Geometry question finding radius and area
Thank You!!

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]

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