Ex 12.3, 15
In figure, 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.
Area of shaded region
= Area of semicircle BEC
(Area of quadrant ABDC Area of ABC)
Area quadrant ABDC
Radius = 14 cm
Area of quadrant ABDC = 1/4 (area of circle)
= 1/4 ( r2)
= 1/4 22/7 (14)^2
= 1/4 22/7 14 14
= 154 cm2
Area triangle ABC
Since ABDC is a quadrant
BAC = 90
Hence ABC is a right triangle
with Base AC & Height AB
So, Area ABC = 1/2 Base Height
= 1/2 AC AB
= 1/2 14 14
= 7 14
= 98 cm2
Area semicircle BEC
Here, Diameter = BC
Finding BC first
Since ABC is a right triangle
Applying Pythagoras theorem in right triangle ABC
(Hypotenuse)2 = (Height)2 + (Base)2
(BC)2 = (AB)2 + (AC)2
(BC)2 = (14)2 + (14)2
(BC)2 = (14)2 2
BC = ("(14)2 2" )
BC = ("(14)2" ) ("2" )
BC = 14 ("2" )
BC = 14 2 cm
So, Diameter = BC = 14 2 cm
Radius = r = /2 = /2 = (14 2)/2 = 7 2 cm
Area of semicircle BEC = 1/2 Area of circle
= 1/2 ( 2)
= 1/2 22/7 (7 2)^2
= 1/2 22/7 7 7 2 2
= 1/2 22 7 2
= 154 cm2
Area of semi circle BEC = 154 cm2
Area of shaded region
= Area of semicircle BEC
(Area of quadrant ABDC Area of ABC)
= 154 (154 98)
= 154 154 + 98
= 98 cm2
Hence, area of shaded region = 98 cm2

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.