A) 30m
B) 40m
C) 50m
D) 20m
Correct Answer: B
Solution :
Let height of Ajay's house be \[GF={{h}_{1}}\] |
Since, \[\Delta HFG\tilde{\ }\Delta ACD\] |
\[\therefore \frac{HF}{HC}=\frac{FG}{CD}\Rightarrow \frac{20}{50}=\frac{{{h}_{1}}}{100}\] |
\[\Rightarrow {{h}_{1}}=\frac{20\times 100}{50}=40m\] |
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