(UFMG) No paralelogramo ABCD, temos AB = DB = CD e AD = 1/2AB. Se AB = 4 cm, então a área do paralel.... Pergunta de ideia deG0mes

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Olá.

$A = 4.y\\\\
x^2 + y^2 = 2^2\\
x^2+y^2 = 4\\\\
(4-x)^2 + y^2 = 4^2\\\\
16-8x+x^2 + y^2 = 16\\
x^2+y^2-8x = 0\\\\
 \left \{ {{x^2+y^2=4} \atop {x^2+y^2-8x=0}} \right. \\\\
y^2 = 4-x^2\\\\
x^2+(4-x^2)-8x=0\\
x^2+4-x^2-8x = 0\\
4 = 8x\\\\
x = \frac{4}{8} = \frac{1}{2}$

$4-x = 4-\frac{1}{2} = \frac{7}{2}\\\\
(\frac{7}{2})^2 + y^2 = 4^2\\\\
\frac{49}{4} +y^2 = 16\\\\
y^2 = 16-\frac{49}{4}\\\\
y^2 = \frac{64-49}{4} = \frac{15}{4}\\\\
y = \sqrt{\frac{15}{4}}\\\\
y =\frac{\sqrt{15}}{2}$

$A = 4.y\\\\
A = 4.\frac{\sqrt{15}}{2}\\\\
A = 2\sqrt{15}cm^2$

Gabarito letra E.

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