A) \[\sqrt{1961}\,m\]
B) \[\sqrt{802}\,m\]
C) \[29\,m\]
D) \[41\,m\]
Correct Answer: C
Solution :
| From the figure we note that |
| \[BC=40\,m-19\,m=21\,m\] |
| and \[AC=20m.\] |
| Therefore, by Pythagoras theorem, |
| \[A{{B}^{2}}=A{{C}^{2}}+B{{C}^{2}}={{20}^{2}}+{{21}^{2}}=400+441=841\]\[\Rightarrow \,\,\,\,\,\,\,\,\,AB=29\,m\] |
| So, option [c] is correct. |
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