SOLUTION: The perimeter of a right triangle is 42 and the sum of the squares of its 3 sides is 722. Determine the area of the right triangle.
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Question 1131514: The perimeter of a right triangle is 42 and the sum of the squares of its 3 sides is 722. Determine the area of the right triangle. Found 2 solutions by Alan3354, ikleyn:Answer by Alan3354(69443) (Show Source):
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The perimeter of a right triangle is 42 and the sum of the squares of its sides is 722. Determine the area of the right triangle.
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a + b + c = 42 (1)
a^2 + b^2 = c^2 (2)
a^2 + b^2 + c^2 = 722 (3)
From (2) and (3)
2c^2 = 722 ====> c^2 = 722/2 = 361
c = = 19.
Now substitute c= 19 and c^2 = 361 into eq(1) and eq(2). You will get
a + b = 23 (1')
a^2 + b^2 = 361 (2')
Now square both sides of (1'). Keep (2') as is. You will get
a^2 + 2ab + b^2 = 23^2 = 529 (1'')
a^2 + b^2 = 361 (2'')
Subtract (2'') from (1''). You will get
2ab = 529 - 361 = 168.
Now the area of the triangle = = = 42. ANSWER
