. In the fig, ABC is a quadrant of a circle of radius 14cm and a semicircle is drawn BC as diameter.Find the area of the shaded region.
Answers
Answer:
add the figure also but I have tried to answer it
Step-by-step explanation:
Area of shaded region
= Area of semicircle BEC
- (Area of quadrant ABDC - Area of A ABC)
Area quadrant ABDC
Radius = 14 cm
Area of quadrant ABDC = (area of circle) 4
= 4 X (r?)
=x (14)? 22 7
22 =x 4 x 14 x 14
= 154 cm2
Area triangle ABC
Since ABDC is a quadrant
Z BAC = 90°
Hence A ABC is a right triangle
with Base AC & Height AB
BA
E
с
14 cm
So, Area A ABC =x Base x Height 2
X AC X AB
=x 14 x 14
= 7 x 14
= 98 cm
Area semicircle BEC
Here, Diameter = BC
Finding BC first
Since A ABC is a right triangle
Applying Pythagoras theorem in right triangle ABC
(Hypotenuse)? = (Height)? + (Base)? %3!
(BC)2 = (AB)2 + (AC)?
(BC)2 = (14)2 + (14)
(BC)2 = (14)2 x 2
BC = /(14)2 x 2 =
BC = /(14)? x V2 =
BC = 14 x V2 =
BC = 14 v2 cm
So, Diameter = BC = 14V2 cm =
Radius = r =
= 2 = 7 V2 cm =
Diameter
2 2
Area of semicircle BEC = X Area of circle
(12)
х 22 x (7v2) 2
Xx7x 7 x V2 x V2
=x 22 x 7 x 2
= 154 cm?
: Area of semi circle BEC = 154 cm
Area of shaded region
= Area of semicircle BEC
- (Area of quadrant ABDC - Area of A A
= 15415498
= 154 - 154 + 98
= 98 cm
Hence, area of shaded region 98 cm
maybe the answer is different because I searched for the figure on google so i have solved this question as per the figure I got.