What is the best solution of ncert class 10 areas related to circles of exercise 12.3 question number 4. Please give me the easiest and simplest solution of this question .Find the best solution of this important question.Find the area of the shaded region in Fig. 12.21, if ABCD is a square of side 14 cm and APD and BPC are semicircles.
It is given that OAB is an equilateral triangle having each angle as 60°
Area of the sector is common in both.
Radius of the circle = 6 cm.
Side of the triangle = 12 cm.
Area of the equilateral triangle = (√3/4) (OA)2= (√3/40×122 = 36√3 cm2
Area of the circle = πR2 = (22/7)×62 = 792/7 cm2
Area of the sector making angle 60° = (60°/360°) ×πr2 cm2
= (1/6)×(22/7)× 62 cm2 = 132/7 cm2
Area of the shaded region = Area of the equilateral triangle + Area of the circle – Area of the sector
= 36√3 cm2 +792/7 cm2-132/7 cm2
= (36√3+660/7) cm2