Question

9) As medidas dos lados do triangulo abaixo formam uma progressão aritmética (PA) de razão 4 (quatro). Qual é a medida de perimetro do triángulo?​

Answer 1

Façamos a PA de razão 4 :

$\text{cateto 1} = \text a-4 \\\\ \text {cateto 2}=\text a \\\\ \text{hipotenusa} = \text a+4 \\\\ \text{P.A }\to (\text a-4,\text a,\text a+4) \\\\ \underline{\text{Aplicando o teorema de Pitagoras}}:\\\\ (\text a+4)^2=\text a^2+(\text a-4)^2 \\\\ \text a^2+8\text a+16=\text a^2+\text a^2-8\text a+16 \\\\ 8\text a = \text a^2-8\text a \\\\ \text a^2-16\text a = 0 \\\\ \text a(\text a-16) =0 \ ; \ \text a\neq 0 \\\\ \text a-16=0 \\\\ \text a =16 \\\\ \underline{\text{Lados do tri{\^a}ngulo}}:$

$\text{cateto 1} = \text a-4 \to 16-4 \to 12 \\\\ \text{cateto 2}= \text a \to 16 \\\\ \text{hipotenusa}=\text a+4 \to 16+4\to20 \\\\ \underline{\text{Per{\'i}metro do tri{\^a}ngulo (2p) }}: \\\\ \text {2p}= 12+16+20 \\\\ \huge\boxed{2\text p =48\ }\checkmark$