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?

• 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
  
• 11 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
• 11 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.
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
• 10 years agoHelpfull: Yes(7) No(0)
• r=3
  area is 36*3.14
• 11 years agoHelpfull: Yes(0) No(1)
• area=1/2*pi*r^2
  cd=r=6
  a=1/2*3.14*6^2=56.52
• 11 years agoHelpfull: Yes(0) No(1)
• 18*pi..
  cd=6 is a radius of the semicircle..
• 11 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)