I got the same answer but got it just before the need to recognize that the short side of the red rectangle was 1/2 of the area of the large square. By that point it was already obvious that the large square has area 2 and half the small square, (two of the triangles together), have area .5. Once you see that all of the small triangles are congruent you can just double that and subtract, 2-1=1. Interesting problem with the different ways it can work out.

Clever

Area is 1. Set x the side of the small triangle. From the Gogu theorem we get the side of the inclined square : sqrt(1+x2) = y Let z be the distance from the top left corner to the intersection point between the top side of the inclined square and the constructed parallel line. Because the bottom left right triangle is similar to the top right angle triangle of hypotenuse x (same two angles), we can write : z/x = x/y Solving for z with respect to x we get: z = x2/sqrt(1+x2) The area we are interested in is : A = y*(y-z) which simplifies quickly to 1 after replacing y and z with our previous formulas.

@ThePhantomoftheMath

13 күн бұрын

Nice! Good job!

Respetando las premisas del trazado de la figura propuesta podemos suponer, sin alterar el resultado final, que el cuadrado original es de dimensiones √3*√3→ El cuadrado grande estará girado 30º respecto al inicial y tendrá dimensiones 2*2→ El triángulo cuya hipotenusa es el lado superior del cuadrado inicial tendrá un cateto largo de longitud (√3√3/2=3/2)→ Área del rectángulo rojo =(2*2)-[(3/2)*2] =4-3=1 ud². Interesante problema. Gracias y un saludo cordial.

@ThePhantomoftheMath

14 күн бұрын

Gracias por tu contribución

Area = 1. My reasoning is as follows. Suppose the small square has side (1 + x) and the large square has side y. Pythagoras using the triangle at the left side of the diagram, tells us that 1^2 + (1 + x)^2 = y^2. The rectangle with sides y and z - in other words the piece of the large square which is not red - has area yz; so the area of the red rectangle is y^2 - yz. The triangle at the left formed by the sides of the two squares, and the triangle at the top formed by the edge of the small square and the part edge of the large square are similar. Looking at side ratios of these two triangles and comparing the hypotenuse to the longer side y / (1 + x) = (1 + x) / z Multiply out gives yz = (1 + x)^2 So the area of the red rectangle is y^2 - yz We know that y^2 = 1 + (1+x)^2 , and we know that yz = (1 + x)^2 So the area of the red rectangle is 1 + (1+x)^2 - (1+x)^2 = 1 I’ll post this and take a look at your answer.

@andrewclifton9772

14 күн бұрын

My answer is the same as your method 1, but basically I did it in a different order but essentially the same procedure.

@ThePhantomoftheMath

13 күн бұрын

Very nice! Good logic! 👍

This is an excellent work! there is another simple method that can be use to solve the given question. See solution to similar question here: kzread.info/dash/bejne/hJejxrJvn6XgqdI.html