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.

Let a be the area of the triangle ABC. Then the value of (64a)² is:

The correct answer is 1008.

Let the angles of triangle $ABC$ be $A$, $B$, and $C$, ordered such that $A>B>C$.
Since $ABC$ is an obtuse-angled triangle, $A$ is the largest angle (which is obtuse, i.e., $A>\frac{\pi }{2}$) and $C$ is the smallest angle.

We are given that the difference between the largest and smallest angle is $\frac{\pi }{2}$:

$A-C=\frac{\pi }{2} \Rightarrow A=\frac{\pi }{2}+C$

We are also given that the sides of the triangle are in arithmetic progression.
By the Law of Sines, the sides $a$, $b$, and $c$ (opposite to angles $A$, $B$, and $C$ respectively) are proportional to $\sin A$, $\sin B$, and $\sin C$.
Therefore, $\sin A$, $\sin B$, and $\sin C$ are also in arithmetic progression, which means:

$\sin A+\sin C=2\sin B$

Since $A+B+C=\pi$, we have $B=\pi -(A+C)$.
Substituting $A=\frac{\pi }{2}+C$:

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

$B=\pi -\left(\frac{\pi }{2}+2C\right)=\frac{\pi }{2}-2C$

Now express $\sin A$ and $\sin B$ in terms of $C$:

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

$\sin B=\sin \left(\frac{\pi }{2}-2C\right)=\cos 2C$

Substitute these into the AP condition $\sin A+\sin C=2\sin B$:

$\cos C+\sin C=2\cos 2C$

Recall that $\cos 2C={\cos }^{2}C-{\sin }^{2}C=(\cos C-\sin C)(\cos C+\sin C)$.
Since $C>0$, $\cos C+\sin C\&neq;0$, so we can divide both sides by $\cos C+\sin C$:

$1=2(\cos C-\sin C) \Rightarrow \cos C-\sin C=\frac{1}{2}$

Squaring both sides:

${\left(\cos C-\sin C\right)}^2=\frac{1}{4}$

$1-2\sin C\cos C=\frac{1}{4} \Rightarrow \sin 2C=\frac{3}{4}$

Also, $\cos 2C=\sqrt{1-{\sin }^{2}2C}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}$.

The circumradius of the circle is given as $R=1$.
The area of the triangle in terms of circumradius $R$ and its angles is:

$\mathrm{Area}=2R^2\sin A\sin B\sin C$

Since $R=1$, the area $a$ is:

$a=2\sin A\sin B\sin C=2\cos C\cos 2C\sin C=\left(2\sin C\cos C\right)\cos 2C=\sin 2C\cos 2C$

Substitute the values of $\sin 2C$ and $\cos 2C$:

$a=\frac{3}{4}\times \frac{\sqrt{7}}{4}=\frac{3\sqrt{7}}{16}$

Now, calculate $64a$:

$64a=64\times \frac{3\sqrt{7}}{16}=4\times 3\sqrt{7}=12\sqrt{7}$

Finally, find the value of ${\left(64a\right)}^2$:

${\left(64a\right)}^2={\left(12\sqrt{7}\right)}^2=144\times 7=1008$