a semicircle is drawn with AB as its diameter from C a point on AB a line parpendicular to AB

radius of circle will be 10 because sq(r-2)+sq(6)=sq(r)
then area of semi circle will be 3.14*10*10/2=50pie
10 years agoHelpfull: Yes(4) No(0)
Let mid point of AB be E
AE=r radius of semicircle
Given D is the point on circum of circle
Then ED = radius of semicircle r
Let C be the point lie in b/w AE
CD perpendicular to AB
CD = 6cm given
AC = 2cm given
then CE = r-2
From the rt angle triangle DCE
r^2=36+(r-2)^2
On solving we get r=10
Area of semicircle = (1/2)(pi)(r)^2

= 50pi answer
10 years agoHelpfull: Yes(3) No(0)
Draw a semicircle with M as a centre(say) and AB is the diameter,therefore,AM=MB=r(say,radius of the semicircle)
Also,draw perpendicular DC on AB such that,CD=6 cm,AC= 2cm
so, CM= AM-AC=r-2.also join M and D and MD= r
In triangle MCD,by pythagoras theorem
MD^2=CM^2 + CD^2
r^2 = (r-2)^2 + 6^2
r^2 = r^2 + 4 -4r + 36
so, r= 10 cm

therefore area of semicircle =1/2*pi*(r^2)
=1/2*pi*(10^2) =50 pi

Hope you are clear with the soluion
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ans is 20pie
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9 years agoHelpfull: Yes(0) No(0)
Ans is 50pi
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a semicircle is drawn with AB as its diameter. from C a point on AB,a line parpendicular to AB is drawn,meeting the circumference of the semicircle at D. Given that AC=2 and CD=6,the area of the semicircle is?