Step 1

$ \Delta ABC$ is a right angled

triangle, right angled at $C$

$ ( \because$ angle in a semicircle is $ 90^{\circ})$

Hence the area of the $\Delta ABC$ is

$A = \large\frac{1}{2} \times AC \times BC$

Let $D$ be the diameter and $BC = x$

$ \therefore AC=\sqrt{D^2-x^2}$

$ \therefore A= \large\frac{1}{2}=\sqrt{D^2-x^2} \times x$

Step 2

Differentiating w.r.t $x$ we get,

$ \large\frac{dA}{dx}=\large\frac{1}{2} \bigg[ \sqrt{D^2-x^2}.1 + x \large\frac{1}{2\sqrt{D^2-x^2}} (-2x) \bigg]$

$ =\large\frac{1}{2} \bigg[ \large\frac{D^2-x^2-x^2}{\sqrt{D^2-x^2}} \bigg] = \large\frac{1}{2} \bigg[ \large\frac{D^2-2x^2}{\sqrt{D^2-x^2}} \bigg]$

To find the maximum or minimum area

$ \large\frac{dA}{dx}=0$

i.e., $ \large\frac{1}{2} \bigg[ \large\frac{D^2-2x^2}{\sqrt{D^2-x^2}} \bigg]=0$

$ \Rightarrow D^2=2x^2$

$ D = x\sqrt 2$

Step 3

To prove that the triangle has maximum area let us differentiate $ \large\frac{dA}{dx}$ again

$\large\frac{d^2A}{dx^2}=\large\frac{\sqrt{D^2-x^2}(-4x)-(D^2-2x^2) \large\frac{1}{2\sqrt{D^2-x^2}}(-2x)}{(D^2-x^2)}$

$ \large\frac{d^2A}{dx^2} = \large\frac{\sqrt{D^2-x^2}[-3x]}{D^2-x^2} <0$

hence this proves that the triangle has maximum value.

$ AC^2=D^2-x^2$

$=2x^2-x^2$

$ AC^2=x^2$

$ \Rightarrow AC=x$

$ AC=BC=x$

Hence the triangle has maximum value when it is an isosceles triangle.