Question

Em um triângulo retângulo de catetos x e y, sabe-se que sua hipotenusa mede 17 cm e o perímetro é de 40 cm. Calcule os valores de x e y. (Dados: x < y)

Answer 1

Devemos lembrar da formula de Pitágoras: $\mathtt{a^2+b^2 = c^2}$

$\mathtt{a = cateto'~~~~b=cateto''~~~~c=hipotenusa} \\ \\ \mathtt{a^2+b^2 = 17^2}\\ \boxed{\mathtt{a^2 + b^2 = 289}}$

Agora formula para achar o perímetro do triângulo retângulo:

$\mathtt{P = a+b+c} \\ \mathtt{40 = a + b + 17}} \\ \mathtt{a + b = 40 -17} \\ \boxed{\mathtt{a + b =23 }}$

Agora vamos criar um sistema:

$\mathtt{a^2 + b^2 = 289} \\ \mathtt{a + b =23 } \\ \\ \mathtt{a + b = 23} \\ \mathtt{a = 23 - b} \\ \\ \mathtt{Substituir~a} \\ \\ \mathtt{a^2+b^2=289} \\ \mathtt{(23-b)^2+b^2=289} \\ \mathtt{(23-b)~.~(23-b)+b^2=289} \\ \mathtt{529 - 23b - 23b + b^2 + b^2 =289} \\ \mathtt{2b^2 - 46b + 529- 289 =0} \\ \mathtt{2b^2-46b+240=0} \\ \\ \mathtt{a = 2} \\ \mathtt{b = -46} \\ \mathtt{c = 240}$

$\\ \mathtt{\Delta =b^2-4~.~a~.~c } \\ \mathtt{\Delta = (-46)^2-4~.~2~.~240} \\ \mathtt{\Delta =2.116-8~.~240 } \\ \mathtt{\Delta =2.116-1.920 } \\ \mathtt{\Delta = 196} \\ \\ \\ \mathtt{\dfrac{-b+-~\sqrt{\Delta}}{2~.~a}~~=~~\dfrac{-(-46)+-~\sqrt{196}}{2~.~2}~~=~~\dfrac{46+-~14}{4}} \\ \\ \\ \mathtt{x'= \dfrac{46+14}{4}~~=~~\dfrac{60}{4}~~=~~15~cm} \\ \\ \\ \mathtt{x'' =\dfrac{46-14}{4}~~=~~\dfrac{32}{4}~~=~~8}~cm$

$\mathtt{Resposta:} \\ \mathtt{x = 8~cm} \\ \mathtt{y = 15~cm}$