A) \[\frac{108}{11}cm\]
B) \[\frac{108}{10}cm\]
C) \[\frac{99}{10}cm\]
D) \[\frac{108}{17}cm\]
Correct Answer: A
Solution :
Area of parallelogram with base AB and attitude AM \[=12\times 9=108\,c{{m}^{2}}\] \[108\,c{{m}^{2}}=AD\times 11\,cm\] \[\Rightarrow \]\[AD=\frac{108}{11}\,cm\]You need to login to perform this action.
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