Question

porfavor, doy coronita ​

Answer 1

PARA EL PROBLEMA "A" =

Area del Cuadrado =

$a = 6cm \times 6cm = 36 {cm}^{2}$

Area de la Seccion de color Blanco =

$a \: = \pi \times {r}^{2} \\ a = \pi \times {6cm}^{2} \\ a = (3.1415..) \times (36 {cm}^{2} ) \\ a = 113.0973355.. {cm}^{2} \\ a = \frac{113.0973355.. {cm}^{2} }{4} \\ a = 28.27433388... {cm}^{2}$

RESTAMOS EL AREA BLANCA A TODO EL CUADRADO, SEA AREA NEGRA = "X"

$x \: = 36 {cm}^{2} - 28.27... {cm}^{2} \\ a = 7.725666.. {cm}^{2}$

PARA EL PROBLEMA "B" =

Area del Cuadrado =

$a = 8cm \times 8cm = 64 {cm}^{2}$

Area de la reguon blanca =

$r = \frac{d}{2} \\ r = \frac{8cm}{2} \\ r = 4cm$

Radio = 4cm

$a = \pi \times {r}^{2} \\ a = \pi \times {4cm}^{2} \\ a = (3.1415..) \times ( {16cm}^{2} ) \\ a = 50.26548246.. {cm}^{2}$

SON DOS SEMICIRCULOS, ASI QUE SUMANDOLOS HACEN UNA CIRCUNFERENCIA DENTRO DEL CUADRADO.

Area Sonbreada =

$a = 64 {cm}^{2} - 50.26548246.. {cm}^{2} \\ a = 13.73451754.. {cm}^{2}$

$saludosssssssssss$