To solve this exercise, let's calculate the lengths $HG$ and $EF$ in the given right triangle using the Pythagorean theorem and properties of similar triangles.

Observations from the Figure

1. We have a right triangle $\triangle EFG$ with a right angle at $H$.
2. $EH = 2$ (horizontal segment).
3. $HF = 8$ (vertical segment).
4. $FG = 10$ (hypotenuse of $\triangle HFG$).

Step 1: Calculating $HG$

Since $\triangle HFG$ is a right triangle with $HF = 8$ and $HG = x$ as the unknown, and $FG = 10$ as the hypotenuse, we can apply the Pythagorean theorem:

$HG^2 + HF^2 = FG^2$

Substitute the known values:

$HG^2 + 8^2 = 10^2$

$HG^2 + 64 = 100$

$HG^2 = 100 - 64$

$HG^2 = 36$

$HG = \sqrt{36} = 6$

So, $HG = 6$.

Step 2: Calculating $EF$

The segment $EF$ can be found using the similarity of triangles $\triangle EFG$ and $\triangle HFG$. Since $\triangle HFG$ is a smaller, similar triangle with $EF$ as the corresponding side of $FG$, we can set up a proportion.

Since $EH = 2$, $HG = 6$, and $FG = 10$, the triangle proportions give:

$\frac{EF}{FG} = \frac{EH}{HG}$

Substitute the known values:

$\frac{EF}{10} = \frac{2}{6}$

Solve for $EF$:

$EF = 10 \times \frac{2}{6}$

$EF = 10 \times \frac{1}{3} = \frac{10}{3} \approx 3.33$

So, $EF \approx 3.33$ (or exactly $\frac{10}{3}$).

Final Answer

• $HG = 6$
• $EF = \frac{10}{3}$ (approximately $3.33$)

Would you like further details on any part of the solution?

Here are 5 related questions for further practice:

1. How would you find the length of $EF$ if only $FG$ and the angle $\angle EFG$ were given?
2. Can you find the area of $\triangle EFG$?
3. What is the perimeter of $\triangle EFG$?
4. How would the calculations change if $\triangle EFG$ were not a right triangle?
5. Can you verify the similarity between $\triangle EFG$ and $\triangle HFG$ using angle-angle similarity?

Tip: In right triangles, the Pythagorean theorem is a powerful tool to find missing side lengths when you know two sides.