Expert: Ahmed Salami - 11/4/2008

Question
What is the largest possible area for a right triangle whose hypotenuse is 5 cm long, and what are its dimensions?

Answer
Hi davis,
Let the right angle have sides x and y. Its area
A = (1/2)xy
but by the pythagoras theorem,
x^2 + y^2 = 5^2
x^2 + y^2 = 25
y^2 = 25 - x^2
y = sqrt(25 - x^2)
Therefore,
A = x.sqrt(25 - x^2)/2
To maximise A, we find dA/dx and equate to zero.
dA/dx = (1/2) [x[1/2sqrt(25 - x^2)].(-2x) + sqrt(25 - x^2).(1)]
     = (1/2) [-x^2/sqrt(25 - x^2) + sqrt(25 - x^2)]
equating to zero,
(1/2) [-x^2/sqrt(25 - x^2) + sqrt(25 - x^2)] = 0
-x^2/sqrt(25 - x^2) + sqrt(25 - x^2) = 0
-x^2/sqrt(25 - x^2) = -sqrt(25 - x^2)
x^2 = sqrt(25 - x^2).sqrt(25 - x^2)
x^2 = 25 - x^2
2x^2 = 25
x^2 = 25/2
x = sqrt(25/2)
 = 5/sqrt2
 = (5sqrt2)/2
From y = sqrt(25 - x^2)
y = sqrt(25 - x^2)
 = sqrt(25 - 25/2)
 = sqrt(25/2)
 = 5/sqrt2
 = (5sqrt2)/2
The largest possible area is therefore,
A(max) = (1/2)xy
      = (1/2)[(5sqrt2)/2][(5sqrt2)/2]
      = (1/2)(25/2)
      = 25/4 sq cm

Regards.