Suhas, a 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 7.5 m from Suhas and 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. Find the height (in m) of the tree.
Show answer & explanation
Correct answer: C
Explanation
The correct answer is option 3 i.e. 3.15.
Let the height of the tree be h meters.
Suhas is 7.5 m from the tree, and the building is 45 m from the tree.
So, total distance between Suhas and the building = 7.5 + 45 = 52.5 m
Given: Suhas’s eye level = 1.8 m
Line joining Suhas’s eye, top of the tree, and top of the building is straight ⇒ similar triangles.
Use section formula (internal division):
The height difference from Suhas’s eye to the top of tree = (h − 1.8)
The height difference from Suhas’s eye to the top of building = (11.25 − 1.8) = 9.45 m
Since the horizontal distances from Suhas to tree and Suhas to building are 7.5 m and
52.5 m respectively,
The ratio of vertical to horizontal change remains same.
Using similar triangles:
(h − 1.8)/7.5 = 9.45/52.5
Cross-multiplied:
52.5(h − 1.8) = 9.45 × 7.5
⇒ 52.5h − 94.5 = 70.875
⇒ 52.5h = 165.375
⇒ h = 165.375 / 52.5 = 3.15 meters