SAT Circles

Correct Answer: C

Solution:

Tip: Area of a circle with radius $r: A_{\bigodot} = \pi r^2.$
We can infer from the diagram that the radius of the big circle equals the diameter of the middle circle, 4; the radius of the middle circle is 2; and the radius of the small circle is 1.

$\begin{array}{r c l} A_{\text{shaded}} &=& A_{\text{big}} - A_{\text{middle}} + A_{\text{small}}\\ &=& \pi\cdot 4^2 - \pi \cdot 2^2 + \pi \cdot 1^2\\ &=& 16 \pi - 4\pi + \pi\\ &=& 13 \pi \end{array}$

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Incorrect Choices:

(A)
If you think the radius of the big circle is 3, you will get this wrong answer. The radius of the big circle is 4.

(B)
If you find $A_{\text{big}} - A_{\text{middle}} - A_{\text{small}}$ , you will get this wrong answer.

(D)
If you find $A_{\text{big}} + A_{\text{middle}} - A_{\text{small}}$ , you will get this wrong answer.

(E)
If you find $A_{\text{big}} + A_{\text{middle}} + A_{\text{small}}$ , you will get this wrong answer.

Big circle = 16 π

Middle circle = 4π

Small circle = π

Shaded = 16π - 4π + π = 13π

The area of the shaded region is equal to the area of the largest circle (A= π⋅4 squared) - the area of the middle circle ( A = π⋅2 squared) + the area of the smallest circle (A = π⋅1 squared).

(π⋅4 squared) - ( π⋅2 squared) + (π⋅1 squared) = solution

16π - 4π + π = solution

12π + π = solution

13π = solution