For a triangle, given length of one side=68 cm, second side= 32 cm. Area of the triangle= 960 sq. cm. Find the exact value of third side?

• we see 68^2-32^2=(68+32)*(68-32)=100*36=60^2
  now (1/2)*60*32=960(match with given options)
  (i.e area of a right angled triangle whose sides are 32,60,68)
  third side=60
• 10 years agoHelpfull: Yes(36) No(1)
• We can solve it by Heron's formula. Here a=68 b=32 and c=? area is given 960
  semi-perimeter s=(a+b+c)/3 and area T =sqrt[s*(s-a)*(s-b)*(s-c)]
  modifying the formula for area we get
  T = (1/4)*sqrt[4*a^2*b^2-(a^2 + b^2 - c^2)]-----eq1
  solving eq1 by putting values of T(=960),a(=68),b(=32) we get
  c=60cm
  
• 10 years agoHelpfull: Yes(8) No(2)
• how the formula is modified to T=(1/4)*sqrt[4a^b^2-(a^2+b^2-c^2)] please explain
• 10 years agoHelpfull: Yes(1) No(0)
• Using the formulas
  A
  =
  s
  (
  s
  ﹣
  a
  )
  (
  s
  ﹣
  b
  )
  (
  s
  ﹣
  c
  )
  s
  =
  a
  +
  b
  +
  c
  2
  There are 2 solutions for
  c
  c
  =
  a
  2
  +
  b
  2
  +
  2
  (
  a
  b
  )
  2
  ﹣
  4
  A
  2
  =
  32
  2
  +
  68
  2
  +
  2
  ·
  (
  32
  ·
  68
  )
  2
  ﹣
  4
  ·
  960
  2
  ≈
  87.72685
  c
  =
  a
  2
  +
  b
  2
  ﹣
  2
  (
  a
  b
  )
  2
  ﹣
  4
  A
  2
  =
  32
  2
  +
  68
  2
  ﹣
  2
  ·
  (
  32
  ·
  68
  )
  2
  ﹣
  4
  ·
  960
  2
  =
  60
  
• 9 years agoHelpfull: Yes(1) No(17)