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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?

 

Simplify your answer as much as you can.

Guest Aug 15, 2017
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 #1
avatar+76160 
+1

 

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

 

 

 

cool cool cool

CPhill  Aug 15, 2017
edited by CPhill  Aug 15, 2017
edited by CPhill  Aug 15, 2017
 #2
avatar+18564 
0

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 \(= A_\text{red}\)

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

 

 

laugh

heureka  Aug 16, 2017

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