Let d1=12 and d2=16 be the length of diagonals and ' a ' be the side of Rhombus and r= radius of inscribed circle.
Using Pythagoras theorem a=(2d1)2+(2d2)2 ⇒a=(212)2+(<16br/>2)2=62+82=100 ⇒a=10 Area of Rhombus =2d1d2 and 2ar ia 2ar is derived using = area of rhombus = 4× area of triangle Area of Rhombus =4×21×a×r=2ar Using (i), we get 2d1d2=2ar 212×16=2×10×r r=2×1016×6⇒r=4.8