Problem

• The perimeter, the area and Pythagora theorems gives three equations as follows
  
  

  

  
  
  a + b + h = 60
  
  (1 / 2) a b = 150    or    a b = 300
  
  a ² + b ² = h ²
  
  
• Rewrite the equation a + b + h = 60 as follows
  
  a + b = 60 - h
  
  
• Square both sides
  
  (a + b)² = (60 - h)²
  
  
• Expand both sides
  
  a² + b² + 2 a b = 60² + h² - 120 h
  
  
• Combine the equation a ² + b ² = h ² with the above equation to obtain
  
  2 a b = 60² - 120 h
  
  
• a b is known to be equal to 300, hence the above equation becomes
  
  600 = 60² - 120 h
  
  
• Solve for h to obtain
  
  h = 25 units
  
  
• Substitute h by 25 in the equation a + b + h = 60 to obtain
  
  a + b = 60 - 25 = 35
  
  
• Since a b = 300, then b = 300 / a which is substituted in the equation a + b = 35 to obtain
  
  a + 300 / a - 35 = 0
  
  
• Multiply all terms by a to obtain a quadratic equation of the form
  
  a ² + 300 - 35 a = 0
  
  
• Solve the above equation to obtain two solutions
  
  a = 20 and a = 15
  
  
• Use the equation a b = 300 to obtain
  
  when a 20 , b = 15 and when a = 15 , b = 20
  
  
• The two sides of the right triangle and the hypotenuse are, respectively, given by
  
  15 units, 20 units and 25 units.
  
  
• As an exercise check the perimeter given in the problem.

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