Thread: Surface Area/Volume Problem: did i do it correctly?

Please check to see if i did this right. Thanks a trillion!

A wooden shed in the shape os a rectangular prism with base length and width of 8' and 6' has a hieght of 10'. The shed is resting on the ground.

a)What is the surface area of the shed that can be painted?

SA=ph+2B (but you cant paint the part resting on the ground)
SA=ph+B
SA=300+48
SA=348ft can be painted (is this squared? i dont think soo....)

b) How many boxes with a volume of 6ft^3 can be stored in the shed?

V=Bh
V=48x10
V=480ft^3
480/6
80 boxes with the volume of 6ft^3 can be stored in the shed

Originally Posted by sarahgasoline

Please check to see if i did this right. Thanks a trillion!

A wooden shed in the shape os a rectangular prism with base length and width of 8' and 6' has a hieght of 10'. The shed is resting on the ground.

a)What is the surface area of the shed that can be painted?

SA=ph+2B (but you cant paint the part resting on the ground)
SA=ph+B
SA=300+48
SA=348ft can be painted (is this squared? i dont think soo....)

b) How many boxes with a volume of 6ft^3 can be stored in the shed?

V=Bh
V=48x10
V=480ft^3
480/6
80 boxes with the volume of 6ft^3 can be stored in the shed

a) p, the perimeter is 2 x 8 + 2 x 6 = 16 + 12 = 28 ft. So the surface area of the walls is 10 x 28 = 280 ft^2, not 300. (And yes, the unit is ft. - squared.)

b) I don't like this question as it should depend on the dimensions of the boxes and not just their volume. However for the apparent spirit of the question this is done correctly.

-Dan

ha ha thanks. i always make mistakes on the easy parts