Question

A vertical pole 4m 80cm high, casts a shadow 2m 60cm long. Find at the same

time (i) the length of shadow casted by a 3m 25 cm high pole. (ii) the height of the

pole which casts a shadow of length 1m 50cm.​

Answer 1

Given:

A vertical pole 4m 80cm high, casts a shadow 2m 60cm long. Find at the same  time

To find:

(i) the length of the shadow cast by a 3m 25 cm high pole.

(ii) the height of the  pole which casts a shadow of length 1m 50cm.​

Solution:

We know,

$\boxed{\bold{1 \:cm = \frac{1}{100} \:m = 0.01 \:m}}$

So,

4 m 80 cm = 4 + 0.8 = 4.8 m

2 m 60 cm = 2 + 0.6 = 2.6 m

3 m 25 cm = 3 + 0.25 = 3.25 m

1 m 50 cm = 1 + 0.50 = 1.5 m

Since it is given that the shadows are cast at the same time

∴ the angle of elevation of the sun is the same in both the cases

Case (i): Finding the length of the shadow cast by a 3m 25 cm high pole:

Let "x" represents the length of the shadow.

Since the value of tan θ is the same in both cases here,

∴ $\frac{4.8 }{2.6} = \frac{3.25}{x}$

$\implies x = \frac{3.25 \times 2.6 }{4.8}$

$\implies \bold{x = 1.76 \:m}$

$\implies \bold{x = 1\:m\: 76 \:cm}$

Thus, the length of the shadow cast by a 3 m 25 cm high pole is → 1 m 76 cm.

Case (ii): Finding the height of the  pole which casts a shadow of length 1m 50cm:

Let "x" represents the height of the pole.

Since the value of tan θ is the same in both cases here,

∴ $\frac{4.8 }{2.6} = \frac{x}{1.5}$

$\implies x = \frac{4.8 \times 1.5 }{2.6}$

$\implies \bold{x = 2.76 \:m}$

$\implies \bold{x = 2\:m\: 76 \:cm}$

Thus, the height of the  pole which casts a shadow of length 1m 50cm is → 2 m 76 cm.

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Answer 2

Given:

A vertical pole 4m 80cm high, casts a shadow 2m 60cm long. Find at the same  time

To find:

(i) the length of the shadow cast by a 3m 25 cm high pole.

(ii) the height of the  pole which casts a shadow of length 1m 50cm.

Solution:

We know,

$\boxed{\bold{1 \:cm = \frac{1}{100} \:m = 0.01 \:m}}$

So,

4 m 80 cm = 4 + 0.8 = 4.8 m

2 m 60 cm = 2 + 0.6 = 2.6 m

3 m 25 cm = 3 + 0.25 = 3.25 m

1 m 50 cm = 1 + 0.50 = 1.5 m

Since it is given that the shadows are cast at the same time

∴ the angle of elevation of the sun is the same in both the cases

Case (i): Finding the length of the shadow cast by a 3m 25 cm high pole:

Let "x" represents the length of the shadow.

Since the value of tan θ is the same in both cases here,

∴ $\frac{4.8 }{2.6} = \frac{3.25}{x}$

$\implies x = \frac{3.25 \times 2.6 }{4.8}$

$\implies \bold{x = 1.76 \:m}$

$\implies \bold{x = 1\:m\: 76 \:cm}$

Thus, the length of the shadow cast by a 3 m 25 cm high pole is → 1 m 76 cm.

Case (ii): Finding the height of the  pole which casts a shadow of length 1m 50cm:

Let "x" represents the height of the pole.

Since the value of tan θ is the same in both cases here,

∴ $\frac{4.8 }{2.6} = \frac{x}{1.5}$

$\implies x = \frac{4.8 \times 1.5 }{2.6}$

$\implies \bold{x = 2.76 \:m}$

$\implies \bold{x = 2\:m\: 76 \:cm}$

Thus, the height of the  pole which casts a shadow of length 1m 50cm is → 2 m 76 cm.

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