Question

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

Answer 1

Answer:

Step-by-step explanation:

Let the girl of height be at point D on the ground from the lamp post after 4 seconds. Therefore,  

AD=1.2 m/sec ×4 sec =4.8 m =480 cm

Suppose the length of the shadow of the girl be x cm when she at position D. Therefore,

 BD=x cm

 In ∆BDE and ∆BAC,

 ∠BDE=∠BAC  (90°)

 ∠DBE=∠ABC   (Common)

 Thus, ∆BDE∼∆BAC   (AA similarity)

 BE /BC=DE/AC=BD/AB

​( Corresponding sides are proportional )

90/360=X/480+X

1/4=X/480+X

4X=480+X

4X-X=480

3X=480

X=160

Hence length of her shadow after 4 seconds is 160 cm.