Question

A girl of height 90 cm is walking away from the base of a lamp post at a speed of 1.2 m/s. If the lamp is 3.6m above the ground find the length of her shadow after 4 second​

Answer 1

Given :-

• Height of the girl = 90 cm = 0.9 m.
• Height of the lamp = 3.6 m.
• Speed of the girl = 1.2 m/s.

To Find :-

• Length of her shadow after 4 seconds.

Solution :-

Figure:-

$\setlength{\unitlength}{1.5mm}\begin{picture}(5,5)\put(0,0){\line(1,0){48}}\put(0,0){\line(0,1){36}}\put(36,0){\line(0,1){9}}\put(48,0){\line(-4,3){48}}\put(-5,36){A}\put(-5,-5){B}\put(36,10){C}\put(36,-5){D}\put(48,-5){E}\footnotesize\put(2,15){3.6\ m}\put(37,2.5){0.9\ m}\put(36,0.8){\framebox{}}\end{picture}$

In figure, AB represents the lamp of height 3.6 m. CD represents the girl of height 0.9 m. DE is the length of her shadow after walking 4 seconds at a speed of 1.2 m/s from the base of the lamp.

$\hrulefill$

We are given that the girl walks at a speed of 1.2 m/s for 4 seconds to reach point D from point B.

Then,

⇒ BD = 1.2 × 4 (distance = speed × time)

⇒ BD = 4.8 m

Now, in △ ABE & △ CDE,

⇒ ∠ E = ∠ E (Common angle)

And,

⇒ ∠ B = ∠ D (Both angles are 90°)

Hence, △ ABE similar to △ CDE (AA similarity). Then,

$\implies\rm{\dfrac{BE}{DE}=\dfrac{AB}{CD}}$ (sides are proportional)

$\implies\rm{\dfrac{BD+DE}{DE}=\dfrac{AB}{CD}}$

$\implies\rm{\dfrac{4.8+DE}{DE}=\dfrac{3.6}{0.9}}$

$\implies\rm{\dfrac{4.8+DE}{DE}=4}$

$\implies\rm{4.8+DE=4DE}$

$\implies\rm{3DE=4.8}$

$\implies\rm{DE=\dfrac{4.8}{3}=}\:\bold{1.6\:m.}$

Hence, length of her shadow = 1.6 metres.