Area of the shaded region is 540 – 65.25π.
Solution:
Length of the rectangle = 12 + 9 + 6 + 3 = 30 in
Width of the rectangle = 18 in
Radius of the larger circle ([tex]r_1[/tex]) = 12 ÷ 2 = 6 in
Radius of the medium circle ([tex]r_2[/tex]) = 9 ÷ 2 = 4.5 in
Radius of the smaller circle ([tex]r_3[/tex]) = 6 ÷ 2 = 3 in
Area of the shaded region = Area of the rectangle – Area of the larger circle – Area of the medium circle – Area of the smaller circle
[tex]=l\times b-\pi r_1^2-\pi r_2^2-\pi r_3^2[/tex]
[tex]=30\times 18-\pi \times 6^2-\pi \times (4.5)^2-\pi \times 3^2[/tex]
[tex]=540-36\pi-20.25\pi-9\pi[/tex]
[tex]=540-65.25\pi[/tex]
Area of the shaded region is 540 – 65.25π.
The area of shaded region is, [tex][540-\pi(65.25)] inch^{2}[/tex]
From given figure, it is observed that unshaded region is sum of area of three circles.
Diameter of first circle = 12 inch , radius = 12/2 = 6 inch
Diameter of second circle = 9 inch , radius = 9/2 = 4.5 inch
Diameter of third circle = 6 inch , radius = 6/2 = 3 inch
Area of circle = [tex]\pi r^{2}[/tex] , where r is radius of circle.
Dimension of rectangle is,
length = 12 + 9 + 6 +3 = 30 inch
width = 18 inch
Shaded area = Area of rectangle - sum of area of three circles.
[tex]=(18*30)-\pi(6^{2} +4.5^{2} +3^{2} )[/tex]
[tex]=540-\pi(36+20.25+9)\\\\=540-\pi(65.25)[/tex]
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