Answer:
(C) 10 units
Step-by-step explanation:
It is given that EFGH is a rhombus and EG = 16 and FH = 12.
We know that the diagonals of the rhombus are the perpendicular bisectors, therefore OF=6, OH=6, OE=8 and OG=8.
Now, using the Pythagoras theorem in ΔOFG, we have
[tex](FG)^{2}=(OG)^{2}+(OF)^{2}[/tex]
[tex](FG)^{2}=(8)^{2}+(6)^{2}[/tex]
[tex](FG)^2=64+36[/tex]
[tex]FG=\sqrt{100}[/tex]
[tex]FG=10[/tex]
Thus, the value of the side of the rhombus will be 10 units.
Hence, option C is correct.
