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From AB as a diameter, a semicircle is drawn, from some random point C on AB, a perpendicular is drawn meeting the semicircle at D. Given AC=2, CD=6, find area of the semicircle?
Read Solution (Total 7)
-
- draw the diagram
Given AC=2, CD=6, so AD=(40)^1/2
let BC=x, BD=y ,so AB=(2+x)
angle of a semicircle=90 degree
find area of triangle ABD & equate
(1/2)*(x+2)*6=(1/2)*(40)^1/2*y
or 10y^2=9x^2+36x+36
or 10(x^2+36)=9x^2+36x+36 (in triangle BCD,6^2+x^2=y^2)
or x^2-36x+324=0
or (x-18)^2=0
or x=18
dia of semicircle=AB=(2+x)=2+18=20
so radius of semicircle=20/2=10
Area of semicircle=1/2*pi*10^2=50*pi=50*3.14=157
- 10 years agoHelpfull: Yes(37) No(0)
- let o be the center,then oA is radius(let it be r)
AC=Ao+oC=r+oC (oC be x)
oCD is right angled
oD^2=oC^2+CD^2
r^2=6^2+(2-r)^2
r=10
area=(pi*r^2)/2=157 - 10 years agoHelpfull: Yes(18) No(3)
- Draw the diagram,
Given,AC=2,CD=6.
Now, use the property,AC*BC=CD^2.
On putting, the value of AC and CD in the above property we get BC=18.
so, now diameter of semicircle(AB)=20.
So, Radius of semicircle=10.
Now,Area of semicircle=1/2(pi(r)^2)
=1/2(pi(10)^2)
=1/2(pi*100)
=(50 pi)Ans.
- 9 years agoHelpfull: Yes(7) No(0)
- r=3
area is 36*3.14 - 10 years agoHelpfull: Yes(0) No(1)
- area=1/2*pi*r^2
cd=r=6
a=1/2*3.14*6^2=56.52 - 10 years agoHelpfull: Yes(0) No(1)
- 18*pi..
cd=6 is a radius of the semicircle.. - 10 years agoHelpfull: Yes(0) No(1)
- draw the diagram
Given AC=2, CD=6, so AD=(40)^1/2
let BC=x, BD=y ,so AB=(2+x)
angle of a semicircle=90 degree
find area of triangle ABD & equate
(1/2)*(x+2)*6=(1/2)*(40)^1/2*y
or 10y^2=9x^2+36x+36
or 10(x^2+36)=9x^2+36x+36 (in triangle BCD,6^2+x^2=y^2)
or x^2-36x+324=0
or (x-18)^2=0
or x=18
dia of semicircle=AB=(2+x)=2+18=20
so radius of semicircle=20/2=10
Area of semicircle=1/2*pi*10^2=50*pi=50*3.14= 157 - 10 years agoHelpfull: Yes(0) No(1)
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