A) 9cm
B) 10cm
C) 8cm
D) 20cm
Correct Answer: B
Solution :
We know that, the diagonals of a rhombus are perpendicular bisector of each other. Given, AC = 16 cm and BD = 12 cm [let] |
\[\therefore \,\,\,AO=8\,cm,\,BO=6\,\,cm\] and \[\angle AOB=90{}^\circ\] |
In right angled \[\Delta AOB\], |
\[A{{B}^{2}}=A{{O}^{2}}+O{{B}^{2}}\] |
[by Pythagoras theorem] |
\[\Rightarrow \,\,\,A{{B}^{2}}={{8}^{2}}+{{6}^{2}}=64+36=100\] |
\[\therefore \,\,\,AB=10\,\,cm\] |
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