Can you find area of the Yellow triangle? | (Fun Geometry Problem) |

[@marioalb9726]

tan α = 6/8 --> α = 36,87° Pytagorean theorem : (2r)²= 6²+8² ---> 2r = 10m c² = 6²+6² ---> c= √72= 3√8m Angle between c and R : β = 360°-90°-(90-α) -135° = 81,87° Radius of quadrant : cosβ = ½c/R --> R = 30m Base of yellow right triangle: b/R = 8/2r --> b = 24m Area of yellow triangle: A = ½R.b.sinα = ½*30*24*sin36,87° A = 216m² ( Solved √ )

[@Mipullo321]

Math and Engineering never disappoint 💯, Interesting Questions thanks 🤗

[@MathandEngineering]

Thanks 👍

[@juanalfaro7522]

Let A= Top Point, O= origin, and B= rightmost point. Let C & D= intersecting points of 6m segment, so CD=6m, and DB=8m. Let E= where both circles intersect., and F= the other point in yellow triangle. So OBA=a -> OAB=b where a+b=90. Now AOF=a and FOB=b. Let ED=x and EC=y. We see AE=R√2. Now ACE = 1/2 * (360-90) = 135 --> ECD=180-135=45. Now EDC=90 --> CED = 180-90-45 = 45 --> ECD=CED --> ED=CD=6m --> CE=6√2 (CED is right isosceles triangle). Now look at ACE triangle: R√2/sin 135 = CE/sin (b-45). Since ED=6, a=arctan (6/8) = 36.87 -> b=90-a = 53.13 -> sin (b-45=8.13) = sin b*cos 45 -cos b * sin 45 = (4/5 - 3/5) * √2/2 = √2/10. Now sin 135 = √2/2. Thus R√2/ (√2/2) = 6√2/ (√2/10) -> R/ (1/2) = 6/ (1/10) --> 2R=60 -> R=30. Now OFA (Yellow) triangle: OF = 30 *cos a = 30*4/5 = 24 and AF = 30 *sin a = 30*3/5 = 18 --> Area = OF*AF/2 = 24*18/2 = 216 m^2.

[@MathandEngineering]

This is a perfect method, thanks for sharing 👍

[@soli9mana-soli4953]

I have very often encountered questions like this with this configuration where there is a move that is truly decisive which consists in drawing two segments from the two points of intersection of the line cutting the two circles and joining them at the point of tangency of the two circles along the diameter line, thus drawing a right-angled triangle. In fact, in the semicircle we will have a right angle with the segment that measures 8 and in the quarter circle instead we will have an angle of 135° (since the central angle is 270) with the chord. This tells us that one of the angles of the right-angled triangle that we have drawn is 45° and therefore it is isosceles. It follows that the sides of the right-angled triangle that rests on the diameter of the semicircle will be equal to 8, 6, 10 (pythagorean triplet) Now we can find the missing data through the similarity between the right-angled triangle inside the semicircle and the larger right-angled triangle 1) R : 6 = (R+10) : 8 R = 30 2) (2h+6+8) : 10 = 30 : 6 h = 18 with Euclid's theorem we can find b b² = 18*(18+6+8)= 18*32 = 3²*8² b = 24 Area = 1/2*24*18 = 216

[@MathandEngineering]

Wow, thanks, I like the way you Explained every single step, you made it even easier

[@giuseppemalaguti435]

Posto R=raggio del cerchio maggiore...risultano due equazioni..(2Rsinα+6+8)sinα=R..tg α=R/(R+8/cosα)..R=30,α=2arctg(1/3)..A=RcosαRsinα/2...

[@MathandEngineering]

Thanks it's Good 👍

[@wasimahmad-t6c]

12×9÷2=54