Question

1. triangles sut and abc are similar. select all proportionalities by placing a check mark

√ in the box at the beginning of each choice.

□ \\(\frac{tu}{ab} = \frac{su}{ac}\\)

□ \\(\frac{bc}{ab} = \frac{su}{tu}\\)

□ \\(\frac{st}{su} = \frac{ac}{bc}\\)

□ \\(\frac{ab}{tu} = \frac{su}{bc}\\)

Explanation:

Step 1: Identify Similar Triangles

Triangles \( SUT \) and \( ABC \) are similar (right triangles with right angles at \( T \) and \( A \), respectively). So, corresponding sides are proportional. Let's identify corresponding sides:
- Right angle at \( T \) (in \( \triangle SUT \)) and right angle at \( A \) (in \( \triangle ABC \)).
- So, \( \angle T = \angle A = 90^\circ \), \( \angle S \) corresponds to \( \angle C \), and \( \angle U \) corresponds to \( \angle B \).
- Thus, sides: \( SU \) (hypotenuse of \( \triangle SUT \)) corresponds to \( BC \) (hypotenuse of \( \triangle ABC \)), \( TU \) (leg of \( \triangle SUT \)) corresponds to \( AB \) (leg of \( \triangle ABC \)), \( ST \) (leg of \( \triangle SUT \)) corresponds to \( AC \) (leg of \( \triangle ABC \)).

Step 2: Check Each Proportion

1. \( \frac{TU}{AB} = \frac{SU}{AC} \): \( TU \) corresponds to \( AB \), \( SU \) corresponds to \( BC \) (not \( AC \)). So, incorrect.
2. \( \frac{BC}{AB} = \frac{SU}{TU} \): \( BC \) (hypotenuse of \( \triangle ABC \)) corresponds to \( SU \) (hypotenuse of \( \triangle SUT \)), \( AB \) (leg of \( \triangle ABC \)) corresponds to \( TU \) (leg of \( \triangle SUT \)). So, \( \frac{BC}{AB} = \frac{SU}{TU} \) (proportionality of hypotenuse to leg). Correct.
3. \( \frac{ST}{SU} = \frac{AC}{BC} \): \( ST \) (leg of \( \triangle SUT \)) corresponds to \( AC \) (leg of \( \triangle ABC \)), \( SU \) (hypotenuse of \( \triangle SUT \)) corresponds to \( BC \) (hypotenuse of \( \triangle ABC \)). So, \( \frac{ST}{SU} = \frac{AC}{BC} \) (leg to hypotenuse ratio). Correct.
4. \( \frac{AB}{TU} = \frac{SU}{BC} \): \( AB \) (leg of \( \triangle ABC \)) corresponds to \( TU \) (leg of \( \triangle SUT \)), \( SU \) (hypotenuse of \( \triangle SUT \)) corresponds to \( BC \) (hypotenuse of \( \triangle ABC \)). So, \( \frac{AB}{TU} = \frac{BC}{SU} \) (not \( \frac{SU}{BC} \)). Wait, no: \( AB \) corresponds to \( TU \), \( SU \) corresponds to \( BC \). So, \( \frac{AB}{TU} = \frac{BC}{SU} \) (cross - multiply: \( AB \cdot SU = TU \cdot BC \)), but the given is \( \frac{AB}{TU} = \frac{SU}{BC} \) (which would be \( AB \cdot BC = TU \cdot SU \)), incorrect. Wait, re - check:
Wait, \( AB \) (leg) ~ \( TU \) (leg), \( SU \) (hypotenuse) ~ \( BC \) (hypotenuse), \( ST \) (leg) ~ \( AC \) (leg).
So, \( \frac{ST}{SU} = \frac{AC}{BC} \): \( ST \) (leg) / \( SU \) (hypotenuse) = \( AC \) (leg) / \( BC \) (hypotenuse). Correct (leg - hypotenuse ratio).
\( \frac{BC}{AB} = \frac{SU}{TU} \): \( BC \) (hypotenuse) / \( AB \) (leg) = \( SU \) (hypotenuse) / \( TU \) (leg). Correct (hypotenuse - leg ratio).

Wait, let's re - express the proportions with correct correspondence:
\( \triangle SUT \sim \triangle CBA \) (since \( \angle T = \angle A = 90^\circ \), \( \angle S = \angle C \), so \( \triangle SUT \sim \triangle CAB \)? Wait, maybe I mixed up the correspondence. Let's use angle - angle similarity:
\( \angle T = 90^\circ \), \( \angle A = 90^\circ \); \( \angle S = \angle C \) (so \( \triangle SUT \sim \triangle CAB \))? Wait, no, the labels: \( \triangle SUT \) has vertices \( S, U, T \) with right angle at \( T \); \( \triangle ABC \) has vertices \( A, B, C \) with right angle at \( A \). So, \( \angle S \) (at \( S \)) corresponds to \( \angle C \) (at \( C \)), \( \angle U \) (at \( U \)) corresponds to \( \angle B \) (at \( B \)), \( \angle T \) (at \( T \)) corresponds to \( \angle A \) (at \( A \)). So, sides:
- \( ST \) (opposite \( \angle U \)) corresponds to \( AC \) (opposite \( \angle B \))
- \( TU \) (opposite \( \angle S \)) corresponds to \( AB \) (opposite \( \angle C \))
- \( SU \) (opposite \( \angle T \)) corresponds to \( BC \) (opposite \( \angle A \))

So, proportionality: \( \frac{ST}{AC} = \frac{TU}{AB} = \frac{SU}{BC} \)

Now, check each option:
1. \( \frac{TU}{AB} = \frac{SU}{AC} \): \( \frac{TU}{AB} \) is 1 (since \( TU \sim AB \)), but \( \frac{SU}{AC} \): \( SU \sim BC \), not \( AC \). Incorrect.
2. \( \frac{BC}{AB} = \frac{SU}{TU} \): \( BC \sim SU \), \( AB \sim TU \), so \( \frac{BC}{AB} = \frac{SU}{TU} \). Correct.
3. \( \frac{ST}{SU} = \frac{AC}{BC} \): \( ST \sim AC \), \( SU \sim BC \), so \( \frac{ST}{SU} = \frac{AC}{BC} \). Correct.
4. \( \frac{AB}{TU} = \frac{SU}{BC} \): \( AB \sim TU \), so \( \frac{AB}{TU} = 1 \), and \( \frac{SU}{BC} = 1 \) (since \( SU \sim BC \)). Wait, \( AB \sim TU \) (so \( AB = k \cdot TU \)), \( SU \sim BC \) (so \( SU = k \cdot BC \)). Then \( \frac{AB}{TU} = k \), \( \frac{SU}{BC} = k \). So, \( \frac{AB}{TU} = \frac{SU}{BC} \). Wait, earlier mistake. Let's re - check:

From \( \frac{ST}{AC} = \frac{TU}{AB} = \frac{SU}{BC} = k \) (scale factor).

So:
- \( \frac{TU}{AB} = k \), \( \frac{SU}{AC} \): \( SU = k \cdot BC \), \( AC = \frac{ST}{k} \), so \( \frac{SU}{AC} = \frac{k \cdot BC}{\frac{ST}{k}} = \frac{k^2 \cdot BC}{ST} eq k \). So, first option incorrect.
- \( \frac{BC}{AB} = \frac{BC}{AB} \), \( \frac{SU}{TU} = \frac{k \cdot BC}{k \cdot AB} = \frac{BC}{AB} \). So, \( \frac{BC}{AB} = \frac{SU}{TU} \). Correct.
- \( \frac{ST}{SU} = \frac{ST}{k \cdot BC} \), \( \frac{AC}{BC} = \frac{\frac{ST}{k}}{BC} = \frac{ST}{k \cdot BC} \). So, \( \frac{ST}{SU} = \frac{AC}{BC} \). Correct.
- \( \frac{AB}{TU} = \frac{AB}{k \cdot AB} = \frac{1}{k} \)? Wait, no: \( TU = k \cdot AB \) (since \( \frac{TU}{AB} = k \)), so \( \frac{AB}{TU} = \frac{1}{k} \). \( SU = k \cdot BC \), so \( \frac{SU}{BC} = k \). So, \( \frac{AB}{TU} eq \frac{SU}{BC} \). Wait, I messed up the correspondence. Let's use correct correspondence:

If \( \triangle SUT \sim \triangle CAB \) (right angle at \( T \) and \( A \), \( \angle S = \angle C \)), then:
- \( SU \) (hypotenuse of \( \triangle SUT \)) corresponds to \( CB \) (hypotenuse of \( \triangle CAB \))
- \( UT \) (leg of \( \triangle SUT \)) corresponds to \( AB \) (leg of \( \triangle CAB \))
- \( ST \) (leg of \( \triangle SUT \)) corresponds to \( AC \) (leg of \( \triangle CAB \))

So, proportionality: \( \frac{SU}{CB} = \frac{UT}{AB} = \frac{ST}{AC} \)

Now, check each option:
1. \( \frac{TU}{AB} = \frac{SU}{AC} \): \( TU \) (UT) corresponds to \( AB \), \( SU \) corresponds to \( CB \) (not \( AC \)). Incorrect.
2. \( \frac{BC}{AB} = \frac{SU}{TU} \): \( BC = CB \), \( AB \) corresponds to \( TU \), \( SU \) corresponds to \( CB \). So, \( \frac{BC}{AB} = \frac{SU}{TU} \) (since \( \frac{CB}{AB} = \frac{SU}{TU} \)). Correct.
3. \( \frac{ST}{SU} = \frac{AC}{BC} \): \( ST \) corresponds to \( AC \), \( SU \) corresponds to \( BC \) (CB). So, \( \frac{ST}{SU} = \frac{AC}{BC} \). Correct.
4. \( \frac{AB}{TU} = \frac{SU}{BC} \): \( AB \) corresponds to \( TU \) (so \( \frac{AB}{TU} = 1 \) if similar with scale 1, but generally \( \frac{AB}{TU} = \frac{BC}{SU} \) (since \( \frac{AB}{TU} = \frac{BC}{SU} \) from \( \frac{UT}{AB} = \frac{SU}{CB} \Rightarrow \frac{AB}{UT} = \frac{CB}{SU} \Rightarrow \frac{AB}{TU} = \frac{BC}{SU} \)). So, \( \frac{AB}{TU} eq \frac{SU}{BC} \). Incorrect.

So, correct proportions are \( \frac{BC}{AB} = \frac{SU}{TU} \) and \( \frac{ST}{SU} = \frac{AC}{BC} \).

Answer:

The correct proportions are:

• \( \frac{BC}{AB} = \frac{SU}{TU} \) (check this box)
• \( \frac{ST}{SU} = \frac{AC}{BC} \) (check this box)