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
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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)
BC
BE
=
AC
DE
=
AB
BD
( Corresponding sides are proportional )
360
90
=
480+x
x
⇒
4
1
=
480+x
x
⇒4x=480+x
⇒4x−x=480
⇒3x=480
⇒x=
3
480
=160
Hence length of her shadow after 4 seconds is 160 cm.
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