In areas related to circle of ncert class 10 what is the is best solution of the question of exercise 12.3 question number 15. Find the solution of this question in this way i solve it quickly give me the easiest and best way to solve this tricky question.In Fig. 12.33, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.
Radius of the quadrant ABC of circle = 14 cm
AB = AC = 14 cm
BC is diameter of semicircle.
ABC is right angled triangle.
By Pythagoras theorem in ΔABC,
BC2 = AB2 +AC2
⇒ BC2 = 142 +142
⇒ BC = 14√2 cm
Radius of semicircle = 14√2/2 cm = 7√2 cm
Area of ΔABC =( ½)×14×14 = 98 cm2
Area of quadrant = (¼)×(22/7)×(14×14) = 154 cm2
Area of the semicircle = (½)×(22/7)×7√2×7√2 = 154 cm2
Area of the shaded region =Area of the semicircle + Area of ΔABC – Area of quadrant
= 154 +98-154 cm2 = 98cm2