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# 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. Q.15

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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.

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AB = AC = 14 cm

BC is diameter of semicircle.

ABC is right angled triangle.

By Pythagoras theorem in Î”ABC,

BC2Â = AB2Â +AC2

â‡’ BC2Â = 142Â +142

â‡’ BC = 14âˆš2 cm

Radius of semicircle = 14âˆš2/2 cm = 7âˆš2 cm

Area of Î”ABC =( Â½)Ã—14Ã—14 = 98 cm2

Area of quadrant = (Â¼)Ã—(22/7)Ã—(14Ã—14) = 154 cm2

Area of the semicircle = (Â½)Ã—(22/7)Ã—7âˆš2Ã—7âˆš2 = 154 cm2

Area of the shaded region =Area of the semicircle + Area of Î”ABC â€“ Area of quadrant

= 154 +98-154 cm2Â = 98cm2

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