x : side length of the square base = base of blue triangle = 2*a=16*sqrt(2) (same calculus as the video) Thales in left right angle triangle : (12/sqrt(2))/(x-base)=(x-12/sqrt(2))/x 6*sqrt(2)/(x-base)=(x-6*sqrt(2))/x 6*sqrt(2)/(x-16*sqrt(2))=(x-6*sqrt(2))/x 6*sqrt(2)*x=(x-16*sqrt(2))*(x-6*sqrt(2)) 6*sqrt(2)*x=x^2-22*sqrt(2)*x+192 x^2-28*sqrt(2)*x+192=0 (x-4*sqrt(2))*(x-24*sqrt(2))=0 x bigger than base = 16*sqrt(2) therefore x=4*sqrt(2) is impossible then x=24*sqrt(2) area blue = base *h/2 = base *(x-16/sqrt(2))/2 = 16*sqrt(2)*(24*sqrt(2)-8*sqrt(2))/2 = (16*sqrt(2))^2/2 = 16^2*2/2 =256
12/16=3/4 → 12 es la diagonal de un cuadrado de dimensiones 6√2*6√2→ 4*4*(6√2)²=1152 ; y 16 es la diagonal de un cuadrado de dimensiones 8√2*8√2→ 3*3*(8√2)²=1152 → Ambas matrices definen cuadrados de la misma superficie→ la diagonal del cuadrado de la figura es 4*12=3*16=48→ lado del cuadrado =48/√2=24√2→ El triángulo isósceles azul de la figura tiene iguales la base y la altura y está incrito en un cuadrado de lado =2*8√2=16√2 → Área triángulo azul =(16√2)²/2 =256 m² . Buen acertijo. Gracias y un saludo cordial.
x : side length of the square
base = base of blue triangle = 2*a=16*sqrt(2) (same calculus as the video)
Thales in left right angle triangle :
(12/sqrt(2))/(x-base)=(x-12/sqrt(2))/x
6*sqrt(2)/(x-base)=(x-6*sqrt(2))/x
6*sqrt(2)/(x-16*sqrt(2))=(x-6*sqrt(2))/x
6*sqrt(2)*x=(x-16*sqrt(2))*(x-6*sqrt(2))
6*sqrt(2)*x=x^2-22*sqrt(2)*x+192
x^2-28*sqrt(2)*x+192=0
(x-4*sqrt(2))*(x-24*sqrt(2))=0
x bigger than base = 16*sqrt(2) therefore x=4*sqrt(2) is impossible
then x=24*sqrt(2)
area blue = base *h/2
= base *(x-16/sqrt(2))/2 = 16*sqrt(2)*(24*sqrt(2)-8*sqrt(2))/2 = (16*sqrt(2))^2/2 = 16^2*2/2
=256
Thanks friend for this amazing method
you've shown us a great and useful probem.Thank you very much sir and keep going.
Thanks sir so much, I am also encouraged by your comment
12/16=3/4 → 12 es la diagonal de un cuadrado de dimensiones 6√2*6√2→ 4*4*(6√2)²=1152 ; y 16 es la diagonal de un cuadrado de dimensiones 8√2*8√2→ 3*3*(8√2)²=1152 → Ambas matrices definen cuadrados de la misma superficie→ la diagonal del cuadrado de la figura es 4*12=3*16=48→ lado del cuadrado =48/√2=24√2→ El triángulo isósceles azul de la figura tiene iguales la base y la altura y está incrito en un cuadrado de lado =2*8√2=16√2 → Área triángulo azul =(16√2)²/2 =256 m² .
Buen acertijo. Gracias y un saludo cordial.
Thank you friend for sharing this method.
Thank you for the problem
Thanks you so much sir, I am also encouraged by your comment