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Question

A boy of height 60 cm is walking away from the base of a building at a speed of 2.2 m/s. If the height of the building is 7.2 m, then find the length of his shadow after 5 seconds.

A
2 m
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B
3 m
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C
1 m
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D
4 m
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Solution

The correct option is C 1 m

Let AB be the building and EC be the boy.
Given: Height of building, AB = 7.2 m
Boy's height = 60 cm = 0.6 m
Boy's speed = 2.2 m/s

Let x be the length of his shadowand y be the distance between building and the boy after 5 seconds.

Distance travelled by the boy in 5 seconds
= Speed × Time
= 2.2×5=11 m

y=11 m

In Δ ABD and Δ ECD,

ABD= ECD=90

ADB= EDC(Common in both the triangles)

Δ ABDΔ ECD(By AA similarity criterion)

ABEC=BDCD (Corresponding sides of similar triangles are in the same ratio)

7.20.6=y+xx

12=11+xx

12x=11+x

x=1 m

So, the length of his shadow after5 seconds is 1 m.

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