Answer: [tex]\bold{a)\ \overline{CD}=6\qquad b)\ \overline{AD}=\dfrac{32}{3}\qquad c)\ \overline{AC}=\dfrac{50}{3}\qquad d)\ \overline{OB}=\dfrac{25}{3}}[/tex]
Step-by-step explanation:
a) First, use the Pythagorean Theorem to find CD:
8² + (CD)² = 10² → CD = 6
d) Next, find OB. Notice that both OB and OC are the radius of the circle.
Using the Segment Addition postulate, we know the radius (r) = CD + DO → DO = r - 6
Use the Pythagorean Theorem for ΔOBD to find OB:
(DO)² + (BD)² = (OB)²
(r - 6)² + 8² = r²
r² - 12r + 36 + 64 = r²
-12r + 100 = 0
100 = 12r
[tex]\dfrac{25}{3}\quad = r[/tex]
c) Now, find AC. Notice that AC is the diameter.
Diameter = 2r
= 2(OB)
[tex]=2\bigg(\dfrac{25}{3}\bigg)\qquad =\dfrac{50}{3}[/tex]
b) Lastly, find AD. Using the Segment Addition Postulate, we know that
AD + CD = AC → AD = AC - CD
[tex]= \dfrac{50}{3}-6\qquad \rightarrow \quad \dfrac{50}{3}-\dfrac{18}{3}\qquad =\dfrac{32}{3}[/tex]
