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:
1)55pi 2)50pi 3)82pi 4)31pi

• Ans 50 pi
  Triangle ACD with CD=6,AC=2,AD=sqrt(40) will be formed.Using Tan(/_CAD)=6/2=3.
  In triangle ADB,Tan(/_BAD)=BD/AD=3
  we have AD value.Then BD=sqrt(40)*3.
  In triangle ADB,
  AB^2=AD^2+BD^2
  AB^2=40+360=400
  AB=20.
  Diameter=20,radius =10.
  Area of semicircle=pi*100/2=50pi
  
• 8 years agoHelpfull: Yes(5) No(0)
• ans-50pi
  AD=sqrt(2^2+6^2) (right angle triangle )
  similarly
  BD=sqrt(40)*3
  AB=20
  r=10
  a=pi*50
  btw,get your punctuation right,from look of question it look like c is the center of the circle
• 8 years agoHelpfull: Yes(3) No(0)
• please send a figure for this question..
  
• 8 years agoHelpfull: Yes(0) No(0)
• https://cracku.in/cat-2006-question-paper-solved/21#
  here you can get diagrammatic answer
• 6 years agoHelpfull: Yes(0) No(0)