Question

A man morning at a speed of 36 kmph
Covered the shadow of a tower of height
75 min 3 secde seconds. If the height
of the person 1.75m, the length of the
Shadow Cast by the man in meter is?​

Answer 1

Given:

• Speed of the man is 36 kmph

• Shadow of a 75 m high tower is covered in 3 seconds

• Height of the person is 1.75 m

To find: The length of the shadow cast by the man

Solution:

Shadow of 75 m high tower is covered in 3 seconds

Then the shadow of the person with 1.75 height will be covered in

= 1.75 * 3/75 seconds

= 0.07 seconds

Speed of the man is 36 kmph

i.e., in 3600 seconds, the man goes 36000 m

Then in 0.07 seconds, the man will go

= 0.07 * 36000/3600 m

= 0.7 m

Answer: The length of the shadow cast by the man is 0.7 m

Answer 2

Given:

Speed of the man = 36 km/hr = $36 * \frac{5}{18} m/s$ = 10 m/s

Height of the tower = 75 m

Time taken to cover the shadow of the tower = 3 seconds

Height of the person = 1.75 m

To find:

The length of the shadow by the man in meters

Solution:

From the figure attached below we have,

AB = height of the tower

BE = shadow cast by the tower

DE = shadow cast by the man

CD = height of the man

We have,

The man covers the shadow of the tree "BE" with a speed of 10 m/s in 3 s

∴ The distance covered by the man is,

= the length of the shadow cast by the tower

= BE

= 10 m/s * 3 s   ..... [Formula used: Distance = Speed * Time]

= 30 m

Consider ΔABE & ΔCDE, we have

∠E = ∠E ...... (common angle)

∠ABE = ∠CDE = 90° ...... (both the tower and the man stand vertical)

∴ ΔABE ~ ΔCDE ...... by AA similarity

We know that the corresponding sides of two similar triangle are proportional to each other

∴ $\frac{BE}{DE} = \frac{AB}{CD}$

substituting the values of BE = 30 m, AB = 75 m & CD = 1.75 m

⇒ $\frac{30}{DE} = \frac{75}{1.75}$

⇒ $DE = \frac{30 \:*\:1.75 }{75}$

⇒ DE = 0.70 m

Thus, the length of the shadow cast by the man in meter is 0.7 m.

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