Can You Figure Out the Orange Shaded Area? | Geometry

[@DrTWG]

Cool. Great animations - I bet that takes many hours to do . Enjoyed the problem .

[@ThePhantomoftheMath]

Thanks a lot! 🫶

[@ronbannon]

Really nice job of visually explaining the steps.

[@ThePhantomoftheMath]

Thank you ❤

[@yurenchu]

The right triangle is a (10, 24, 26)-triangle (i.e. a scaled (5, 12, 13)-triangle). Furthermore, the midpoint of the hypotenuse is also the center of the semi-circle. Therefore, the radius of the circle equals half of the hypotenuse, 26/2 = 13 . So the dimensions of the outer rectangle are (13 + 10/2)-by-(13+24/2) = 18-by-25 . So the orange-shaded area equals A = {rectangle area} - {right triangle area} - {semi-circle area} = 18*25 - (10*24)/2 - (π*13²)/2 = 450 - 120 - (84.5)π = 330 - (84.5)π ≈ 64,53542

[@marioalb9726]

Radius of semicircle: r = 13 cm Area of rectangle: A = b.h = (10/2+r).(24/2+r) A = 450 cm² Orange shaded area: A = A₃ - A₁ - A₂ A = A₃ - ½πr² - ½b.h A = 450 - ½169π - ½10*24 A = 64,535 cm² ( Solved √ )

[@bkp_s]

Respected professor, In search of other options, if available,I worked on the downside of the given diagram.joined the point of tangency on the longer and left side of the rectangle to the centre of the circle and extended this line to the right side longer side of the rectangle straight away.Then applied triangle similarity theorem and found the remaining sides of lower triangle as 5 and 12 units respectively which helped me ti find the width and length of rectangle. And rest others the same as you calculated in final steps. But the whole credit goes only to you for your inspirational teachings.🙏

[@ThePhantomoftheMath]

@@bkp_s Excelent! Graet job! 💪

[@KipIngram]

Nice one. I worked this out primarily in my head, though I did use a calculator for some of the arithmetic. But I got the 18x25 dimensions of the rectangle in my head. It's 63.53542, if you want more decimals.

[@gelbkehlchen]

Solution: r = radius of the semicircle. According to the Pythagorean theorem: 2r = √(24²+10²) = 26 ⟹ r = 13 O = center of the semicircle, M = center of the 24-line, A = corner point of the rectangle bottom left, B = corner point of the rectangle bottom right, C = corner point of the rectangle top right, D = corner point of the rectangle top left, E = left corner point of the 10-line, F = lower corner point of the 24-line, T1 = lower point of contact of the semicircle with the rectangle, T2 = left point of contact of the semicircle with the rectangle. According to the ray theorem: OM is parallel to EC. Because there is always a right angle between the tangent and the radius, the following applies: T2O is horizontal and T2OM is a straight line. In addition, OT1 is vertical and equal to the radius r = 13. Therefore, CB = CM+OT1 = 12+13 = 25. And DC = T2O+OM = 13+5 = 18, because OM is 5 according to the second theorem of ray. The rectangle is therefore 18 wide and 27 high. Orange area = area of ​​the rectangle - area of ​​the semicircle - area of ​​the triangle FEC = 18*25-π*13²/2-10*24/2 = 450-84.5π-120 = 330-84.5π ≈ 64.5354

[@ThePhantomoftheMath]

@@gelbkehlchen Nice. Good job! 👍

[@Eren_Sumer]

hey good question! Love you from türkiye! 🇹🇷

[@ThePhantomoftheMath]

❤❤❤

[@JP-re3bc]

It is not apparent nor given that the line is the diameter of the circle. That demands proof or be a given.

[@ThePhantomoftheMath]

Check 0:10 pls! 😊

[@Grizzly01-vr4pn]

At 0:08 the shape is described as a half circle. Therefore the hypotenuse of the right triangle also being the diameter of the circle is a given.

[@santiagoarosam430]

ɸ²=10²+24²→ ɸ=26→ Radio =r =26/2=13 → Área del rectángulo = Base * Altura =b*h→ Potencia de "P", vértice superior izquierdo del rectángulo, respecto a la circunferencia =(24/2)²=(b-10)b→ b=18 → Área sombreada =18*[(24/2)+r] -(10*24/2) -(πr²/2) =(18*25)-(120)-(169π/2) =330-(169π/2)=64,5354....ud². Gracias y saludos.

[@ThePhantomoftheMath]

Good job!

[@colina64]

[@AslamSheikh-c2r]

❤

[@ThePhantomoftheMath]

@@AslamSheikh-c2r ❤️

[@rybosny]

There's a hidden assumption here that the half circle and the rectangle have exactly four common points, one on each side of the rectangle (the drawing represents the exact problem). Without it, you couldn't solve it that way (if ever).

[@yurenchu]

Yes, it only works with rectangles whose ratio of sides is between 1:2 and 2:1 .