Consider an obtuse angled triangle $ABC$ in which the difference between the largest and the smallest angle is $\frac{\pi }{2}$ and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on a circle of radius $1$.
(There are two questions based on $\mathrm{PARAGRAPH} " 1 "$, the question given below is one of them)
Let $\alpha$ be the area of the triangle $ABC$. Then the value of $(64\alpha {)}^2$ is

Let angle $A$ be obtuse angle and let sides be $a-d, a, a+d$

So, plotting the diagram we get,

Now let sides be, $a-d,a,a+d$

Also given angle $A-C=\frac{\pi }{2}$

And circum radius $R=1$

Now by sine rule we get,

$\frac{a+d}{\sin A}=\frac{a}{\sin B}=\frac{a-d}{\sin C}=2$

$\because A=\frac{\pi }{2}+C$

$\Rightarrow \sin A=\sin \left(\frac{\pi }{2}+C\right)$

$\Rightarrow \sin A=\cos C$

$\Rightarrow \frac{a+d}{2}=\sqrt{1-{\sin }^{2}C}$

$\Rightarrow {\left(\frac{a+d}{2}\right)}^2=1-{\left(\frac{a-d}{2}\right)}^2$

$\Rightarrow \frac{2\left(a^2+d^2\right)}{4}=1$

$\Rightarrow a^2+d^2=2 ...\left(1\right)$

Now using cosine rule we get,

$\cos B=\frac{(a-d{)}^2+(a+d{)}^2-a^2}{2\left(a^2-d^2\right)}$

$\Rightarrow \sqrt{1-{\sin }^{2}B}=\frac{2\left(a^2+d^2\right)-a^2}{2\left(a^2-d^2\right)}$

$\Rightarrow \sqrt{1-\frac{a^2}{4}}=\frac{4-a^2}{2\left(a^2-d^2\right)} \left(\because a^2+d^2=2\right)$

$\Rightarrow {\left(a^2-d^2\right)}^2=4-a^2 ...\left(2\right)$

From $\left(1\right) \& \left(2\right)$
$a^2=\frac{7}{4},d^2=\frac{1}{4}$
Area of triangle
$\Delta =\frac{a\left(a^2-d^2\right)}{4}$

$\alpha =\frac{\sqrt{7}}{2}\times \frac{6}{4\times 4}$

Now ${\left(64\alpha \right)}^2=64\times 64\times \frac{7}{4}\times \frac{36}{16\times 16}=4\times 252=1008$