McGraw Hill Integrated II, 2012
MH
McGraw Hill Integrated II, 2012 View details
4. Special Products
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Exercise 55 Page 32

Practice makes perfect
a To find the area of the large circle, we first need to write an expression for its radius. Since the radius of the large circle is 9 ft longer than the radius of the inner circle we can add 9 to the radius of the inner circle to get its radius, r+9.
A= π r^2
A = π( r+9)^2
A= π (r^2 + 2 * 9r + 9^2)
A= π (r^2 + 18r + 81)
A = π r^2 +18π r + 81π
If we approximate π ≈ 3.14, we get 3.14r^2 +56.52r + 254.34 ft^2.
b To find the area of the portion outside the outer circle, we need to subtract our answer from Part A from the area of the square. Let's find the area of the square.
A = s^2
A = 38^2
A = 1444
Now. we can subtract the outer circle's area that we derived in Part A from 1444 ft^2. 1444-(3.14r^2 +56.52r + 254.34) ⇕ 1444-3.14r^2 -56.52r - 254.34 When we combine like terms, we get -3.14r^2 -56.52r +1189.66 ft^2.