12) a semicircle is drawn with ab as its diameter. From c, a point on ab, a line perpendicular to ab is drawn, meeting the circumference of the semicircle at d. Given that ac=2 cm and cd=6 cm, the area of the semicircle (in sq cm) will be:
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Hlw mate
Solution:-
To find area of semi circle
firstly we have to find the length of radius which would denoted as ‘R’
In other words, R = AB/2 which is mid point of AB ,
the mid point can be named as “o” i.e, AO=BO=R
Now AC=2 , & CD =6
Hence CO= (R-2)
If the point ‘O’ and ‘D’ , OD=R, Now right angled triangle CDO
Now using Pythagoras formula, (OD)^2 = (CO)^2 +(CD)^2——-(1)
OD= R, CO= (R-2) and CD=6
Substituting the above in equation (1)
R^2=(R-2)^2 +6^2
R^2= R^2 +4 -4R+36
Now 4R = (36+4)
So, R= 40/4=10
Diameter AB=2R=10x2 =20
Area of Semicircle = (22/7)(20)^2/8=157.08 Sq. Units
Hope it helpful
Solution:-
To find area of semi circle
firstly we have to find the length of radius which would denoted as ‘R’
In other words, R = AB/2 which is mid point of AB ,
the mid point can be named as “o” i.e, AO=BO=R
Now AC=2 , & CD =6
Hence CO= (R-2)
If the point ‘O’ and ‘D’ , OD=R, Now right angled triangle CDO
Now using Pythagoras formula, (OD)^2 = (CO)^2 +(CD)^2——-(1)
OD= R, CO= (R-2) and CD=6
Substituting the above in equation (1)
R^2=(R-2)^2 +6^2
R^2= R^2 +4 -4R+36
Now 4R = (36+4)
So, R= 40/4=10
Diameter AB=2R=10x2 =20
Area of Semicircle = (22/7)(20)^2/8=157.08 Sq. Units
Hope it helpful
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