Question : Suhas, a 3.15 m tall tree, and an 11.25 m tall building are positioned such that their feet on the ground are collinear and the tree is located between Suhas and the building. The tree is located at a distance of 45 m from the building. Further. the eyes of Suhas, the top of the tree, and the top of the building fall in one line, and the eyes of Suhas are at a height of 1.8 m from the ground, At what distance (in m) from Suhas is the tree located?
Option 1: 6
Option 2: 5
Option 3: 7.5
Option 4: 9
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Correct Answer: 7.5
Solution :
Given, Subhas' eyes (EF) are at a height of 1.8 m.
A 3.15 m tall tree (CD), stands 45 m to an 11.25 m tall building (AB). Their heights align.
Observe that $\triangle ABG\sim \triangle CDG\sim \triangle EFG$
According to the question,
$\frac{3.15}{\text{DG}} = \frac{11.25}{ 45+\text{DG}}$
$⇒3.15\times 45+3.15\text{DG}=11.25\text{DG}$
$⇒8.1\text{DG}=3.15\times 45$
$⇒\text{DG} = 17.5$
Also, $\frac{1.8}{\text{FG}} = \frac{3.15}{17.5}$
$⇒\text{FG} = 10$
Now, $\text{DF=DG}-\text{FG}=17.5-10=7.5$
Hence, the correct answer is 7.5.
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