A girl of height 90 cm is walking away from the base of a lamp post at a speed of 1.2m/s. If the lamp is 3.6 m above ground, find the length of her shawdow after 4 secs.
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Step-by-step explanation:
Given, speed=1.2m/sec
time=4sec
speed=
time
distance
1.2=
4
distance
distance=AD=12×4=4.8m
Consider △ABC and △DBE
∠CAB=∠EDB=90
∘
∠CBA=∠EDB (common angle)
So by AA, △ABC∼△DBE
Hence
DE
AC
=
DB
AB
DE
AC
=
DB
AD+DB
0.9
3.6
=
DB
4.8+DB
4DB=4.8+DB
3DB=4.8
DB=1.6
Hence the length of the shadow 1.6m
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