show that of all right triangles inscribed in a circle, the triangle with maximum perimeter is isoceles
Answers
Given : right triangles inscribed in a circle
To Find : triangle with maximum perimeter is isosceles
Solution:
right triangles inscribed in a circle
Hence hypotenuse will be diameter = c ( constant)
Let say one perpendicular side = x
then other perpendicular side = √(c² - x²)
Perimeter P = x + √(c² - x²) + c
dP/dx = 1 + (-2x)/2√(c² - x²) + 0
=> dP/dx = 1 - x/√(c² - x²)
dP/dx = 0
=> 1 - x/√(c² - x²) = 0
=> x/√(c² - x²) = 1
=> x = √(c² - x²)
=> x² = c² - x²
=> 2x² = c²
=> x² = c²/2
=> x = c/√2
dP/dx = 1 - x/√(c² - x²)
d²P/dx² = 0 - 1/√(c² - x²) - x (-2x)(-1/2)/(c² - x²)√(c² - x²)
= - 1/√(c² - x²) - x²/(c² - x²)√(c² - x²)
=> d²P/dx²< 0
Hence P is max for x = c/√2
√(c² - x²) = c/√2
Hence sides are c/√2, c/√2 , c
=> triangle with maximum perimeter is isosceles
QED
Hence proved
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