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
Solution:
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
☛ Check: NCERT Solutions Class 10 Maths Chapter 12
Video Solution:
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
Summary:
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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