Answer:
16π
Step-by-step explanation:
Area of the larger circle:
πr²
π(5)²
25π
Area of the smaller circle:
πr²
π(3)²
9π
Subtract the two:
25π-9π=16π
For this case we have that the area of the shaded region is given by the subtraction of areas of both circles. That is to say:[tex]A_ {s} = \pi * (r_ {1}) ^ 2- \pi * (r_ {2}) ^ 2[/tex]
Where:
[tex]r_ {1}:[/tex] It is the radius of the major circle
[tex]r_ {2}[/tex]: It is the radius of the smaller circle
According to the data we have:
[tex]A_ {s} = \pi * (5) ^ 2- \pi * (3) ^ 2\\A_ {s} = \pi * 25- \pi * 9\\A_ {s} = 16 \pi[/tex]
Taking [tex]\pi = 3.14[/tex]
[tex]A_ {s} = 50.24[/tex]
So, the area of the shaded region is [tex]50.24 \ cm ^ 2[/tex]
Answer:
[tex]50.24 \ cm ^ 2[/tex]
