Triangle ABC is a 30-60-90 triangle, where A is 30º, B is 90º, and C is 60º. A segment BD is the drawn altitude to AC, forming ΔADB, then DE is the altitude drawn to AB,forming new ∆ADE. Then, EF is the altitude drawn to AC, forming new ∆AFE. 78Then, FG is the altitude drawn to AB, forming new ∆AFG.Then, GH is the altitude drawn to AC, forming new ∆AHG. Then, HI is the altitude drawn to AB, forming new ∆AHI. Find the ratio of the area of ∆AIH to the area of ∆ABC?
I found that the resulting ratio 729√(3)/8192 : √3/2 or .154 : .866
Note that all triangles are 30-60-90 right triangles.
In a 30-60-90 triangle, the side opposite 30° is half the hypotenuse.
The side opposite 60° is times the shortest side.
Let , then
The area of is: . .
. . The area of
We have: .
Drawing two altitudes, we have two similar right triangles.
. . The area of the smaller triangle is of the larger.
This procedure is performed three times.
Hence, the area of the final triangle is: . the area of