Question

Geometry question finding radius and area Thank You!!

Answer 1

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:

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

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:

$AK=\sqrt{BD^2-1}=\sqrt{(8)^2-1}=\sqrt{63}=3\sqrt{7}$

Hence:

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

Therefore, the perimeter of ΔABC is:

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