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, we identify the sides of triangle ABC. Since it is a right triangle, we can label the sides as follows:

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

Next, we find the ratio of the side opposite \( \angle C \) (which is side \( AB = 8 \)) to the hypotenuse \( AC = 17 \):

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

Now, we apply this ratio to triangle \( A'B'C' \). In triangle \( A'B'C' \), we need to identify the side opposite to point \( C' \).

The side lengths of triangle \( A'B'C' \) are:

- \( A'B' = 34 \) (one leg)
- \( B'C' = 30 \) (the other leg)
- \( A'C' = 16 \) (the hypotenuse)

In this triangle, the side opposite \( C' \) corresponds to the side \( A'B' \) (34). We use the ratio of triangle \( ABC \) to find the length of the side opposite \( C' \):

Let \( x' \) be the side opposite \( C' \):

\[
\frac{x'}{A'C'} = \frac{8}{17}
\]

where \( A'C' = 16 \), so:

\[
\frac{x'}{16} = \frac{8}{17}
\]

Now we cross-multiply to solve for \( x' \):

\[
17x' = 8 \times 16
\]

\[
17x' = 128
\]

\[
x' = \frac{128}{17} \approx 7.53
\]

Thus, the length of the side opposite \( C' \) is

\[
\frac{128}{17} \text{ units}.
\]