Question

A girl of height 90 cm is walking away from the base of lamp post at a speed of

1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her shadow

after 4 seconds.
​

Answer 1

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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Answer 2

Answer:

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.