i) Apparently, we are to assume that AX and BX are tangent to the circle. Then OX bisects angle AOB and by the definition of trig functions,
AX = AO*tan(π/6)
AX = (12 cm)*(√3)/3
AX = 4√3 cm
ii) The area of triangle AXO is
triangle area = (1/2)(12 cm)(4√3 cm) = 24√3 cm²
The area of quadrilateral AXBO is twice that, 48√3 cm².
The area of sector ABO is
sector area = (1/2)r²θ
sector area = (1/2)(12 cm)²(π/3)
sector area = 24π cm²
Then the shaded area is the difference between the area of the quadrilateral and the sector.
shaded area = (48√3 -24π) cm²
