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