Two concentric circles share a centre. The outer circle has circumference 132 inches and the inner circle has radius 17.5 inches. Find the area of the ring between them, taking pi to be 22/7.

Step-by-step solution

1. 1

   Find the outer radius from its circumference.

   $R=\frac{132}{2\pi}=\frac{132}{\frac{44}{7}}=132\cdot\frac{7}{44}=3\cdot 7=21\ \text{in}.$

   With $\pi=\tfrac{22}{7}$ the $44$ divides $132$ exactly three times, so no decimals appear.

2. 2

   Write the ring area as a difference of two circle areas. The ring (annulus) is the big disc with the small one removed:

   $A_{\text{ring}}=\pi R^{2}-\pi r^{2}=\pi\left(R^{2}-r^{2}\right).$

3. 3

   Factor the difference of squares instead of squaring both radii.

   $R^{2}-r^{2}=(R-r)(R+r)=(21-17.5)(21+17.5)=3.5\times 38.5.$

   This avoids computing $441-306.25$ and keeps the numbers small — the point of the factorisation.

4. 4

   Multiply out.

   $3.5\times 38.5=134.75,$

   so

   $A_{\text{ring}}=\frac{22}{7}\times 134.75.$

5. 5

   Finish with exact fractions. Writing $3.5=\tfrac72$ makes the $7$ cancel:

   $A_{\text{ring}}=\frac{22}{7}\cdot\frac{7}{2}\cdot 38.5=11\times 38.5=423.5\ \text{in}^2.$

6. 6

   Cross-check the direct way. $\pi R^2=\tfrac{22}{7}(441)=1386$ and $\pi r^2=\tfrac{22}{7}(306.25)=962.5$; their difference is $1386-962.5=423.5\ \text{in}^2$ ✓.