A boat is anchored in the water. The anchor lies at point A. At low tide, the boat is floating 20 feet above the seafloor and is a horizontal distance of 40 feet away from its anchor. At high tide, the boat is a horizontal distance of 35 feet away from its anchor.
An image shows two right-angled triangles, named low tide and high tide. The length of the opposite and adjacent sides of low tide triangles is 20 ft. and 40 ft. The length of the adjacent side of the high tide triangle is 35 ft.
Approximately how high is the boat floating above the seafloor at high tide? Assume the anchor rope remains tight, without any slack.
To solve this problem, we can use the concept of similar triangles.
Let's label the length of the adjacent side of the low tide triangle as x, and the length of the opposite side of the high tide triangle as h.
In the low tide triangle, we have a ratio of adjacent to opposite sides of 40:20. Therefore, the ratio of the adjacent side at high tide (35 ft) to the opposite side at high tide (h) is also 40:20.
This can be written as:
35 ft / h = 40 ft / 20 ft
Cross-multiplying, we get:
35 ft * 20 ft = 40 ft * h
700 ft^2 = 40 ft * h
Dividing both sides by 40 ft:
700 ft^2 / 40 ft = h
17.5 ft = h
Therefore, the boat is floating approximately 17.5 feet above the seafloor at high tide.