one side of triangle is 50cm and other is 10cm and area given is 150 cm^2. then find the third side of the bermuda triangle?

a)30*(root(2)) c)(root(30))*2 b) root(1860) d)none of these

• Area of a triangle is given as when the sides are giver as,
  (s(s-a)(s-b)(s-c))^1/2, where s= a+b+c/2
  Let the third side be x...then,
  area = [(60+x)/2(60+x-2x)/2(60+x-100/2)(60+x-20)/2]^1/2
  or, [(60+x)/2(60+x-2x)/2(60+x-100/2)(60+x-20)/2]^1/2= 150
  now put the value of x from option....
  let x= 30*(root 2)
  = [(30+15(root 2))(30-15(root 2))(15(root 2)-20(15(root 2)+20)]^1/2
  = [(900-450)(450-400)]^1/2
  = [450*50]^1/2
  = [225*100]^1/2
  = 150 and is area of triangle (given).
• 9 years agoHelpfull: Yes(10) No(0)
• lets say ABC is a triangle;
  
  s=(50+10+x)/2;
  
  area of triangle ABC=(S(S-50)*(S-10)*(S-X))^1/2=150;
  
  solve it
  
• 9 years agoHelpfull: Yes(3) No(2)
• lets take ABC as triangle BC=50,AC=10 and AB=?
  let AD be the altitude from A to D
  1/2*BC*AD=150
  1/2*50*AD=150
  AD=6
  in right triang. ADC AD^2+CD^2=AC^2
  CD^2=64
  CD=8
  CD+BD=AD
  8+BD=50
  BD=42
  
  so in right triang.ABD AB^2=AD^2+BD^2
  =42^2+6^2
  1800
  AB=sqrt 1800
  i think C will be answer in it is 1800 instead of 1860
• 9 years agoHelpfull: Yes(2) No(2)
• area of triangle =1/2 b*h assuming ABC is tringle AB is 10 BC is 50
  150=1/2*50*h D is point on height on BC
  h=3 cm
  now in triangle ABD
  AB= 10 AD= 3 then by pythagorus BD= root 91
  now in triangle ADC
  AD= 3 DC = 50-root 91
  so we can calculate AC as AC= root (sqr 9 +sqr (50-root 91)
  a is answer
  
  
  
  
• 9 years agoHelpfull: Yes(1) No(2)
• 30*(root(2))
• 9 years agoHelpfull: Yes(0) No(0)
• one side of triangle=50 cm
  other side of triangle =10 cm
  let the third side of triangle=x cm
  area of triangle =(1/2)*x*10=150
  therefore, x=30
  and area of triangle=(1/2)*x*50=150
  therefore,x=6
  then, answer is (d)none of these
• 9 years agoHelpfull: Yes(0) No(2)