Área del cuadrado dentro de un triángulo rectángulo | Teorema de Tales y Pitágoras

We have a right triangle with a square inside it. The hypotenuse is divided into two segments of lengths 2 and 3, and the goal is to calculate the area of ​​the square. The key to this exercise is converting the geometric problem into an algebraic one. To do this, we call the side of the square x. Since the area of ​​the square is x², our real objective is to directly find the value of x². The first important element is Thales' theorem. Thanks to the similarity of triangles, we can relate the small segments around the square to side x. This gives us two fundamental relationships: a = 2x/3 d = 3x/2 Then the Pythagorean theorem comes into play. The hypotenuse of the large triangle measures 2 + 3 = 5, and the legs of the large triangle can be written in terms of x: x + d = 5x/2 x + a = 5x/3 Applying the Pythagorean theorem to the large right triangle: 5² = (5x/2)² + (5x/3)² Simplifying the equation, we finally obtain: x² = 36/13 Therefore, the area of ​​the square is 36/13 square units. This exercise is perfect for practicing triangle similarity, Thales' theorem, the Pythagorean theorem, and algebraic formulation using a geometric figure. It is a very good example of how a seemingly simple figure can conceal a beautiful mathematical idea. VIDEO CHAPTERS 00:00 Introduction to the geometric problem 00:06 Right triangle and 90-degree angle 00:10 The hypotenuse measures 2 + 3 = 5 00:19 Objective: Calculate the area of ​​the square 00:29 The four sides of the square are equal 00:36 Pause to try to solve it 01:19 We begin the solution 01:31 We want to calculate the area of ​​the square 01:41 We name auxiliary segments 01:52 We call x the side of the square 02:02 The area will be x² 02:19 We look for relationships in the figure 02:26 Thales' theorem comes into play 02:35 First proportion: a is to x as 2 is to 3 02:58 We obtain a = 2x/3 03:06 Second proportion: d is to x as 3 is to 2 03:27 We get d = 3x/2 03:38 Thales gives us the necessary relationships 03:42 The Pythagorean theorem comes into play 03:55 We apply Pythagoras to the large triangle 04:20 We convert geometry to algebra 04:35 We substitute the data into the equation 05:08 We simplify x + 3x/2 05:30 We simplify x + 2x/3 05:57 We square both sides 06:16 We factor out a common factor 06:37 We divide by 25 07:03 We eliminate denominators by multiplying by 4 and 9 07:48 We get 36 = 13x² 08:38 We solve for x² 08:55 Final result: area = 36/13 09:21 Final comment on the beauty of the problem 09:32 Proposed exercise for Practice 09:55 Goodbye More problems where we calculate the area:    • Cálculo del Área de Figuras Planas📏   #geometry #mathematics #mathwithjuan