Considering ΔADE and ΔACB as similar triangles, the distance across the lake is, DE = 107.1 ft.
Similar Triangles
- Similar triangles are formed if a line segment that is parallel to a one of the sides of a triangle joins the two other sides of the triangle.
- For triangles that are similar to each other, the ratio of their corresponding sizes are congruent.
The image shows two similar triangles, ΔADE and ΔACB. Therefore:
AD/AC = DE/CB
75/(75 + 100) = DE/250
75/175 = DE/250
DE = 107.1 ft
Therefore, considering ΔADE and ΔACB as similar triangles, the distance across the lake is, DE = 107.1 ft.
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