resuelvanlo y doy 10 puntos
Respuestas a la pregunta
Respuesta:
seria necesario coronita por eso
Respuesta:
1)
podemos ajustar esa area sombreada colocando el semicirculo sombreado en el lugar donde no esta sombreado y asi formar 2 cuadrados completos
el area sombreada quedaria como
As = 2(Acuadrado)
As = 2L²
As = 2(2cm)²
As = 2*4cm² = 8cm²
2)
As = triangulo rectangulo
As = (b*h)/2
As = (1m*6m)/2
As = 6m²/2
As = 3m²
3)
As = Area de semicirculo grande - 2area de semicirculo pequeño
As = πr²/2 - (2*πr²/2)
As = πR²/2 - *πr²
As = (8m)²π/2 - (4m)²π
As = 64πm²/2 - 16πm²
As = 32πm² - 16πm²
As = 32πm² ≈ 100.53m²
4)
nos dice que EFGH = perimetro del cuadrado
perimetro = 20cm = 4*L
L = 20cm/4 = 5cm
ya con este dato es halla el triangulo grande sombreado y dos veces el triangulo pequeño sombreado
As = Atriangulo grande + 2Atriangulo pequeño
As = (b*h/2) + 2(B*H/2)
As = (b*h/2) + B*H
As = [ (5cm*2.5cm)/2 ] + [2.5cm*2.5cm]
As = 6.25cm² + 6.25cm²
As = 12.5cm²