Respuesta :

aosera
area of semicircle minus area of the triangle

Area of a circle :
[tex]\pi {r}^{2} [/tex]
So, area of a SEMIcircle:
[tex] \frac{1}{2} \pi \: r {}^{2} [/tex]
Area of a right-angled triangle:
[tex] \frac{1}{2} \times base \times height[/tex]
Area of the shaded region :
[tex] \frac{1}{2} \times 3.14 \times 6 {}^{2} - \frac{1}{2} \times 12 \times 6 \\ = 56.52 - 36 \\ = 20.52 \: {yd}^{2} [/tex]

Base of the triangle is 6+6 = 12.