Question

Si en la figura siguiente d¹=2d² , ¿Qué fracción del círculo está sombreada ( Azul ) ?​

Answer 1

Respuesta:

$\frac{3\pi d_2^2}{8}$

Explicación paso a paso:

$\frac{\pi(\frac{2d_2+d_2}{2})^2}{2}-\frac{\pi\frac{(2d_2)^2}{2}}{2}+\frac{\pi\frac{(d_2)^2}{2}}{2}$

$=\frac{\pi \left(\frac{2d_2+d_2}{2}\right)^2-\pi \frac{\left(2d_2\right)^2}{2}+\pi \frac{d_2^2}{2}}{2}$

$=\frac{\pi \left(\frac{3d_2}{2}\right)^2-\pi \frac{\left(2d_2\right)^2}{2}+\pi \frac{d_2^2}{2}}{2}$

$\pi \left(\frac{3d_2}{2}\right)^2 =\frac{9\pi d_2^2}{4}\\\\\pi \frac{\left(2d_2\right)^2}{2} =2\pi d_2^2\\\\\pi \frac{d_2^2}{2} =\pi \frac{d_2^2}{2}$

$\frac{9d_2^2\pi }{4}-\pi 2d_2^2+\frac{d_2^2\pi }{2}=\frac{9d_2^2\pi }{4}-\frac{\pi 2d_2^2}{1}+\frac{d_2^2\pi }{2}\\\\\text{Sacamos minimo comun mltiplo que es 4}\\\\\frac{9d_2^2\pi }{4}-\frac{8\pi d_2^2}{4}+\frac{d_2^2\pi 2}{4}\\\\\frac{9d_2^2\pi -8\pi d_2^2+d_2^2\pi 2}{4} = \frac{3\pi d_2^2}{4}\\\\\frac{\frac{3\pi d_2^2}{4}}{2}\\\\= \frac{3\pi d_2^2}{8}$