Question

Ab is a diameter of a circle and c is any point on the circle. show that the area of δ abc is maximum, when it is isosceles.

Answer 1

ab is diameter of circle , Let length of diameter is d , and c is any point in the circle . we know, diameter subtends right angle at any point on the circle .
so, it is clear ∆abc is right angle triangle.
now, Let one side of triangle is x
Then, other side will be $\bold{\sqrt{d^2 - x^2}}$ [ from Pythagoras theorem , d is hypotenuse , of abc triangle ]
now, area of triangle abc = 1/2 height × base
If height is x then, base = √{d² - x²) and vice -versa

Now, area of ∆abc , A = 1/2x √(d²-x²)
⇒ A = √{d²x² - x⁴}/2
now, differentiate with respect to x
dA/dx = 1/2√{d²x² - x⁴} × (2d²x - 4x³) = 0
⇒4x³ = 2d²x
⇒ 2x² = d² ⇒ √2x = d
at x = d/√2 , check d²A/dx² , it will be negative hence, at x = d/√2 , A{area } will be maximum .

Now, x = d/√2 put in other side of triangle e.g., √(d² - x²) = √(d² - d²/2) = d/√2
two sides of triangle are same .
Hence, triangle is an isosceles triangle .

Hence, it is clear that area of abc will be maximum when it is isosceles.