One side of a triangle is 20 and one of its others sides is 10.Given the area is 80 what is the third side?

• Area of triangle A=sqrt of s(s-a)(s-b)(s-c).
  where s=(a+b+c)/2.
  here a=20,b=10,A=80.
  therefore s=15+c/2.
  80^2=(15+c/2)(c/2-5)(c/2+5)(15-c/2)
  c=16.12
• 11 years agoHelpfull: Yes(30) No(0)
• let ABD be any tri. of sides AC=20, BD=10 and AD=x.
  Draw perpendicular from A meeting B on extending BD to C.
  area of tri=1/2*b*h
  = 1/2*AC*BD
  =1/2*AC*10
  AC=16
  Calc. BD using pytho. thrm as AB^2=AC^2+BC^2, giving BD=12 now in ABD, AD= (260)^1/2, using pytho thrm.
• 11 years agoHelpfull: Yes(10) No(13)
• sorry...80=0.5*a*b*sinC=0.5*20*10*sinC
  =>sinC=0.8 =>cosC=0.6=(a^2+b^2-c^2)/2ab
  putting a=20 & b=10 we get,
  c=root(260)=16.12
• 11 years agoHelpfull: Yes(3) No(0)
• a=20,b=10
  therefore area=1/2*(absinC)=1/2*(20*10*sinC)=80
  or, sinC=80/100
  therefore other side is=60
  
• 11 years agoHelpfull: Yes(2) No(13)
• the third side is 12.16.....
• 11 years agoHelpfull: Yes(0) No(3)