=64

@santiagoarosam430 3 ай бұрын

Radio del semicírculo =r → Altura del rectángulo =Radio del cuarto de círculo =2r → 8²=(r+2r)²-r²=8r²→ r²=8 → r=2√2→ 2r=4√2 → Área azul =(8*4√2) -8π/2 -8π =32√2 -12π =7,5557 ud². Gracias y saludos.

@ThePhantomoftheMath 3 ай бұрын

Si eso es! Gracias por tu contribución!

@arkdev9 3 ай бұрын

how can we be sure that the line ab meets at the connection of the two circles and also starts / ends at the midpoint of smaller circle?

@NeatCrown 3 ай бұрын

Because they're (sectors of) circles. A circle is, by definition, a shape whose points are all equidistant from its centre.

@KSZLegion 3 ай бұрын

@@NeatCrown better question is how do we know that line AB is straight and lines: AO, OB not intersecting at 0,1degres for example

@ThePhantomoftheMath 3 ай бұрын

Point O represents the intersection point between two circular sectors. At point O, a common tangent line can be drawn for both segments. Lines AO and BO are radii of these circular sectors. Since the angle between a tangent and a radius is always 90 degrees, both AO and BO are perpendicular to the tangent at point O. Therefore, the line passing through points A, O, and B (line AB) must be straight.

@Math_oma 2 ай бұрын

@@ThePhantomoftheMath Alternately, a Euclid-style proof. Suppose two circles, the first with center A, are externally tangent at a point P. Now draw AP through the second circle. Suppose (with a view to a contradiction), that the center of the second circle B is off this line. Now draw AB, which will now touch the two circles at two new points, C and D. Also draw BP, forming triangle ABP. By inspection of the diagram: AP + PB = AC + BD < AB (since AP,AC and PB,BD are pairs of radii) so AP+PB < AB But we know by the triangle inequality that: AP+PB > AB Hence we have a contradiction, and so the center of the second circle, B, is on the line through AP. Hence A,P,B are collinear.

@Mathe_erklaert 3 ай бұрын

Thank you for this nice example. Which software do you use to animate this ?

@ThePhantomoftheMath 3 ай бұрын

Hi. I use a combination of Sparkol VideoScribe Pro and PowerPoint. Of course, everything is edited with Adobe Premiere. Trust me when I say... creating these animations takes a LOT of effort and time to get them just right.

@DrTWG 2 ай бұрын

@@ThePhantomoftheMath I can just imagine how long it takes . You have to account for every little thing happening on the screen . I make fractals but at least they are generated . Great work my friend .

@ThePhantomoftheMath 2 ай бұрын

@@DrTWG Then my friend, you above all have a great idea of how long it takes to make this kind of video! 😃

@andrewclifton9772 3 ай бұрын

I did it slightly differently but the major point is the same: a line from the SE corner to the mid point of the West edge is a straight line through the interaction of the two circle segments. (A radius from the center of a circle to a tangent is perpendicular to the tangent.). So if the height of the rectangle is R then Pythagoras tells us that (R + R/2)^2 = 64 + (R/2)^2 ==> R^2 +(R/2)^2 + 2 R^2/2 = 64 + (R/2)^2 ==> R = √32. So the area of the rectangle is 8 x √32 = 45.255. The area of the segments is (pi x R^2)/4 and (pi x (R/2)^2)/2 . R^2 = 32 so substituting that in ==> area of segments = 8 x pi + 4 x pi = 12 x pi = 37.699. Subtracting to get the shaded area ==> 7.554

@ThePhantomoftheMath 3 ай бұрын

Great job!!!

@quigonkenny 2 ай бұрын

Let O be the center of the quarter circle (lower right corner) and P be the center of the semicircle (midpoint of left side). Let T be the point of tangency between the two circles, and A be the lower left vertex of the rectangle. By observation, the radius of the semicircle is half the height of the rectangle. Let's call that r. As the radius of the quarter circle is the entire height of the rectangle, then that radius is 2r. Draw a line from O to P. For any two circles tangent to each other, the point of tangency is collinear with the two centers. This means that the length of OP is equal to OT, which is the radius of the quarter circle, or 2r, plus PT, the radius of the semicircle, or r. Therefore OP = 3r. As point A lies at the circumference of the semicircle, PA = r. And OA is given as 8. Triangle ∆OPA: PA² + OA² = OP² r² + 8² = (3r)² r² + 64 = 9r² 8r² = 64 r² = 64/8 = 8 r = √8 = 2√2 The Shaded area will be equal to the total area of the rectangle minus the areas of semicircle P and quarter circle O. A = hw - π(2r)²/4 - πr²/2 A = (4√2)(8) - π(4√2)²/4 - π(2√2)²/2 A = 32√2 - 32π/4 - 8π/2 A = 32√2 - 12π ≈ 7.556 sq units

@ThePhantomoftheMath 2 ай бұрын

Nice job!! 👍

@hocinechaouadi9108 3 ай бұрын

bravo pour ces exercices d’intelligence

@ThePhantomoftheMath 3 ай бұрын

Merci, mon ami

@Straight_Talk 2 ай бұрын

Which theorem says AB passes through the point of tangency of the two circles?

@ThePhantomoftheMath 2 ай бұрын

Hi friend. The theorem that states AB passes through the point of tangency of the two circles is known as the Radical Axis Theorem. The Radical Axis Theorem states that the radical axis of two circles (the locus of points where the power of the point with respect to both circles is equal) is the line passing through their points of tangency when the circles are externally tangent. In this video, AB is the radical axis of the two circles, and O is the point of tangency.

@Straight_Talk 2 ай бұрын

@@ThePhantomoftheMath Thanks. I'll check it out.

@Straight_Talk 2 ай бұрын

@@ThePhantomoftheMathSorry to be difficult, but AB isn't the radical axis of the two circles. Rather, the radical axis is the line perpendicular to the line connecting the circles' centers. I believe that according to the radical axis theorem, as the circles intersect at one point, the tangent line is the radical axis. Further, the centres of the two circles and the point of tangency are colinear.

@ThePhantomoftheMath 2 ай бұрын

@@Straight_Talk You’re right about the definition and properties of the radical axis. AB represents the line through the centers of the two tangent circles. Here’s the clarification: The radical axis of two circles is the line where the power of any point on it with respect to both circles is equal. This line is perpendicular to the line connecting the centers of the two circles. When two circles are tangent to each other, the line connecting their centers passes through the point of tangency. In the image, the line AB is this line, passing through the centers of the two circles and the point of tangency O. So, the property in question is that the line through the centers of two tangent circles passes through their point of tangency. I hope this helps clarify the concept!

@user-vg4cg4uw9c Ай бұрын

That is how I did it too (with the correct result) 😎

@ThePhantomoftheMath Ай бұрын

@@user-vg4cg4uw9c Well done friend! 👍

@douglassmith1466 2 ай бұрын

You found the area of the blue, which contains 2 parts. Is it possible to find the areas of each blue section?

@ThePhantomoftheMath 2 ай бұрын

Yes. You can calculate the area of the purple triangle and then subtract the areas of the two red circular segments inside that triangle. Since you already have the radii of those segments, the only additional information you need is the central angle for each segment. You can find these angles by calculating the arccosine: arccos(1/3) for the angle at point A and arccos(8 / (6√2)) for the angle at point B.