The five circles making up this archery target have diameters of length 2,4,6,8, and 10.
What is the total red area?
Simplify your answer as much as you can.
Let A = area of a circle \(= \pi r^2\)
Ler r = radius of a circle
\(\begin{array}{|rclrcl|} \hline A_1 &=& \text{ area of 1st red circle and radius } & r_1 &=& 1 \\ &=& \pi \cdot 1^2 \\ &=& 1\pi \\ A_2 &=& \text{ area of 1st white circle and radius } & r_2 &=& 2 \\ &=& \pi \cdot 2^2 \\ &=& 4\pi \\ A_3 &=& \text{ area of 2nd red circle and radius } & r_3 &=& 3 \\ &=& \pi \cdot 3^2 \\ &=& 9\pi \\ A_4 &=& \text{ area of 2nd white circle and radius } & r_4 &=& 4 \\ &=& \pi \cdot 4^2 \\ &=& 16\pi \\ A_5 &=& \text{ area of 3rd red circle and radius } & r_5 &=& 5 \\ &=& \pi \cdot 5^2 \\ &=& 25\pi \\ \hline \end{array} \)
The total red area \(= A_\text{red}\)
\(\begin{array}{|rcll|} \hline A_\text{red} &=& A_5-A_4+A_3-A_2+A_1 \\ &=& 25\pi-16\pi+9\pi-4\pi+1\pi \\ &=& 15\pi \\ &=& 47.1238898038\ \text{units}^2 \\ \hline \end{array} \)