A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/sec. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.

We have


Height of girl = 90cm = 0.9m


Height of lamp post = 3.6m


Speed of girl = 1.2 m/sec


So, Distance moved by girl (CQ) = Speed x time


= 1.2 x 4 = 4.8m


Let length of shadow (Ac) = xcm


In ΔABC and ΔAPQ


<ACB = <AQP (Each 90 °)


<BAC = <PAQ (Common)


Then , ΔABC ~ ΔAPQ (By AA similarity)


So, AC/AQ = BC/ PQ(Corresponding parts of similar triangle are proportional)


Or, x/x +4.8 = 0.9/3.6


Or, x/x +4.8 = 1/4


Or, 4x – x = 4.8


Or, 3x = 4.8


Or x = 4.8/3


Or x = 1.6m


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