Question

the smallest of the three circles with center d has a radius of 8 inches and cb = ba = 4 inches. what is the sum of the areas of all three circles? 80π in.² 96π in.² 208π in.² 464π in.²

Explanation:

Step1: Find radii of three circles

The smallest circle (radius \( r_1 \)) has \( r_1 = 8 \) inches.
For the middle circle, radius \( r_2 = 8 + 4 = 12 \) inches (since \( CB = 4 \)).
For the largest circle, radius \( r_3 = 12 + 4 = 16 \) inches (since \( BA = 4 \)).

Step2: Calculate area of each circle

Area of a circle is \( A = \pi r^2 \).

• Smallest: \( A_1 = \pi(8)^2 = 64\pi \)
• Middle: \( A_2 = \pi(12)^2 = 144\pi \)
• Largest: \( A_3 = \pi(16)^2 = 256\pi \)

Wait, no—wait, the smallest circle’s radius is 8? Wait, no, wait: Wait, the smallest circle (center D, radius DC? Wait, no, the problem says “the smallest of the three circles with center D has a radius of 8 inches” and \( CB = BA = 4 \) inches. So DC is 8? Wait, no, maybe I misread. Wait, the smallest circle: radius is 8. Then the middle circle: radius is \( 8 + 4 = 12 \) (since CB is 4, so from D to C is 8, D to B is 8+4=12, D to A is 12+4=16. Wait, no—wait, the three circles: smallest (radius 8), middle (radius 8+4=12), largest (radius 12+4=16)? Wait, no, that can’t be. Wait, maybe the smallest circle has radius 8, then the next circle (middle) has radius 8 + 4 = 12, and the largest has radius 12 + 4 = 16? Wait, but then sum of areas would be \( \pi(8^2 + 12^2 + 16^2) = \pi(64 + 144 + 256) = \pi(464) \)? No, that’s one of the options, but that seems big. Wait, maybe I messed up the radii. Wait, maybe the smallest circle is DC = 8, then CB = 4, so DB = 8 + 4 = 12 (middle circle radius), and BA = 4, so DA = 12 + 4 = 16 (largest circle radius). Then areas:
\( 8^2\pi + 12^2\pi + 16^2\pi = 64\pi + 144\pi + 256\pi = 464\pi \)? But wait, the options include 464π. Wait, but maybe I misread the smallest radius. Wait, the problem says “the smallest of the three circles with center D has a radius of 8 inches” – so smallest radius \( r_1 = 8 \), middle \( r_2 = 8 + 4 = 12 \), largest \( r_3 = 12 + 4 = 16 \). Then sum of areas: \( \pi(8^2 + 12^2 + 16^2) = \pi(64 + 144 + 256) = 464\pi \). Wait, but let me check again. Wait, maybe the smallest circle is DC = 8, then CB = 4, so the middle circle is from D to B, which is 8 + 4 = 12, and the largest is from D to A, which is 12 + 4 = 16. So yes, that’s correct. So the sum is \( 64\pi + 144\pi + 256\pi = 464\pi \) square inches.

Answer:

\( 464\pi \) in.² (the option with \( 464\pi \) in.²)