A local public school encourages, but does not require, students to wear uniforms. The principal of the school compares the gr
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I hope this makes sense, if not please feel free to let me know and I'll explain it.
FOUND THE COMPLETE QUESTION IN ANOTHER SOURCE.ATTACHED IMAGE.
For this case what we have is the following:
For the two semicircles we can model it as a complete circle.
We have to then:
Perimeter:
P = 2 * pi * r
or
P = pi * d
Where,
r = radius
d = diameter
Therefore the perimeter is:
P = 10 * pi
For the largest circle we have:
radius = 10
Perimeter:
P '= 2pi10
P '= 20pi
1/4 since 1/4 circle:
P '' = 20pi / 4 = 5pi
Then, the total perimeter of the source is:
Pt = P + P '' = 10pi + 5pi = 15pi
Pt = 15 * (3.141592)
Pt = 47.1239
round
Pt = 47.1 ft
Area:
The total area will be:
A = A (two semicircles) + A (quarter big circle)
A = (pi / 4) * (d ^ 2) + (1/4) * pi * r ^ 2
A = (pi / 4) * ((10) ^ 2) + (1/4) * pi * (5) ^ 2
A = 98.17477042 feet ^ 2
Round:
A = 98.2 feet ^ 2
Answer:
Perimeter of the source:
Pt = 47.1 ft
Area of the source:
A = 98.2 feet ^ 2
For this case what we have is the following:
For the two semicircles we can model it as a complete circle.
We have to then:
Perimeter:
P = 2 * pi * r
or
P = pi * d
Where,
r = radius
d = diameter
Therefore the perimeter is:
P = 10 * pi
For the largest circle we have:
radius = 10
Perimeter:
P '= 2pi10
P '= 20pi
1/4 since 1/4 circle:
P '' = 20pi / 4 = 5pi
Then, the total perimeter of the source is:
Pt = P + P '' = 10pi + 5pi = 15pi
Pt = 15 * (3.141592)
Pt = 47.1239
round
Pt = 47.1 ft
Area:
The total area will be:
A = A (two semicircles) + A (quarter big circle)
A = (pi / 4) * (d ^ 2) + (1/4) * pi * r ^ 2
A = (pi / 4) * ((10) ^ 2) + (1/4) * pi * (5) ^ 2
A = 98.17477042 feet ^ 2
Round:
A = 98.2 feet ^ 2
Answer:
Perimeter of the source:
Pt = 47.1 ft
Area of the source:
A = 98.2 feet ^ 2