Question

115. A girl of height 100 cm is walking away from the base of a lamp post at a speed of 1.9 m/sec. If the lamp is 5 m above the ground, the length of the shadow after 4 sec. is : (1) 160 cm (2) 180 cm (3) 200 cm (4) 190 cm​

Answer 1

Answer:

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.