Question

section 6.1
for questions 1 through 6, find the indicated m

1. find \\(\overline{ab}\\)

Explanation:

Step1: Identify the properties

From the diagram, \( CB = 10 \), \( \angle CBA = 90^\circ \) (right angle), and the segments \( CD \) and \( DA \) seem to have markings indicating \( CB = AB \) (since the perpendicular from \( C \) to \( AB \) and the congruency markings on the sides). Also, the line \( DB \) is a perpendicular bisector? Wait, actually, since \( \triangle CDB \) and \( \triangle ADB \) might be congruent? Wait, more clearly, the segment \( CB = 10 \), and \( B \) is the midpoint? Wait, no, the right angle at \( B \) for \( CB \), and the markings on \( CD \) and \( DA \) suggest that \( CB = AB \). Wait, actually, in the diagram, \( CB = 10 \), and \( B \) is the foot of the perpendicular, and since the sides \( CD \) and \( DA \) are marked equal? Wait, no, the key is that \( CB = AB \) because \( DB \) is a perpendicular bisector? Wait, no, the length \( CB = 10 \), and since \( \angle CBA = 90^\circ \), and the segments \( CD \) and \( DA \) have the same marking, so \( CB = AB \). So \( AB = CB \).

Step2: Determine \( AB \)

Since \( CB = 10 \) and \( AB = CB \) (from the diagram's congruency or perpendicular bisector property, or maybe isosceles triangle? Wait, actually, the right angle at \( B \), and \( CB = AB \) because the segments from \( D \) to \( C \) and \( D \) to \( A \) are equal (marked with the same tick), so \( \triangle CDB \cong \triangle ADB \) by HL (right angle, hypotenuse \( CD = DA \), leg \( DB = DB \)), so \( CB = AB \). Therefore, \( AB = CB = 10 \).

Answer:

\( AB = 10 \)