1.A girl of height 90cm 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 the ground, find the length of her shadow after 4 seconds.
Answers
Hey !
Solution :
So refer to the attached picture for reference.
So let the girl be standing at point D on the triangle ABC.
So she travels 1.2 m / s. So she is travelling for 4 seconds.
Hence distance traveled = 4 * 1.2 = 4.8 m or 480 cm.
So let the shadow be denoted as 's' which is BD according to attachment.
So we know that the lamp post is 3.6 m. This can be written as 360 cm.
So According to the diagram,
AC = Lamp post ( Height of lamp post )
DE = Girl's Position ( Height of Girl )
BD = Shadow of girl
AB = Total length traveled + Shadow length.
Now let us consider Δ EDB and Δ CAD.
So we know that,
Girl's height is Parallel to the lamp post.
So ∠ BED = ∠ BCA ( Corresponding angles )
Also Lamp post and the girl are perpendicular to the ground. Hence,
∠ CAD = ∠ EDB ( 90° each )
Hence Δ CAB ∼ Δ EDB by AA similarity criteria.
Hence The corresponding sides are in equal proportions.
=> CA / ED = AB / BD
=> 360 / 90 = 480 + x / x
Cross multiplying we get,
=> 360 ( x ) = 90 ( 480 + x )
=> 360 x = 43200 + 90 x
=> 360 x - 90 x = 43200
=> 270 x = 43200
=> x = 43200 / 270
=> x = 160 m
Hence the length of shadow ( BD ) = 160 cm.
Hope my answer helped !
Height of the girl = CE = 90cm = 0.9m
Speed of girl = 1.2 m/s
Distance travelled by her in one second ,
= ( BC ) = 1.2 mts
Let the distance travelled by girl after
4 sec be ( DC ) = x m
From the diagram ∆CDE ~ ∆DBA
[ By AA similarity ]
DC/DB = CE/AB
x/( DC + CD ) = 0.9/3.6
x/(x + 1.2 ) = 1/4
4x = x + 1.2
4x - x = 1.2
3x = 1.2
x = 1.2/3
x = 0.4 m
Therefore ,
Length of girl shadow after 4 seconds
BD = DC + CB
= x + 1.2
= 0.4 + 1.2
= 1.6 m
I hope this helps you.
: )
