Question
Surface area of a composite object.
A cone with a slant height of 8.9cm and a radius of 4cm. The cones base is attached on top of a rectangular prism whose length is 41cm, width is 20cm and height is 40cm. It is attached with a hemisphere whose radius is 4cm.
Find the surface area of this composite object.
A cone with a slant height of 8.9cm and a radius of 4cm. The cones base is attached on top of a rectangular prism whose length is 41cm, width is 20cm and height is 40cm. It is attached with a hemisphere whose radius is 4cm.
Find the surface area of this composite object.
Answers
Steve
surely you have the formulas for finding the volumes of these solids. Just add 'em up.
What do you get? Show work if you get stuck.
What do you get? Show work if you get stuck.
Maryam
Rectangular prism:
Sa= 2lw+2wh+2lh
2 (41)(20)+ 2 (20)(40)+ 2(41)(40)
1640+1600+3280
SA= 6520cm2
There's no overlaps for rectangular prism so it stays the same.
Cone:
SA= πrs+πr^2
π4×8.94427191+π4^2
112.3970357+50.26548246= 162.6625182cm2
Hemisphere:
SA= 3πr^2
3π4^2
SA= 150.7964474cm2
Sa= 2lw+2wh+2lh
2 (41)(20)+ 2 (20)(40)+ 2(41)(40)
1640+1600+3280
SA= 6520cm2
There's no overlaps for rectangular prism so it stays the same.
Cone:
SA= πrs+πr^2
π4×8.94427191+π4^2
112.3970357+50.26548246= 162.6625182cm2
Hemisphere:
SA= 3πr^2
3π4^2
SA= 150.7964474cm2
Maryam
Now I have to find the overlap. So:
Cone πr2= 50.26548246
And I don't get how to subtract the overlap of tye hemisphere
Cone πr2= 50.26548246
And I don't get how to subtract the overlap of tye hemisphere
Maryam
And since the circle of the cone is placed on top of the cylinder, do I do 2πr^2 since it covers 2 surfaces?
Also for the hemisphere, I'm not understanding what formula to use since it is at the bottom of the rectabgular prism. They're connected.do I still use 3πr^2?
Also for the hemisphere, I'm not understanding what formula to use since it is at the bottom of the rectabgular prism. They're connected.do I still use 3πr^2?
Maryam
My total surface area I found is 6833.458965cm2
But I need to subtract the overlapping or hidden sides. I need help on that
But I need to subtract the overlapping or hidden sides. I need help on that
Steve
Oops - my bad. I was looking for volumes.
The overlap is the area of the circular base of the cone (pi r^2) and the circular base of the hemisphere (pi r^2).
Subtract it twice, once for each solid.
The overlap is the area of the circular base of the cone (pi r^2) and the circular base of the hemisphere (pi r^2).
Subtract it twice, once for each solid.