Puntos notables en el perímetro del cuadrado: A, F, B, C, D y E. Diagonales del cuadrado: (B, G, O, K, D) y (A, M,O,C). Paralela a BD: (F, M, E) → Ángulos de los triángulos interiores: EDC=FBC→75º/90º/15º ; EAF→45º/90º/45º ; EFC→60º/60º/60º→ EFC es triángulo equilátero de lado "a" y altura =MC=h→ KGC es semejante al anterior de lado "b" y altura =OC=k →→→ AC=10√2 → Si FE=a→ FM=ME=AM=a/2→ MC=h=10√2-(a/2) = a√3/2→ a=10(√6-√2)=10,35276...→ h=a√3/2=5√3(√6-√2)=8,96575... →→ OC=k=10√2/2=5√2 →→ Razón de semejanza entre los dos triángulos equiláteros =s =k/h=(3+√3)6 → Razón entre sus áreas =s² =(2+√3)/6 → Razón entre el área azul EFGK y el triángulo EFC =1-s² =(4-√3)/6=0,37799... →→→ Área azul EFGK =(a*h/2)*(1-s²) =(550-300√3)/√3=17,54264...m². Interesante problema. Gracias y un saludo cordial.
I obtained the same answer as 100(11/6*√3) - 300 = 17.5426 m^2. I did this by noting that the top and bottom right triangles are equilateral: Let AB be AEB and AD be AFG where EF is the outer side of the blue area. EA=AF since both sides suspended by 45 angles, and AB=AD --> BE=DF --> tan (15) = tan (p) --> p=15. Then CF=CE and ECF = 90-2*15=60 --> CEF is an equilateral triangle with all sides equal to 10/cos (15) = 10(√6-√2). Then the diagonal BD be represented by BGHD where GH is the inner side of the blue area. Then BG/sin (75) = BE/sin (60) -> BG = 10 * tan (15) * sin (75) / sin (60) = 10*sin (15) / sin (60) = 10 * [(√6-√2)/4 * 2√3/3) = 10 * (3√2-√6)/6 = DH --> GH = AD - 2*BG = 10√2 - 10(3√2-√6)/3 = 10√6/3. Then [EGHF] = [CEF] - [CGH]. Now [CEF] = √3/4* EF^2 = [10(√6-√2)] ^2 * √3/4 = 100(8-4√3) * √3/4 = 100(2√3 - 3) and [CGH] = [10√6/3] ^2 * √3/4 = 100*6/9*√3/4 = 100√3/6 --> [EGHF] = 100(2√3 - √3/6 - 3) = 100(11/6*√3) - 300 (go to beginning)
The side length of the large equilateral triangle is equal to the length of the hypotenuse of a 15°-75°-90° right triangle. The ratio of sides, short, long, hypotenuse for this triangle is (√3 - 1):(√3 + 1):2√2, so the side length is 20√2/(√3 + 1). Multiply numerator and denominator by (√3 - 1) and simplify to get the form in the video: (10√2)(√3 - 1). The square's diagonal has length 10√2. The height of the small equilateral triangle, when a side is considered the base, is half the length of the square's diagonal, or (1/2)(10√2) = 5√2. To find the side length, multiply the height by 2/(√3): (5√2)(2)/(√3). Multiply numerator and denominator by √3 and simplify to get the form in the video: (10√6)/3. These are the side lengths that Math and Engineering found. Proceed to compute the area as done in the video.
I solved this on my phone’s calculator app but I didn’t use your exact steps. I found the area of the half square to be 50. Then I found the height of the segment from the corner to the parallel lines to be 10tan15°≈2.679492. Finally I solved for the two small triangle areas (sum is 5.662433). So the blue area is 50-26.794919-5.662433 ≈ 17.542648
Puntos notables en el perímetro del cuadrado: A, F, B, C, D y E. Diagonales del cuadrado: (B, G, O, K, D) y (A, M,O,C). Paralela a BD: (F, M, E) → Ángulos de los triángulos interiores: EDC=FBC→75º/90º/15º ; EAF→45º/90º/45º ; EFC→60º/60º/60º→ EFC es triángulo equilátero de lado "a" y altura =MC=h→ KGC es semejante al anterior de lado "b" y altura =OC=k →→→
AC=10√2 → Si FE=a→ FM=ME=AM=a/2→ MC=h=10√2-(a/2) = a√3/2→ a=10(√6-√2)=10,35276...→ h=a√3/2=5√3(√6-√2)=8,96575... →→ OC=k=10√2/2=5√2 →→ Razón de semejanza entre los dos triángulos equiláteros =s =k/h=(3+√3)6 → Razón entre sus áreas =s² =(2+√3)/6 → Razón entre el área azul EFGK y el triángulo EFC =1-s² =(4-√3)/6=0,37799... →→→ Área azul EFGK =(a*h/2)*(1-s²) =(550-300√3)/√3=17,54264...m².
Interesante problema. Gracias y un saludo cordial.
I obtained the same answer as 100(11/6*√3) - 300 = 17.5426 m^2. I did this by noting that the top and bottom right triangles are equilateral: Let AB be AEB and AD be AFG where EF is the outer side of the blue area. EA=AF since both sides suspended by 45 angles, and AB=AD --> BE=DF --> tan (15) = tan (p) --> p=15. Then CF=CE and ECF = 90-2*15=60 --> CEF is an equilateral triangle with all sides equal to 10/cos (15) = 10(√6-√2). Then the diagonal BD be represented by BGHD where GH is the inner side of the blue area. Then BG/sin (75) = BE/sin (60) -> BG = 10 * tan (15) * sin (75) / sin (60) = 10*sin (15) / sin (60) = 10 * [(√6-√2)/4 * 2√3/3) = 10 * (3√2-√6)/6 = DH --> GH = AD - 2*BG = 10√2 - 10(3√2-√6)/3 = 10√6/3. Then [EGHF] = [CEF] - [CGH]. Now [CEF] = √3/4* EF^2 = [10(√6-√2)] ^2 * √3/4 = 100(8-4√3) * √3/4 = 100(2√3 - 3) and [CGH] = [10√6/3] ^2 * √3/4 = 100*6/9*√3/4 = 100√3/6 --> [EGHF] = 100(2√3 - √3/6 - 3) = 100(11/6*√3) - 300 (go to beginning)
Ok yes this is perfect, thanks for sharing
The side length of the large equilateral triangle is equal to the length of the hypotenuse of a 15°-75°-90° right triangle. The ratio of sides, short, long, hypotenuse for this triangle is (√3 - 1):(√3 + 1):2√2, so the side length is 20√2/(√3 + 1). Multiply numerator and denominator by (√3 - 1) and simplify to get the form in the video: (10√2)(√3 - 1). The square's diagonal has length 10√2. The height of the small equilateral triangle, when a side is considered the base, is half the length of the square's diagonal, or (1/2)(10√2) = 5√2. To find the side length, multiply the height by 2/(√3): (5√2)(2)/(√3). Multiply numerator and denominator by √3 and simplify to get the form in the video: (10√6)/3. These are the side lengths that Math and Engineering found. Proceed to compute the area as done in the video.
Math and Engineering, cool video it was really entertaining
Thanks you so much ❤️
I solved this on my phone’s calculator app but I didn’t use your exact steps. I found the area of the half square to be 50. Then I found the height of the segment from the corner to the parallel lines to be 10tan15°≈2.679492. Finally I solved for the two small triangle areas (sum is 5.662433). So the blue area is 50-26.794919-5.662433
≈ 17.542648
Wow sir, this is perfect. If you simplify 50√3(11-6√3)/3 it's 17.54264805429417048003182927607660060618096