Question

what is the length of \\(\overline{sa}\\)?
2.43 ft
1.89 ft
7 ft
3 ft

Explanation:

Step1: Identify Similar Triangles

Triangles \( \triangle ACT \) and \( \triangle BST \) are similar (right triangles with shared angles, so AA similarity). Thus, \( \frac{AC}{BS} = \frac{CT}{ST} = \frac{AT}{BT} \). Wait, actually, \( \triangle ACT \sim \triangle BST \), and also \( \triangle AST \) and \( \triangle BCT \)? Wait, no, let's check the segments. \( AB = 10 \) ft, \( BT = 9 \) ft? Wait, no, \( AB \) is 10 ft, \( BT \) is 9? Wait, \( ST = 6.3 \) ft, \( CT = 2.7 \) ft. Let's denote \( SA = x \), so \( AB = SA + SB = x + SB = 10 \), so \( SB = 10 - x \). Since \( \triangle ACT \sim \triangle BST \) (right angles, and \( \angle ATC = \angle BTS \) vertical angles? Wait, no, \( \angle ACT = \angle BST = 90^\circ \)? Wait, the diagram has right angles at \( C \) and \( T \)? Wait, \( AC \perp CT \), \( ST \perp BT \), and \( AC \parallel ST \)? So \( \triangle AAC \)? Wait, maybe \( \triangle ACT \sim \triangle BST \) by AA (both right, and \( \angle A = \angle B \)? No, better to use the Angle-Angle similarity for right triangles. Wait, \( \angle ACT = \angle BST = 90^\circ \), and \( \angle ATC = \angle BTS \) (vertical angles), so \( \triangle ACT \sim \triangle BST \). Thus, \( \frac{AC}{BS} = \frac{CT}{ST} = \frac{AT}{BT} \). Wait, but we can also use \( \triangle AST \) and \( \triangle BCT \)? Wait, maybe another approach: Let \( SA = x \), \( SB = 10 - x \). Then, since \( \triangle ACT \sim \triangle BST \), \( \frac{CT}{ST} = \frac{AC}{BS} \), but we don't know \( AC \). Wait, alternatively, \( \triangle AST \sim \triangle BCT \)? Wait, \( \angle AST = \angle BCT = 90^\circ \), and \( \angle A = \angle B \)? No, maybe the triangles \( \triangle ACT \) and \( \triangle BST \) have \( \frac{CT}{ST} = \frac{2.7}{6.3} = \frac{3}{7} \). So the ratio of similarity is \( \frac{3}{7} \). Then, \( \frac{AT}{BT} = \frac{3}{7} \). Let \( AT = y \), \( BT = 9 \)? Wait, no, \( BT \) is 9? Wait, \( AB = 10 \), so \( AT + BT = AB = 10 \)? Wait, no, \( AT = SA + ST \)? No, the diagram: \( A \) to \( S \) to \( B \) is \( AB \), length 10. \( S \) to \( T \) is 6.3, \( T \) to \( C \) is 2.7, \( C \) to \( A \) has a right angle, \( T \) to \( B \) has a right angle. So \( \triangle ACT \) and \( \triangle BST \) are similar, so \( \frac{CT}{ST} = \frac{AC}{BS} = \frac{AT}{BT} \). \( \frac{2.7}{6.3} = \frac{3}{7} \). Let \( SA = x \), so \( BS = 10 - x \). Let \( AT = x + ST \)? No, \( AT \) is from \( A \) to \( T \), \( BT \) is from \( B \) to \( T \). Wait, maybe \( \frac{SA}{BT} = \frac{ST}{CT} \)? Wait, no, let's use the ratio \( \frac{CT}{ST} = \frac{2.7}{6.3} = \frac{3}{7} \). So the ratio of \( CT \) to \( ST \) is 3:7. Then, the ratio of \( SA \) to \( BT \) should be the same? Wait, \( BT = 9 \)? Wait, the options include 3, 7, etc. Wait, if \( SA = 3 \), then \( SB = 7 \), and \( \frac{CT}{ST} = \frac{2.7}{6.3} = \frac{3}{7} \), which matches \( \frac{SA}{SB} = \frac{3}{7} \). Ah! So \( \frac{SA}{SB} = \frac{CT}{ST} \). So \( \frac{SA}{SB} = \frac{2.7}{6.3} = \frac{3}{7} \). Let \( SA = 3k \), \( SB = 7k \). Then \( SA + SB = 3k + 7k = 10k = 10 \), so \( k = 1 \), so \( SA = 3 \) ft.

Step2: Verify the Ratio

Check \( \frac{CT}{ST} = \frac{2.7}{6.3} = \frac{3}{7} \), and \( \frac{SA}{SB} = \frac{3}{7} \) (if \( SA = 3 \), \( SB = 7 \), since \( 3 + 7 = 10 \)). This satisfies the similarity of triangles (proportional sides, right angles), so \( SA = 3 \) ft.

Answer:

3 ft