Two poles of heights 6 m, and 11 m stand vertically on a plane ground....
Two poles of heights 6 m, and 11 m stand vertically on a plane ground....
Given:
Height of first pole (P1) = 6 m
Height of second pole (P2) = 11 m
Distance between their feet = 12 m
To find:
Distance between their tops
Approach:
We can use the concept of similar triangles to solve this problem.
Let's consider:
The top of the first pole as point A
The top of the second pole as point B
The feet of the first pole as point C
The feet of the second pole as point D
Now, we have a right-angled triangle ADC, where AD represents the height of the first pole (6 m) and CD represents the distance between their feet (12 m).
Similarly, we have a right-angled triangle BDC, where BD represents the height of the second pole (11 m) and CD represents the distance between their feet (12 m).
Since the triangles ADC and BDC share the same base CD, they are similar triangles. Thus, the ratio of their corresponding sides will be equal.
Using the concept of similar triangles, we can write the following proportion:
AD/CD = BD/CD
Simplifying this equation, we get:
AD = BD
Therefore, the distance between the tops of the two poles, i.e., the distance between points A and B, is equal to the height of the second pole (11 m).
Answer:
Hence, the distance between the tops of the two poles is 11 m.
Therefore, the correct answer is option A) 13 m.