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Bunuel
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My approach would be as follows;
First I will find the opposite \(\angle\) between the height and base of the green pole. \(tan x = \frac{1}{2}\)
This could be used in the red pole;

\(\frac{6}{x} = \frac{1}{2} = tan x \)

\(x=12\)

Now consider another triangle with x and (12-4) as a side with the same angle extended.

\(\frac{x}{8 }= \frac{1}{2} \)

\(x= 4\)

Ans E
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The light source is placed in such a way that the shadow of the vertical bars get elongated to double.

If there were no wall, the shadow of the red bar would have been 6x2 = 12 units on the floor.
Of the 12 units, 4 units are on the floor, but remaining units fall on the wall. The equivalent of 8 units length (that would have been on floor) is 4 units length (length on wall = half of the length on floor .

E
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Going for option E.


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Given: In the image above, the green bar has a height of 1 and casts a shadow of 2 on the floor and the red bar has a height of 6 and casts a shadow of 4 on the floor which continues to a height of x on a parallel to the bars wall.
Asked: What is the value of x? (Assume that the light source is distant, the floors and walls are perfectly flat, and the floor is perpendicular to the bars and walls)

Shadow of height 1 is 2 on the floor and 1 on the wall.
Since shadow of 4 is on the floor, height of 2 is consumed and balance 4 (6-2) will be shown on the wall.

IMO E
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My answer is E (4)

If the wall were not there, the shadow of the 6 foot pillar would be 12 foot long
Because the shadow of the 1 foot pillar is 2 foot

The wall is 4 feet away from the pillar

By using the laws of similar triangles, we can see that x is 4
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