Question

A 5 m 60 cm high vertical pole casts a shadow 3
m 20 cm long. Find at the same time
(i) the length of the shadow cast by another pole
10 m 50 cm high,
(ii) the height of a pole which casts a shadow 5 m
long.​

Answer 1

Answer:

Let x m be the length of the pole whose shadow is of length 10m50cm.

Let y m be the length of the pole whose shadow is 5m long.

Clearly, it is a case of direct variation.

i)

3.20

5.60

=

x

10.50

x=

5.60

10.50×3.20

=6

Hence, the length of the shadow is 6m.

ii)

3.20

5.60

=

5

y

x=

3.20

5.60×5

=8.75

Hence, the length of the pole is 8.75 meters.

Answer 2

$(i) \frac{x_{!}}{y_{1}} = \frac{x_{2}}{y_{2}}$

$⇒ \frac{5.6}{3.2} = \frac{10.5}{y_{2}}$

$⇒y_{2}=6 cm$

$(ii) \frac{x_{1}}{y_{1}} = \frac{x_{3}}{y_{3}}$

$⇒ \frac{5.6}{3.2} = \frac{x_{3}}{5}$

$⇒x_{3} =8.75 m$