It probably helps that the diagonal goes through the midpoint of the top line of the marked square. If the diagram were a full rectangle of 3 by 2 squares, I think it would show that the diagonal also passes though that square's vertical line one third of the way down. If that square's sides are a, the area of the omitted triangular piece is ((a/2)(a/3))/2 = a^2/12. a^2 - a^2/12 = 11 12a^2 - a^2 = 132 11a^2 = 132 a^2 = 12. Therefore, the missing piece is 12-11=1, and the red part is the same. However, I did this more quickly in my head.
From Thales' proportions we can see that the height of the red triangle is 1/3 of the square's side and its base is equal to half the side. Therefore, the area of the square is equal to the area of 12 triangles. Therefore, 12*(area of the triangle)=11+(area of the triangle) area of the triangle =11/11=1
Faster to figure it out with some mental arithmetic (using the first method entirely) than to explain it. Also, the way the problem was presented, my gut told me it could be 1 immediately. Turned out my gut was right 😊
Super easy: the area of the red triange is 1/12 the are of the square containing the blue are, and equal to the white triangle in the square containing the blue area. Therefore tge area of the red triangle is 1.
I solved it by seeing that the long diagonal had a slope of -2/3 -> the vertical went 2 distances vertical while going a horizontal distance of 3. That means that the long diagonal cut through the square with the blue at the mid point of the top side so the length of the red triangle was 1/2 and its height was 1/3, and I figured that it was 1 part of 12 parts of a square, so the blues equaled 11 so the red triangle is 1 unit area I just start viewing your channel, I think it was earlier today (?) and enjoying it
От квадрата отделён сегмент с катетами 2 на 3. При этом больший катет, он же основание, составляет половину стороны. Таким образом, 11+2х*3х/2=(6х)². Разворачиваем. 11+3х²=36х²⇔11=36х²-3х²=33х²⇔х²=11/33. Нужно вновь умножить это значение на 3, чтобы получить искомое. Таким образом S=3*11/33=33/33=1, а общая площадь квадрата, стало быть, 12. Я догадывался, что автор загадал именно это число.
Si el lado de los cuadrados es "a"→ La línea inclinada tiene una pendiente =2a/3a=2/3→ corta al cuadrado central formando en su esquina superior derecha un triángulo rectángulo congruente con el rojo, cuyos catetos miden (a/2) y (a/3)→ la zona azul se compone de 11 triángulos de área 11/11=1 u², que son congruentes con el definido anteriormente→ Área triángulo sombreado rojo =1 u². Gracias y un saludo.
You have a line going 3 across and 2 up on a grid, your line intersects the middle of the bottom center cube, and cuts through at the 1/3 like on the side of the cube. If we divide the cube vertically in half, and we divide it into 3rds horizontally, we get 6 rectangles, with the unshaded section being exactly half of a rectangle. This is exactly the same area as half of a rectangle, which has an area of 2, so the red triangle is an area of 1, and the cubes have an area of 12, no complex math. Just apply basic observation, and logic. I'll watch the video now, and see what unneeded math is used. Yeah, got to 2 mins, and the math over complicates the solution, at least for this problem.
Simpler solution. The diagonal slope is 2/3 so it intersect with the second square at 2/3, implying that the triangle height is 1/3s. The trinagle base is 1/2s. The triangle area is 1/12s^2. The square area s^2 minus the triangle area is equal to 11. You solve the equation and find that s^2 is equal to 12 so the trinagle area us 1.
Yes, or geometrically each large square can be split into stacked thirds, which each third divided into 4 congruent triangles. In all there are 12 such triangles with the red area covering 11 of them with an area of 11. So the area of one triangle is 1.
Labelling sides of the two similar triangles using ratios to create algebra formula seemed a bit quicker to find the solution.
Very good question - now let's calculate!
It probably helps that the diagonal goes through the midpoint of the top line of the marked square.
If the diagram were a full rectangle of 3 by 2 squares, I think it would show that the diagonal also passes though that square's vertical line one third of the way down.
If that square's sides are a, the area of the omitted triangular piece is
((a/2)(a/3))/2 = a^2/12.
a^2 - a^2/12 = 11
12a^2 - a^2 = 132
11a^2 = 132
a^2 = 12.
Therefore, the missing piece is 12-11=1, and the red part is the same.
However, I did this more quickly in my head.
@@MrPaulc222 That's it! Good job!
From Thales' proportions we can see that the height of the red triangle is 1/3 of the square's side and its base is equal to half the side. Therefore, the area of the square is equal to the area of 12 triangles. Therefore, 12*(area of the triangle)=11+(area of the triangle)
area of the triangle =11/11=1
@@07Pietruszka1957 Great job! 👍
"This is EASY! Right?" Right. The problem just tessellated me.
Faster to figure it out with some mental arithmetic (using the first method entirely) than to explain it.
Also, the way the problem was presented, my gut told me it could be 1 immediately. Turned out my gut was right 😊
@@AbeIJnst 👑
You have a very intelligent gut!! 😂
Super easy: the area of the red triange is 1/12 the are of the square containing the blue are, and equal to the white triangle in the square containing the blue area. Therefore tge area of the red triangle is 1.
I solved it by seeing that the long diagonal had a slope of -2/3 -> the vertical went 2 distances vertical while going a horizontal distance of 3. That means that the long diagonal cut through the square with the blue at the mid point of the top side so the length of the red triangle was 1/2 and its height was 1/3, and I figured that it was 1 part of 12 parts of a square, so the blues equaled 11 so the red triangle is 1 unit area
I just start viewing your channel, I think it was earlier today (?) and enjoying it
@@ernestboston7707 Yes, that logic also works perfectly.
Thanks for watching, and I'm glad that you are enjoying the content, means a lot! ❤️
Option 2 was much easier for me.
От квадрата отделён сегмент с катетами 2 на 3. При этом больший катет, он же основание, составляет половину стороны. Таким образом, 11+2х*3х/2=(6х)². Разворачиваем. 11+3х²=36х²⇔11=36х²-3х²=33х²⇔х²=11/33. Нужно вновь умножить это значение на 3, чтобы получить искомое. Таким образом S=3*11/33=33/33=1, а общая площадь квадрата, стало быть, 12. Я догадывался, что автор загадал именно это число.
Si el lado de los cuadrados es "a"→ La línea inclinada tiene una pendiente =2a/3a=2/3→ corta al cuadrado central formando en su esquina superior derecha un triángulo rectángulo congruente con el rojo, cuyos catetos miden (a/2) y (a/3)→ la zona azul se compone de 11 triángulos de área 11/11=1 u², que son congruentes con el definido anteriormente→ Área triángulo sombreado rojo =1 u².
Gracias y un saludo.
¡Muchas gracias por tu análisis detallado! Me encanta cómo desglosaste cada paso del problema. 😊Gracias por compartirlo, un saludo cordial.
You have a line going 3 across and 2 up on a grid, your line intersects the middle of the bottom center cube, and cuts through at the 1/3 like on the side of the cube.
If we divide the cube vertically in half, and we divide it into 3rds horizontally, we get 6 rectangles, with the unshaded section being exactly half of a rectangle.
This is exactly the same area as half of a rectangle, which has an area of 2, so the red triangle is an area of 1, and the cubes have an area of 12, no complex math.
Just apply basic observation, and logic.
I'll watch the video now, and see what unneeded math is used.
Yeah, got to 2 mins, and the math over complicates the solution, at least for this problem.
Option 2.
Simpler solution. The diagonal slope is 2/3 so it intersect with the second square at 2/3, implying that the triangle height is 1/3s. The trinagle base is 1/2s. The triangle area is 1/12s^2. The square area s^2 minus the triangle area is equal to 11. You solve the equation and find that s^2 is equal to 12 so the trinagle area us 1.
@@Bokery1000 That also works great! 👍
Yes, or geometrically each large square can be split into stacked thirds, which each third divided into 4 congruent triangles. In all there are 12 such triangles with the red area covering 11 of them with an area of 11. So the area of one triangle is 1.
11÷8.25×9=12
11 / (11/12) · 1/12 = 1.