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# The five circles making up this archery target have diameters of length 2,4,6,8, and 10. What is the total red area?

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The five circles making up this archery target have diameters of length 2,4,6,8, and 10. What is the total red area?

Guest Aug 15, 2017
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#1
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Area of 1st inner red  circle  = pi * ( diameter / 2) ^2  = pi * (2 / 2) ^2  =  pi (1)^2  = pi (1)

Area of 1st inner white ring  =    pi * ( 4/2 ) ^2  - (1)    = pi [ 4 - 1]  =  3pi    (2)

Area of 2nd red inner ring  =   pi * ( 6 / 2) ^2  -  [area of (1) + (2)]  =   9pi  - 3pi - pi  = 5pi      (3)

Area of 2nd inner white ring  =   pi * ( 8 / 2)^2  - [area of (1) + (2) + (3) ]   = 16pi  - 9pi  =  7 pi     (4)

Area of outer red ring   = pi * (10 /2 ) ^2  - area of  [(1) + (2) + (3) + (4)]  =  25 pi - 16pi  = 7pi

Sum of three red areas = pi  + 5pi +  7pi  = 13 pi  units^2  ≈  40.84 units^2

CPhill  Aug 15, 2017
edited by CPhill  Aug 15, 2017
edited by CPhill  Aug 15, 2017
#2
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The five circles making up this archery target have diameters of length 2,4,6,8, and 10.

What is the total red area?

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 $$= A_\text{red}$$

$$\begin{array}{|rcll|} \hline A_\text{red} &=& A_5-A_4+A_3-A_2+A_1 \\ &=& 25\pi-16\pi+9\pi-4\pi+1\pi \\ &=& 15\pi \\ &=& 47.1238898038\ \text{units}^2 \\ \hline \end{array}$$

heureka  Aug 16, 2017

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