Question

61. Se o perímetro de um retângulo é 14 cm e a sua diagonal mede (raiz de 29)cm, então sua área, em cm?, é igual a: (A) 7 (B) 8 (C) 10 (D) 12 (E) 20 62 A​

Answer 1

$\large\boxed{\begin{array}{l}\underline{\rm Per\acute imetro~do~ret\hat angulo}\\\sf 2p=2\cdot( x+y)\\\sf 2p\longrightarrow per\acute imetro\\\sf x\longrightarrow comprimento\\\sf y\longrightarrow largura\\\underline{\rm diagonal~do~ret\hat angulo:}\\\sf d=\sqrt{x^2+y^2}\\\underline{\rm \acute Area~do~ret\hat angulo}\\\sf A=x\cdot y\end{array}}$

$\large\boxed{\begin{array}{l}\sf 2p=2\cdot(x+y)\\\sf 14=2\cdot(x+y)\\\sf x+y=\dfrac{14}{2}\\\sf x+y=7\\\sf d=\sqrt{x^2+y^2}\\\sf \sqrt{29}=\sqrt{x^2+y^2}\\\sf (\sqrt{29})^2=(\sqrt{x^2+y^2})^2\\\sf x^2+y^2=29\\\sf (x+y)^2=x^2+y^2+2xy\\\sf xy=\dfrac{(x+y)^2-(x^2+y^2)}{2}\\\\\sf xy=\dfrac{7^2-29}{2}\\\\\sf xy=\dfrac{49-29}{2}\\\\\sf xy=\dfrac{20}{2}\\\\\sf xy=10\\\sf A=xy\\\sf A=10~cm^2\\\huge\boxed{\boxed{\boxed{\boxed{\sf\maltese~alternativa~C}}}}\end{array}}$