Consider an obtuse angled triangle ABC in which the difference...

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 $A - C = \dfrac{\pi}{2}$

And circum radius $R = 1$

Now by sine rule we get,

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

$∵ A = \dfrac{\pi}{2} + C$

$⇒ sin A = \sin \left(\dfrac{\pi}{2} + C\right)$

$⇒ sin A = cos C$

$⇒ \dfrac{a + d}{2} = \sqrt{1 - \sin^{2} C}$

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

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

$⇒ {a}^{2} + {d}^{2} = 2 . . . ( 1 )$

Now using cosine rule we get,

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

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

$⇒ \sqrt{1 - \dfrac{{a}^{2}}{4}} = \dfrac{4 - {a}^{2}}{2\left({a}^{2} - {d}^{2}\right)}\left(∵ {a}^{2} + {d}^{2} = 2\right)$

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

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

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

Now using the formula of inradius we get,

$r = \dfrac{\text{area of triangle}}{\text{semi perimeter}} = \dfrac{\Delta}{S}$

$⇒ r = \dfrac{\dfrac{\sqrt{7}}{2} \times \dfrac{6}{16}}{\dfrac{3}{2} \times \dfrac{\sqrt{7}}{2}} = \dfrac{4}{16} = \dfrac{1}{4} = 0 . 25$