Answer:
Step-by-step explanation:
d=13
r=6.5
A= πr2
A= π(6.5) to the power of 2
A= 42.25π
A=132.73
that is the big circle, now we subtract the 2 small circles area from it.
d=6
r=3
A= πrto the power of 2
A= π(3) to the power of 2
A=9π
A=28.27
since there is 2, 28.27x2=56.54
132.73-56.54=76.19
A=76.19
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for perimeter, it would be
C=2πr
C=2π6.5
C=13π
C=40.84
if your teacher wants it with the inner circles as well then just add all 3.
small inner circle:
C=2πr
C=2xπx3
C=6π
C=18.84
x2
(part 1)
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page 2
lets first get the circle out of the way
same process
d=5
r=2.5
A= πrto the power of 2
A= π(2.5) to the power of 2
A=6.25π
A=19.63
divide that by 2, since its a half circle
A=9.82
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C=2πr
C=2π(2.5)
C=5π
C=15.70
once again, divide by 2
C=7.85
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now lets get the rest
A=wl
A=6x3
A=18
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P=a+b+2c
P=5+7+2(3.16)
P=15.16
then we subtract the opening from the circle, which would be the r
15.16-5=10.16
plus the C for circle,
10.16+7.85=23.01
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add the 2 area
18+9.82=27.82
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A=27.82unit square
C=23.01
part 2 finish
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part 3
again, circle first
d=5
r=2.5
A= πrto the power of 2
A= π(2.5) to the power of 2
A=6.25π
A=19.63
since its 2 half, we dont have to divide by 2, its just 1 circle
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C=2πr
C=2π(2.5)
C=5π
C=15.70
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A=wl
A=15x5
A=75
subtract the circle
75-19.63=55.63
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P=w+2(l)+5
P=15+10+5
P=30
add the circle
30+15.70=45.7
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A=55.63 unit square
P=45.7
(part 3 done)
