The inradius of a circle inscribed in an isos | vedclass

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(A) Let the sides of the isosceles right-angled triangle be $a, a,$ and $a\sqrt{2}$.The inradius $r$ of a triangle is given by $r = \frac{	ext{Area}}

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$3 + 2\sqrt{2}$

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(A) Let the sides of the isosceles right-angled triangle be $a, a,$ and $a\sqrt{2}$.The inradius $r$ of a triangle is given by $r = \frac{	ext{Area}}{	ext{semi-perimeter}} = \frac{\Delta}{s}$.Here,$\Delta = \frac{1}{2} a^2$ and $s = \frac{a + a + a\sqrt{2}}{2} = \frac{a(2 + \sqrt{2})}{2}$.Given $r = 1$,we have $1 = \frac{\frac{1}{2} a^2}{\frac{a(2 + \sqrt{2})}{2}} = \frac{a}{2 + \sqrt{2}}$.Thus,$a = 2 + \sqrt{2}$.The area of the triangle is $\Delta = \frac{1}{2} a^2 = \frac{1}{2} (2 + \sqrt{2})^2$.$\Delta = \frac{1}{2} (4 + 2 + 4\sqrt{2}) = \frac{1}{2} (6 + 4\sqrt{2}) = 3 + 2\sqrt{2}$ sq. units.

The inradius of a circle inscribed in an isosceles right-angled triangle,where one of the angles is $\frac{\pi}{2}$,is $1$. Find the area of the triangle (in sq. units).

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