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
We use the formula for areas of semi-circles and sector of circles to solve the problem.
To find the area of semi-circle, we need to find the radius or diameter (BC) of the semicircle.
ΔABC is a right-angled triangle, right-angled at A (ABC being a quadrant).
AB = AC = 14 cm [Radius of the circle]
Using Pythagoras theorem, we can find the hypotenuse (BC) of ΔABC.
BC2 = AB2 + AC2
= (14 cm)2 + (14 cm)2
BC = √2 × (14 cm)²
= 14√2 cm
∴ Radius of semicircle BDC, r = BC/2 = 14√2/2 cm = 7√2 cm
Area of the shaded region = Area of semicircle - (Area of quadrant ABC - Area ΔABC)
= πr2/2 - [90°/360° × π(14)2 - 1/2 × AC × AB]
= π(7√2)2/2 - [π(14)2/4 - 1/2 × 14 × 14]
= [(22 × 7 × 7 × 2)/(7 × 2)] - [(22 × 14 × 14)/(7 × 4) - 7 × 14]
= 154 - (154 - 98)
= 98 cm2
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.
NCERT Solutions Class 10 Maths Chapter 12 Exercise 12.3 Question 15
The area of the shaded region where ABC is a quadrant of a circle of radius 14 cm and a semicircle drawn with BC as the diameter is 98 cm2.
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