Explanation:
Let \( \angle A=\theta, \ \angle B=\theta+\frac{\pi}{2} \)
\( \Rightarrow \angle C=\frac{\pi}{2}-2\theta \)
Let \( p,q,r \) denote lengths of side \( BC, CA \) and \( AB \). Since \( \angle A \) is smallest angle and \( \angle B \) is the largest, \( p \) and \( q \) are the smallest and largest side of the \( \triangle \).
\( \Rightarrow p+q=2r \)
Also, \( R=\text{Circum–Radius of } \triangle ABC=1 \)
\( \Rightarrow p=2\sin\theta,\ q=2\cos\theta,\ r=2\cos2\theta \)
\( \Rightarrow 2\cos2\theta=\sin\theta+\cos\theta \Rightarrow \cos\theta-\sin\theta=\frac{1}{2} \)
Since \( \sin\theta+\cos\theta\neq0 \)
\( \Rightarrow \sin\!\left(\frac{\pi}{4}-\theta\right)=\frac{1}{2\sqrt{2}} \Rightarrow \theta=\frac{\pi}{4}-\sin^{-1}\!\left(\frac{1}{2\sqrt{2}}\right) \)
\( a=\text{Area}=\frac{1}{2}pq\sin C=\frac{1}{2}\sin4\theta \)
\( 4\theta=\pi-4\sin^{-1}\!\left(\frac{1}{2\sqrt{2}}\right) \Rightarrow \sin4\theta=\sin\!\left(4\sin^{-1}\!\left(\frac{1}{2\sqrt{2}}\right)\right) \)
\( \Rightarrow a=\frac{1}{2}\times4\times\frac{1}{2\sqrt{2}}\times\frac{\sqrt{7}}{2\sqrt{2}}\times\frac{3}{4}=\frac{3\sqrt{7}}{16} \Rightarrow 64a=12\sqrt{7} \)
In radius \( =\frac{\Delta}{s}=\frac{\frac{3\sqrt{7}}{16}}{3\cos2\theta}=\frac{\sqrt{7}}{16\cos2\theta} \)
\( 2\theta=\frac{\pi}{2}-2\sin^{-1}\!\left(\frac{1}{2\sqrt{2}}\right) \)
\( \cos2\theta=\sin\!\left(2\sin^{-1}\!\left(\frac{1}{2\sqrt{2}}\right)\right)=2\times\frac{1}{2\sqrt{2}}\times\frac{\sqrt{7}}{2\sqrt{2}}=\frac{\sqrt{7}}{4} \)
\( \Rightarrow 16\cos2\theta=4\sqrt{7} \)
\( \Rightarrow \text{In-radius}=0.25 \)

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