A girl of height 100 cm …
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Sia ? 5 years, 11 months ago
Let AB be the lamp-post, DE be the girl and D be the position of girl after 4s.
Again, let DC = xm be the length of shadow of the girl.
Given, DE = 100 cm =1 m, AB =5 m and speed of the girl= 1.9 m/s
∴ Distance of the girl from lamp-post after 4 s = BD = 1.9×4=7.6m [∵ Distance = speed × time ]
In △ABC and △EDC, ∠B=∠D
[∵each 90o]
∠C=∠C [ ∵common angle ]
∴△ABC∼△EDC [∵ by AA similarity criterion ]
⇒BCDC=ABDE ∵[since, corresponding sides of similar triangles are proportional]...(i)
On substituting all the values in Eq(i), we get
7.6+xx=51
⇒7.6+x=5x
⇒7.6=5x−x
⇒7.6=4x
⇒x=7.64
⇒x=1.9m
Hence the length of her shadow after 4 s is 1.9 m.
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