A) 18m
B) 26m
C) 21m
D) 20.4m
Correct Answer: D
Solution :
Let BC= 18m be the flag pole and its shadow be AB = 9.6 m. The distance of the top of the pole, C from the far end i.e. A of the shadow is AC. |
In right angled \[\Delta ABC\], \[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\] |
[by Pythagoras theorem] |
\[\Rightarrow A{{C}^{2}}={{\left( 9.6 \right)}^{2}}+{{\left( 18 \right)}^{2}}\] |
\[A{{C}^{2}}=92.16+324\] |
\[\Rightarrow A{{C}^{2}}=416.16\] |
\[\therefore AC=\sqrt{416.16}=20.4m\] |
Hence, the required distance is 20.4 m. |
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