Answer:
1.89cm²
Step-by-step explanation:
Area of the shaded segment = area of the sector OMN - Area of the triangle OMN
Area of the sector OMN = [tex]\frac{\theta}{360} * \pi r^{2}[/tex]
r= radius of the sector
[tex]\theta[/tex] = angle substended by the sector
Area of the sector OMN
[tex]= \frac{50}{360}*\pi (6)^{2}\\ = \frac{50}{360}*\pi 36\\=\frac{50\pi }{10}\\= 5\pi cm^{2}\\= 15.71cm^{2}[/tex]
Area of triangle ABC = [tex]\frac{1}{2} * base * height\\[/tex]
To get the height, we will use SOH CAH TOA trigonometry identity
According to SOH;
sin 25° = opp/6
opp = 6sin25°
opp = 2.54cm
The base of the triangle = 2(2.54) = 5.08cm
Th height can be gotten using the pythagoras theorem. The hypotenuse = 6cm and the opposite = 2.54cm
[tex]hyp^{2}=adj^{2}+opp^{2}\\ adj= \sqrt{6^{2}-2.54^{2} } \\adj = \sqrt{29.55}\\ adj = 5.44cm[/tex]
the adjacent = height = 5.44cm
Area of the triangle OMN
[tex]\frac{5.44*5.08}{2}\\ = 13.82cm^{2}[/tex]
Area of the shaded region = 15.71-13.82
Area of the shaded region = 1.89cm² to 3sf
