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Mathematics 21 Online
OpenStudy (anonymous):

Which is closest to the total area in square feet of the foundaton? (diagram in question))

OpenStudy (anonymous):

OpenStudy (anonymous):

2328

OpenStudy (anonymous):

how'd you get that answer?

OpenStudy (anonymous):

@k.rajabhishek

OpenStudy (anonymous):

\[c=2 \pi r,\pi r=40,r=\frac{ 40 }{ \pi }\] \[area=25*16+32.5*\frac{ 80 }{ \pi }+\frac{ \pi \left( \frac{ 40 }{ \pi } \right)^2 }{ 2 }\] calculate it.

OpenStudy (anonymous):

\[area=400+\frac{ 2600 }{ \pi }+\frac{ 800 }{ \pi }=400+\frac{ 3400 }{ \pi }=?\]

OpenStudy (anonymous):

can you explain it because im confused. @surjithayer

OpenStudy (bibby):

in order to get the length of the 2nd rectangle, you need to find the diameter of the semicircle on the right.

OpenStudy (anonymous):

length=diameter=2*radius \[=2*\frac{ 40 }{ \pi }=\frac{ 80 }{ \pi }\] circumference of circle\[=2 \pi r\] circumference of semi circle\[=\frac{ 2~\pi~r }{ 2 }=\pi~r\]

OpenStudy (anonymous):

now any problem.

OpenStudy (anonymous):

i cant figure out the area of the semi circle

OpenStudy (acxbox22):

you have to find the diameter first since it is a semicircle then we can find the diameter by using pi*d=80 (40 x 2=80) d=80/pi

OpenStudy (anonymous):

so 80^2 x 3.14?

OpenStudy (acxbox22):

80/pi=diameter so radius=40/pi now we plug that into the equation A=(pi x r^2)/2 since it is a semicircle

OpenStudy (acxbox22):

so the area of the semicircle would be |dw:1422811253194:dw|

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