14. Find the area of the segment of a circle of radius 12cm whose corresponding sector has a central angle of 90°.
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It is clear that Area ACB= Area OACB-Area OAB
Asegment = Asector - Atriangle
Given that the radius is 12 cm and cnetral angle 60°
8
60° 22
A sector= x πr² = 360" 360° Triangle OAB is an isosceles triangle. x 12 x 12 75.43 cm²
Let us drop a perpendicular, OM from O to AB.
Thus OM divides the triangle in to exactly two right trianlges, with angles 30-60-90 That is OMB and OMA are right triangles. Since it is a right triangle we can apply trigonometric ratios.
Let OM=h and MB=x
sin60" opp h 12
hyp
2 12
⇒h=6√3 cm
Similarly, cos60° = adj hyp
⇒ x = 6 cm
AB=2MB 2x6= 12 cm Area of OAB = xbase x height
X ABX OM
x 12x6√3
= 62.35 cm²
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