A) 3.2m
B) 4.8m
C) 1.6m
D) 3.6m
Correct Answer: C
Solution :
Let AB be the lamp-post, CD be the girl and D be the position of girl after 4s. |
Again, let DE = x m be the length of shadow of the girl. |
Given, CD = 90 cm = 03 m, AS = 3.6 m and speed of the girl =1.2 m/s |
\[\therefore\]Distance of the girl from lamp-post after 4s, |
\[BD=1.2\,\times 4=4.8\,\,m\] |
[\[\because\]distance = speed \[\times\] time] |
In \[\Delta ABE\] and \[\Delta CDE\], |
\[\angle B=\angle D\] [each \[90{}^\circ\]] |
\[\angle E=\angle E\] [common angle] |
\[\therefore \Delta ABE\tilde{\ }\Delta CDE\] |
[by AA similarity criterion] |
\[\Rightarrow \frac{BE}{DE}=\frac{AB}{CD}\] … (i) |
[since, corresponding sides of similar triangles are proportional] |
On substituting all the values in Eq. (i), we get |
\[\frac{4.8+x}{x}=\frac{3.6}{0.9}\] |
\[\left[ \because \,\,\,BE=BD+DE=4.8+x \right]\] |
\[\Rightarrow \,\,\,\frac{4.8+x}{x}=4\] |
\[\Rightarrow \,\,\,\,4.8+x=4x\] |
\[\Rightarrow \,\,\,\,3x=4.8\] |
\[\Rightarrow \,\,\,\,=\frac{4.8}{3}=1.6\,\,\,m\] |
Hence, the length of her shadow after 4s is 1.6 m. |
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