Question

Qual o valor de x no triângulo abaixo? O gabarito é √5. Obrigada

Answer 1

Utilizando a relação métrica $b^2=am$ no triângulo $\text{ABC}$:

$b^2=am$

$b=\text{AB}=6 \ \ \ \ \ \ m=\text{AH} \ \ \ \ \ \ a=\text{AC}=m+5$

$6^2=(m+5)\cdot m \iff 36=m^2+5m \iff m^2+5m-36=0$

$\Delta=5^2-4\cdot1\cdot(-36)=25+144=169$

$m=\dfrac{-5+\sqrt{169}}{2\cdot1}=\dfrac{-5+13}{2}=\dfrac{8}{2}=4$

Assim, $\text{AH}=4$ e $\text{AC}=4+5=9$

Aplicando o Teorema de Pitágoras no triângulo $\text{ABC}$:

$\text{AC}^2=\text{AB}^2+\text{BC}^2$

$9^2=6^2+\text{BC}^2 \iff 81=36+\text{BC}^2$

$\text{BC}^2=45 \iff \text{BC}=\sqrt{45}$

$\text{BC}=3\sqrt{5}$

Os triângulos $\text{BCH}$ e $\text{ECD}$ são semelhantes, pois seus ângulos internos são iguais.

Logo:

$\dfrac{\text{CH}}{\text{BC}}=\dfrac{\text{CD}}{\text{CE}}$

$\dfrac{5}{3\sqrt{5}}=\dfrac{x}{3}$

$(3\sqrt{5})\cdot x=15 \iff x=\dfrac{15}{3\sqrt{5}} \iff x=\dfrac{5}{\sqrt{5}}$

$x=\dfrac{5}{\sqrt{5}}\cdot\dfrac{\sqrt{5}}{\sqrt{5}}=\dfrac{5\sqrt{5}}{5} \iff \boxe{x=\sqrt{5}}$