Question

Geometry question finding radius and area
Thank You!!

Answer 1

The radius of Circle D is 8 cm.

The perimeter of ΔABC is (24 + 6√7) cm.

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:

`AB^2=(BD+1)^2+AK^2`

Since AB = 12:

`144=(BD+1)^2+AK^2`

For ΔADK, by the Pythagorean Theorem:

`AD^2=1^2+AK^2`

Since AD = BD:

`BD^2=1+AK^2`

Subtract the second equation into the first:

`144-(BD^2)=(BD+1)^2+AK^2-(1+AK^2)`

Simplify:

`144-BD^2=BD^2+2BD+1-1`

Hence:

`2BD^2+2BD-144=0`

Simplify:

`BD^2+BD-72=0`

Factor:

`(BD+9)(BD-8)=0`

By the Zero Product Property:

`BD=-9\text{ or } BD=8`

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:

`BD^2=1+AK^2`

Thus:

`AK=√(BD^2-1)=√((8)^2-1)=√(63)=3√(7)`

Hence:

`AC=2(AK)=2(3\sqrt7)=6\sqrt7`

Therefore, the perimeter of ΔABC is:

`P=(12)+(12)+(6\sqrt7)=24+6\sqrt7\text{ cm}`