Question

A boy of height 90cm is walking away from the base of a lamp post at a speedof 1.2m/sec. If the lamppost is 3.6m above the ground, find the length of his shadow cast after 4 seconds.

Answer 1

Solution :-

The height of the lamp is 3.6m

A height of the boy = 90cm = 0.9m

The boy covers distance BD in 4 seconds.

The boy is walking towards lamp post at a speed of 1.2m/s

As we know that,

Speed = Distance / Time

1.2 = BD / 4

BD = 1.2 * 4 = 4.8 m

Now, In ΔABE and ΔCDE

Angle E = Angle E ( Common)

Angle B = Angle D

( Each 90° , Because everything standing on the earth considered perpendicular to the earth)

By AA criteria ,

ΔABE similar to ΔCDE

Now, By BPT Theorem

AB/CD = BE/DE

Here, BE = BD + DE. eq( 1 )

Let consider ED be x

Therefore,

AB/CD = BD + DE / DE ( From eq( 1 ))

3.6 / 0.9 = 4.8 + x / x

4x = 4.8 + x

4x - x = 4.8

3x = 4.8

x = 4.8/3

x = 1.6 m

Hence, The length of the shadow is 1.6m $.$

Answer 2

Answer:

$\huge \mathfrak \purple{Answer}$

The height of the lamp is 3.6m

A height of the boy = 90cm = 0.9m

The boy covers distance BD in 4 seconds.

The boy is walking towards lamp post at a speed of 1.2m/s

As we know that,

Speed = Distance / Time

1.2 = BD / 4

BD = 1.2 * 4 = 4.8 m

Now, In ΔABE and ΔCDE

Angle E = Angle E ( Common)

Angle B = Angle D

( Each 90° , Because everything standing on the earth considered perpendicular to the earth)

By AA criteria ,

ΔABE similar to ΔCDE

Now, By BPT Theorem

AB/CD = BE/DE

Here, BE = BD + DE. eq( 1 )

Let consider ED be x

Therefore,

AB/CD = BD + DE / DE ( From eq( 1 ))

3.6 / 0.9 = 4.8 + x / x

4x = 4.8 + x

4x - x = 4.8

3x = 4.8

x = 4.8/3

x = 1.6 m

▶ Hence, The length of the shadow is 1.6m