Answer:
The radius of Circle D is 8 cm.
The perimeter of ΔABC is (24 + 6√7) cm.
Step-by-step explanation:
First, let the intersection point below D be K and let the intersection point between A and B be J.
Since segment BK, which passes through the center of the circle, is perpendicular to chord AC, BK also bisects AC. Hence, AK = CK.
Connect points A and D to create radius AD. Note that BD is also a radius. Hence, AD = BD.
For ΔABK, by the Pythagorean Theorem:
[tex]AB^2=(BD+1)^2+AK^2[/tex]
Since AB = 12:
[tex]144=(BD+1)^2+AK^2[/tex]
For ΔADK, by the Pythagorean Theorem:
[tex]AD^2=1^2+AK^2[/tex]
Since AD = BD:
[tex]BD^2=1+AK^2[/tex]
Subtract the second equation into the first:
[tex]144-(BD^2)=(BD+1)^2+AK^2-(1+AK^2)[/tex]
Simplify:
[tex]144-BD^2=BD^2+2BD+1-1[/tex]
Hence:
[tex]2BD^2+2BD-144=0[/tex]
Simplify:
[tex]BD^2+BD-72=0[/tex]
Factor:
[tex](BD+9)(BD-8)=0[/tex]
By the Zero Product Property:
[tex]BD=-9\text{ or } BD=8[/tex]
Since the radius must be positive, the radius is 8 cm.
Since we already know AB and BC, we need to find AC to find the perimeter.
Note that AC = AK + CK = 2AK.
From the second equation:
[tex]BD^2=1+AK^2[/tex]
Thus:
[tex]AK=\sqrt{BD^2-1}=\sqrt{(8)^2-1}=\sqrt{63}=3\sqrt{7}[/tex]
Hence:
[tex]AC=2(AK)=2(3\sqrt7)=6\sqrt7[/tex]
Therefore, the perimeter of ΔABC is:
[tex]P=(12)+(12)+(6\sqrt7)=24+6\sqrt7\text{ cm}[/tex]
