Question

EJERCICIOS: Hallar el valor del lado desconocido en cada triangulo, aplicando el teorema. de. pitagoras​

Answer 1

Explicación paso a paso:

• Ejercicio 1:

${x}^{2} = {5}^{2} + {12}^{2} \\ {x}^{2} = 25 + 144 \\ {x}^{2} = 169 \\ x = \sqrt{169} \\ x = 13$

• Ejercicio 2:

${x}^{2} = {10}^{2} + {24}^{2} \\ {x}^{2} = 100 + 576 \\ {x}^{2} = 676 \\ x = \sqrt{676} \\ x = 26$

• Ejercicio 3:

${34}^{2} = {x}^{2} + {30}^{2} \\ 1156 = {x}^{2} + 900 \\ 1156 - 900 = {x}^{2} \\ {x}^{2} = 256 \\ x = \sqrt{256} \\ x = 16$

• Ejercicio 4:

${39}^{2} = {x}^{2} + {36}^{2} \\ 1521 = {x}^{2} + 1296 \\ 1521 - 1296 = {x}^{2} \\ {x}^{2} = 225 \\ x = \sqrt{225} \\ x = 15$

• Ejercicio 5:

${x}^{2} = {8}^{2} + {15}^{2} \\ {x}^{2} = 64 + 225 \\ {x}^{2} = 289 \\ x = \sqrt{289} \\ x = 17$

Espero que te sirva !)