a circle whose diameter is 18cm,
a right angled triangle is inserted in that circle,
and a small circle whose diameter is 6cm inscribed in right angled triangle.
find the perimeter of triangle

• triangle is right angled
  so, diameter of circle is the hypotensue of triangle.
  
  let other two sides of triangle are x,y
  area of triangle = r*s where,r= inradius, s =semiperimeter
  
  1/2 *xy = (x+y+18)/2 * 6/2
  => 2xy = 6(x+y) + 108
  
  x^2 + y^2 = 18^2
  => (x+y)^2 - 2xy - 324 = 0
  => (x+y)^2 - 6(x+y) - 108 - 324 = 0
  => (x+y)^2 - 6(x+y) - 432 = 0
  => (x+y-24)(x+y+18)= 0
  => (x+y)= 24
  
  perimeter = x+y+18 = 24 + 18 = 42
  
  
• 9 years agoHelpfull: Yes(31) No(1)
• let the sides of the triangle be x,y,z.
  Now,let the diamter of outer circle be the largest side of the right angled triangle i.e it's hypotenuse.x=18
  we know dat (z+y-x)/2=inradius which is 3 in the given question
  therefore,(y+z-18)/2=3,
  above eqn gives y+z=24
  the perimeter of the triangle is x+y+z=24+18=42
• 9 years agoHelpfull: Yes(10) No(0)