A triangle has sides of lengths $\sqrt { 5 } , 2 \sqrt { 2 } , \sqrt { 3 }$ units. Then, the radius of its inscribed circle is:\\
(A) $\frac { \sqrt { 5 } + \sqrt { 3 } + 2 \sqrt { 2 } } { 2 }$\\
(B) $\frac { \sqrt { 5 } + \sqrt { 3 } + 2 \sqrt { 2 } } { 3 }$\\
(C) $\sqrt { 5 } + \sqrt { 3 } + 2 \sqrt { 2 }$\\
(D) $\frac { \sqrt { 5 } + \sqrt { 3 } - 2 \sqrt { 2 } } { 2 }$