# Triangle Ratios

• Nov 11th 2009, 04:33 AM
warriors837
Triangle Ratios
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
• Nov 11th 2009, 06:25 AM
Soroban
Hello, warriors837!

Quote:

Triangle $\displaystyle ABC$ is a 30-60-90 triangle, where $\displaystyle A = 30^o,\:B = 90^o,\:C = 60^o.$
Altitude BD is drawn to AC, forming $\displaystyle \Delta ADB$
Altitude DE is drawn to AB, forming $\displaystyle \Delta DE.$
Altitude EF is drawn to AC, forming $\displaystyle \Delta AFE.$
Altitude FG is drawn to AB, forming $\displaystyle \Delta AFG.$
Altitude GH is drawn to AC, forming $\displaystyle \Delta HG.$
Altitude HI is drawn to AB, forming $\displaystyle \Delta AHI.$

Find the ratio of the area of $\displaystyle \Delta AIH$ to the area of $\displaystyle \Delta ABC.$

I found that the resulting ratio $\displaystyle \frac{729\sqrt{3}}{8192}$
. .
Close, but incorrect.

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 $\displaystyle \sqrt{3}$ times the shortest side.

Let $\displaystyle BC = 8$, then $\displaystyle AB = 8\sqrt{3}$
The area of $\displaystyle \Delta ABC$ is: .$\displaystyle A \:=\:\tfrac{1}{2}(8\sqrt{3})(8) \:=\:32\sqrt{3}$ .[1]

In $\displaystyle \Delta BDC\!:hyp = 8,\; CD =4,\;BD = 4\sqrt{3}$

In $\displaystyle \Delta DEB\!:\;hyp = 4\sqrt{3},\;BE = 2\sqrt{3},\;DE = 6$

In $\displaystyle \Delta DEA\!:\;DE = 6,\;EA \:=\: 8\sqrt{3} - 2\sqrt{3} \:=\:6\sqrt{3}$
. . The area of $\displaystyle \Delta DEA \:=\:\tfrac{1}{2}(6\sqrt{3})(6) \:=\:18\sqrt{3}$

We have: .$\displaystyle \frac{\text{area of }\Delta DEA}{\text{area of }\Delta ABC} \;=\;\frac{18\sqrt{3}}{32\sqrt{3}} \;=\;\frac{9}{16}$

Drawing two altitudes, we have two similar right triangles.
. . The area of the smaller triangle is $\displaystyle \tfrac{9}{16}$ of the larger.

This procedure is performed three times.
Hence, the area of the final triangle $\displaystyle (\Delta AIH)$ is: .$\displaystyle \left(\tfrac{9}{16}\right)^3$ the area of $\displaystyle \Delta ABC.$

Therefore: .$\displaystyle \frac{\text{area of }\Delta AIH}{\text{area of }\Delta ABC} \;=\;\frac{729}{4096}$