Today i am solving the ncert class 10 chapter areas related to circles of exercise 12.3 question no. 12 . Please give me the best and easiest solution of this question ,find the simplest solution of this question i want to solve this tricky question easily. In Fig. 12.30, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the (i) quadrant OACB, (ii) shaded region.
Radius of the quadrant = 3.5 cm = 7/2 cm
(i) Area of quadrant OACB = (πR2)/4 cm2
= (22/7)×(7/2)×(7/2)/4 cm2
= 77/8 cm2
(ii) Area of triangle BOD = (½)×(7/2)×2 cm2
= 7/2 cm2
Area of shaded region = Area of quadrant – Area of triangle BOD
= (77/8)-(7/2) cm2 = 49/8 cm2
= 6.125 cm2