Today i am solving the ncert class 10 question of chapter areas related to circles its very hard to solve the exercise 12.3 question no. 16 . Give me the easiest solution of this question also give me the best solution of this question .Calculate the area of the designed region in Fig. 12.34 common between the two quadrants of circles of radius 8 cm each.
AB = BC = CD = AD = 8 cm
Area of ΔABC = Area of ΔADC = (½)×8×8 = 32 cm2
Area of quadrant AECB = Area of quadrant AFCD = (¼)×22/7×82
= 352/7 cm2
Area of shaded region = (Area of quadrant AECB – Area of ΔABC) = (Area of quadrant AFCD – Area of ΔADC)
= (352/7 -32)+(352/7- 32) cm2
= 2×(352/7-32) cm2
= 256/7 cm2