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the smallest of the three circles with center d has a radius of 8 inche…

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 the radii of all circles

The radius of the smallest circle, $r_1 = 8$ inches. The radius of the middle - sized circle, $r_2=8 + 4=12$ inches. The radius of the largest circle, $r_3=12 + 4 = 16$ inches.

Step2: Recall the formula for the area of a circle

The formula for the area of a circle is $A=\pi r^{2}$.

Step3: Calculate the areas of each circle

The area of the smallest circle, $A_1=\pi r_1^{2}=\pi\times8^{2}=64\pi$ square inches. The area of the middle - sized circle, $A_2=\pi r_2^{2}=\pi\times12^{2}=144\pi$ square inches. The area of the largest circle, $A_3=\pi r_3^{2}=\pi\times16^{2}=256\pi$ square inches.

Step4: Calculate the sum of the areas

$A = A_1+A_2+A_3=64\pi+144\pi + 256\pi=(64 + 144+256)\pi=464\pi$ square inches.

Answer:

$464\pi$ in.$^{2}$