Question

Sabendo que na figura a seguir a medida do lado AB = 6 cm e AC = 2 cm, qual a área do quadrado CDEF?

Answer 1

Explicação passo-a-passo:

Pelo Teorema de Pitágoras:

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

$6^2=2^2+\overline{BC}^2$

$36=4+\overline{BC}^2$

$\overline{BC}^2=36-4$

$\overline{BC}^2=32$

$\overline{BC}=\sqrt{32}$

$\overline{BC}=4\sqrt{2}~\text{cm}$

Seja $l$ a medida do lado do quadrado CDEF

Assim, $\overline{AD}=2-l$ e $\overline{BF}=4\sqrt{2}-l$

Os triângulos ADE e EFB são semelhantes

$\dfrac{\overline{DE}}{\overline{AD}}=\dfrac{\overline{BF}}{\overline{EF}}$

$\dfrac{l}{2-l}=\dfrac{4\sqrt{2}-l}{l}$

$l\cdot l=(2-l)\cdot(4\sqrt{2}-l)$

$l^2=8\sqrt{2}-2l-4l\sqrt{2}+l^2$

$2l+4l\sqrt{2}=8\sqrt{2}$

$l\cdot(2+4\sqrt{2})=8\sqrt{2}$

$l=\dfrac{8\sqrt{2}}{4\sqrt{2}+2}\cdot\dfrac{4\sqrt{2}-2}{4\sqrt{2}-2}$

$l=\dfrac{64-16\sqrt{2}}{(4\sqrt{2})^2-2^2}$

$l=\dfrac{64-16\sqrt{2}}{32-4}$

$l=\dfrac{64-16\sqrt{2}}{28}$

$l=\dfrac{16-4\sqrt{2}}{7}~\text{cm}$

A área do quadrado CDEF é:

$S=l^2$

$S=\left(\dfrac{16-4\sqrt{2}}{7}\right)^2$

$S=\dfrac{256-128\sqrt{2}+32}{49}$

$S=\dfrac{288-128\sqrt{2}}{49}~\text{cm}^2$