Discussion
There is a tree between houses of A and B. If the tree leans on A’s House, the tree top rests on his window which is 12 m from ground. If the tree leans on B’s House, the tree top rests on his window which is 9 m from ground. If the height of the tree is 15 m, what is distance between A’s and B’s house?
Answer: A
The leaning tree is the hypotenuse of a right triangle in each case, with the tree's base fixed and its 15 m length unchanged.
Towards A: the top rests 12 m up the wall, so the horizontal gap is
\[ \sqrt{15^{2} - 12^{2}} = \sqrt{225 - 144} = \sqrt{81} = 9 \text{ m} \]
Towards B: the top rests 9 m up, so
\[ \sqrt{15^{2} - 9^{2}} = \sqrt{225 - 81} = \sqrt{144} = 12 \text{ m} \]
The tree stands between the two houses, so the distance between them is the sum:
\[ 9 + 12 = 21 \text{ m} \]
Both are 9-12-15 triangles, the 3-4-5 triple scaled by three.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development