The answer is: " 125.6 sq units " ; or, write as: " 125.6 units² " .
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Explanation:
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To solve for the "area of the shaded region" :
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1) We find the area of the: "circle with radius of 11" ; that is, the entire circle that includes the entire outer circle—incorporating the shaded region AND non-shaded region. Record that value.
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2) Then we find the area of the: "circle with radius of 9" ; that is; the entire "inner circle" that is "inside the shaded region". Record that value.
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3) Then, we take the [area of the "circle with radius of 11"] ; and from that value, subtract the: "[area of the circle with radius of 9"] ; to get the value of the "area of the shaded region" {that is, the "area of the blue ring"} .
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Let us begin:
1) Find the area of the: "circle with radius of 11" ;
The formula for the area, "A" of a circle is:
→ " A = [tex] \pi [/tex] r² " ;
in which: " A = area of the circle" (in "units² " ; or "square units" ; or "sq units") ;
"[tex] \pi [/tex] = 3.14 " (approximation) ;
" r = radius = 11 " ;
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→ Plug in these values to solve for the "area of the circle with radius of 11" :
→ A = [tex] \pi [/tex] r² ;
A = (3.14) * 11² ;
A = (3.14) * 11 * 11 ;
A = (3.14) * 121 ;
A = 379.94 units²
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2) Now, let us find the area of the: "circle with radius of 9" ;
The formula for the area, "A" of a circle is:
→ " A = [tex] \pi [/tex] r² " ;
in which: " A = area of the circle" (in "units² " ; or "square units" ; or "sq units") ;
"[tex] \pi [/tex] = 3.14 " (approximation) ;
" r = radius = 9 " ;
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→ Plug in these values to solve for the "area of the circle with radius of 9" :
→ A = [tex] \pi [/tex] r² ;
A = (3.14) * 9² ;
A = (3.14) * 9 * 9 ;
A = (3.14) * 81 ;
A = 254.34 units²
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2) Now, let us subtract the "area of the circle with radius of 9" ; FROM:
the "area of the circle with radius of 11" ;
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3 7 9. 94 units²
— 2 5 4. 34 units²
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1 2 5 . 60 units²
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The answer is: " 125.6 sq units " ; or, write as: " 125.6 units² " .
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