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, CD= 6cm, the are of the semicircle ( in sq. cm ) will be:
1. 82pi
2. 50pi
3. 31pi
4. 55pi

• 2. 50 pi is the answer
  Let O be the centre of semicirle. Draw the figure as per the question and join OD, n let OD=x.
  then OD=OA=x radius of the semicircle
  AC=2, so oc= x-2
  Now in triangle OCD
  CD=6, OD=x, OC=x-2,
  solving OD^2=OC^2+CD^2
  we get x=10
  so area will b 100pi/2=50pi ;) :p
  
• 10 years agoHelpfull: Yes(15) No(0)
• In triangle ACD
  angle ACD is of 90
  Let angle A=theta
  this implies tan(90-theta)=AC/CD ......(1)
  
  In triangle ADB
  angle D is of 90
  Hence angle B will be 90-theta(as angle A is theta we assumed)
  
  In triangle BCD
  angle C is of 90
  angle B is 90-theta
  therefor tan(90-theta)=CD/CB .....(2)
  
  
  Solvind 1 and 2
  AC/CD=CD/CB
  CB=18cm
  
  hence Diameter of circle that is AB=20cm
  
  =>radius=10
  area of semicirle=100pi/2
  =50pi
  
  
  Hence answer is 50pi
  option 2 is correct
  
  
  Correct me if I am Wrong
• 10 years agoHelpfull: Yes(14) No(1)
• AC=2
  CD=6
  SO accordingly the hypotenus with centre and D is radius hit and trial radius can be 10
  so area =50pi
  
• 10 years agoHelpfull: Yes(0) No(1)