Question

Right triangle ABC has side lengths AB=8 , BC=15 , and AC=17 . A second right triangle, A′B′C′ , has side lengths of 34, 30, and 16. First, find the ratio of the side opposite ∠C to the hypotenuse of triangle ABC . Then, use this ratio to identify the side length opposite point C′ on the second right triangle.(1 point) The length of the side opposite C′ is units.

Answer 1

First, let's analyze triangle ABC. In triangle ABC, we have:

- \( AB = 8 \) (one leg)
- \( BC = 15 \) (the other leg)
- \( AC = 17 \) (the hypotenuse)

The side opposite angle \( C \) is \( AB \), which is 8. The hypotenuse \( AC \) is 17.

Now, we find the ratio of the side opposite \( C \) to the hypotenuse:

\[
\text{Ratio} = \frac{AB}{AC} = \frac{8}{17}
\]

Next, we apply this ratio to triangle \( A'B'C' \). The sides of triangle \( A'B'C' \) are given as:

- \( A'B' = 34 \)
- \( B'C' = 30 \)
- \( A'C' = 16 \)

In triangle \( A'B'C' \), we also need to identify which side corresponds to the hypotenuse. It is \( A'C' = 34 \).

To find the length of the side opposite point \( C' \), we can set up a proportion using the ratio determined from triangle \( ABC \):

\[
\text{Side opposite } C' = \text{Ratio} \times \text{Hypotenuse of } A'B'C'
\]

Therefore, we have:

\[
\text{Side opposite } C' = \frac{8}{17} \times 34
\]

Calculating this, we get:

\[
\text{Side opposite } C' = \frac{8 \times 34}{17} = \frac{272}{17} = 16
\]

Thus, the length of the side opposite \( C' \) is \( \boxed{16} \) units.