Given perimeter of isolated triangle is 6+3root2.find area of triangle?

• area of trangle
  when sides are given is A=sqrtof(s(s-a)(s-b)(s-c))----(1)
  where perimeter s=a+b+c
  in our case a,b,c=b ;bcoz isolated trangle
  6+3root2=a+2b
  a=(6+3root2)-2b
  
  substitute in equ--(1)
  and get solution
• 10 years agoHelpfull: Yes(0) No(0)
• area of isoceles triangle=1/4(b*sqt 0f 4a^2-b^2).---(1)
  since isoceles a=c,P=2a+b (since Perimeter= a+b+c)
  Also b=aroot2 (sice b^2=a^2+a^2)
  Therefore P=2a+aroot2
  given P=6+3root2 ,6+3root2=2a+aroot2 => a=3
  therefore b= 3root2
  subsititute a & b in (1), we get Area=9/2
• 10 years agoHelpfull: Yes(0) No(0)