Question

Alguém me ajuda?

As medidas dos lados de um quadrilátero ABCD inscrito na circunferência são AB = BC = 5/2cm, CD= 4cm e AD= 3/2cm. Determine a medida X da diagonal BD.

Answer 1

Observe a figura em anexo ao final desta resposta.

Usando o Teorema do quadrilátero inscritível, temos que $\alpha$ e $\beta$ são ângulos suplementares:

$\beta=180^\circ-\alpha$

______________

$\bullet\;\;$ Aplicando a Lei dos Cossenos ao triângulo BCD:

$(BD)^2=(BC)^2+(CD)^2-2\cdot (BC)\cdot (CD)\cdot \cos \alpha\\\\\\ x^2=\left(\dfrac{5}{2}\right )^{\!\!2}+4^2-2\cdot \left(\dfrac{5}{2}\right )\cdot 4\cdot \cos \alpha\\\\\\ x^2=\dfrac{25}{4}+16-20\cos \alpha\\\\\\ 4x^2=25+64-80\cos \alpha\\\\\\ 4x^2=89-80\cos \alpha\\\\\\ 80\cos\alpha=89-4x^2\\\\\\ \cos \alpha=\dfrac{89-4x^2}{80}~~~~~~\mathbf{(i)}$


$\bullet\;\;$ Aplicando a Lei dos cossenos ao triângulo DAB:

$(BD)^2=(AB)^2+(AD)^2-2\cdot (AB)\cdot (AD)\cdot \cos \beta\\\\\\ x^2=\left(\dfrac{5}{2}\right )^{\!\!2}+\left(\dfrac{3}{2}\right )^{\!\!2}-2\cdot \left(\dfrac{5}{2}\right )\cdot \left(\dfrac{3}{2}\right )\cdot \cos \beta\\\\\\ x^2=\dfrac{25}{4}+\dfrac{9}{4}-\dfrac{30}{4}\cos \beta\\\\\\ 4x^2=25+9-30\cos \beta\\\\ 4x^2=34-30\cos \beta\\\\ 4x^2=34-30\cos (180^\circ-\alpha)\\\\ 4x^2=34-30\cdot (-\cos \alpha)\\\\ 4x^2=34+30\cos \alpha$


Substituindo $\cos \alpha$ pelo que foi dado por $\mathbf{(i)},$ temos

$4x^2=34+30\cdot \left(\dfrac{89-4x^2}{80} \right )\\\\\\ 4x^2=34+\dfrac{30\cdot (89-4x^2)}{80}\\\\\\ 4x^2=34+\dfrac{3\cdot (89-4x^2)}{8}\\\\\\ 32x^2=272+3\cdot (89-4x^2)\\\\ 32x^2=272+267-12x^2\\\\ 32x^2+12x^2=539\\\\ 44x^2=539\\\\ x^2=\dfrac{539}{44}\\\\\\ x^2=\dfrac{\diagup\!\!\!\!\! 11\cdot 49}{\diagup\!\!\!\!\! 11\cdot 4}\\\\\\ x^2=\dfrac{49}{4}\\\\\\ x=\sqrt{\dfrac{49}{4}}\\\\\\ \boxed{\begin{array}{c}x=\dfrac{7}{2}\mathrm{~cm} \end{array}}$


A medida da diagonal BD é $\dfrac{7}{2}\mathrm{~cm}.$


Bons estudos! :-)